What Is The Probability That The Sample Mean For A Sample Of Size 34 Will Be More Than 32?

Affiliate 6 Sampling Distributions

A statistic, such as the sample mean or the sample standard divergence, is a number computed from a sample. Since a sample is random, every statistic is a random variable: it varies from sample to sample in a way that cannot be predicted with certainty. As a random variable it has a hateful, a standard deviation, and a probability distribution. The probability distribution of a statistic is called its sampling distributionThe probability distribution of a sample statistic when the statistic is viewed as a random variable. . Typically sample statistics are not ends in themselves, only are computed in order to gauge the corresponding population parameters, as illustrated in the grand picture of statistics presented in Figure 1.one "The Thousand Picture of Statistics" in Affiliate 1 "Introduction".

This chapter introduces the concepts of the hateful, the standard divergence, and the sampling distribution of a sample statistic, with an emphasis on the sample mean $\stackrel{-}{\mathit{x}}.$

6.one The Mean and Standard Departure of the Sample Hateful

Learning Objectives

1. To become familiar with the concept of the probability distribution of the sample mean.
2. To sympathize the meaning of the formulas for the mean and standard divergence of the sample mean.

Suppose we wish to guess the mean μ of a population. In bodily practice we would typically take simply i sample. Imagine nevertheless that we take sample after sample, all of the same size n, and compute the sample mean $\stackrel{-}{\mathit{x}}$ of each one. We volition likely get a dissimilar value of $\stackrel{-}{\mathit{10}}$ each time. The sample hateful $\stackrel{-}{\mathit{x}}$ is a random variable: it varies from sample to sample in a way that cannot exist predicted with certainty. Nosotros volition write $\stackrel{-}{\mathit{X}}$ when the sample mean is thought of as a random variable, and write $\stackrel{-}{\mathit{10}}$ for the values that it takes. The random variable $\stackrel{-}{\mathit{X}}$ has a meanThe number well-nigh which ways computed from samples of the same size center. , denoted ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ , and a standard divergenceA measure of the variability of means computed from samples of the same size. , denoted ${\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}.$ Here is an example with such a small population and small sample size that we can really write down every single sample.

Instance ane

A rowing team consists of four rowers who weigh 152, 156, 160, and 164 pounds. Find all possible random samples with replacement of size two and compute the sample mean for each 1. Use them to find the probability distribution, the mean, and the standard deviation of the sample hateful $\stackrel{-}{\mathit{10}}.$

Solution

The post-obit table shows all possible samples with replacement of size two, along with the mean of each:

Sample	Mean		Sample	Hateful		Sample	Mean		Sample	Mean
152, 152	152		156, 152	154		160, 152	156		164, 152	158
152, 156	154		156, 156	156		160, 156	158		164, 156	160
152, 160	156		156, 160	158		160, 160	160		164, 160	162
152, 164	158		156, 164	160		160, 164	162		164, 164	164

The table shows that in that location are 7 possible values of the sample mean $\stackrel{-}{\mathit{X}}.$ The value $\stackrel{-}{\mathit{x}}=152$ happens only ane fashion (the rower weighing 152 pounds must be selected both times), as does the value $\stackrel{-}{\mathit{x}}=164$ , only the other values happen more than one way, hence are more likely to be observed than 152 and 164 are. Since the 16 samples are as probable, we obtain the probability distribution of the sample mean just by counting:

$$\begin{array}{cccccccc}\stackrel{-}{\mathit{10}}& 152& 154& 156& 158& 160& 162& 164\\ P\left(\stackrel{-}{\mathit{ten}}\right)& \frac{1}{\mathrm{sixteen}}& \frac{\mathrm{ii}}{16}& \frac{3}{16}& \frac{\mathrm{iv}}{16}& \frac{3}{\mathrm{xvi}}& \frac{2}{16}& \frac{\mathrm{one}}{16}\end{array}$$

Now nosotros utilize the formulas from Department iv.ii.two "The Mean and Standard Deviation of a Discrete Random Variable" in Chapter 4 "Discrete Random Variables" for the hateful and standard deviation of a discrete random variable to $\stackrel{-}{\mathit{X}}.$ For ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ we obtain.

$$\begin{array}{ll}\hfill {\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}& ={\displaystyle \mathrm{\Sigma }\stackrel{-}{\mathit{x}}P\left(\stackrel{-}{\mathit{ten}}\right)}\hfill \\ \hfill & =152\left(\frac{1}{\mathrm{xvi}}\right)+154\left(\frac{2}{\mathrm{sixteen}}\right)+156\left(\frac{3}{16}\right)+158\left(\frac{4}{16}\right)+160\left(\frac{3}{16}\right)+162\left(\frac{2}{16}\right)+164\left(\frac{1}{16}\right)\hfill \\ \hfill & =158\hfill \end{array}$$

For ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}$ we first compute ${\displaystyle \mathrm{\Sigma }{\stackrel{-}{\mathit{x}}}^{2}P\left(\stackrel{-}{\mathit{x}}\right)}$ :

$${152}^2\left(\frac{\mathrm{ane}}{16}\right)+{154}^2\left(\frac{\mathrm{ii}}{16}\right)+{156}^2\left(\frac{3}{\mathrm{sixteen}}\right)+{158}^{\mathrm{ii}}\left(\frac{\mathrm{four}}{16}\right)+{160}^{\mathrm{two}}\left(\frac{\mathrm{iii}}{\mathrm{xvi}}\right)+{162}^2\left(\frac{2}{16}\right)+{164}^2\left(\frac{1}{16}\right)$$

which is 24,974, so that

$${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\sqrt{{\displaystyle \mathrm{\Sigma }{\stackrel{-}{\mathit{x}}}^{2}P\left(\stackrel{-}{\mathit{x}}\right)-{\mathit{\mu }}_{\stackrel{-}{\mathit{x}}}^{\mathrm{ii}}}}=\sqrt{\mathrm{24,974}-{158}^{\mathrm{ii}}}=\sqrt{10}$$

The mean and standard deviation of the population {152,156,160,164} in the example are μ = 158 and $\mathit{\sigma }=\sqrt{20}.$ The hateful of the sample mean $\stackrel{-}{\mathit{X}}$ that we accept just computed is exactly the mean of the population. The standard deviation of the sample mean $\stackrel{-}{\mathit{X}}$ that we have just computed is the standard deviation of the population divided by the square root of the sample size: $\sqrt{10}=\sqrt{20}/\sqrt{2}.$ These relationships are not coincidences, but are illustrations of the following formulas.

Suppose random samples of size n are fatigued from a population with mean μ and standard deviation σ. The hateful ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ and standard departure ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}$ of the sample mean $\stackrel{-}{\mathit{X}}$ satisfy

$${\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}=\mathit{\mu }\text{and}{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\frac{\mathit{\sigma }}{\sqrt{\mathrm{northward}}}$$

The outset formula says that if we could take every possible sample from the population and compute the corresponding sample mean, so those numbers would center at the number we wish to approximate, the population mean μ.

The 2d formula says that averages computed from samples vary less than individual measurements on the population do, and quantifies the human relationship.

Example 2

The mean and standard deviation of the tax value of all vehicles registered in a certain country are $\mathit{\mu }=\$\mathrm{13,525}$ and $\mathit{\sigma }=\$\mathrm{four,180}.$ Suppose random samples of size 100 are drawn from the population of vehicles. What are the hateful ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ and standard departure ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}$ of the sample mean $\stackrel{-}{\mathit{X}}$ ?

Solution

Since due north = 100, the formulas yield

$${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=\mathit{\mu }=\$\mathrm{13,525}\text{and}{\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}=\frac{\mathit{\sigma }}{\sqrt{\mathrm{northward}}}=\frac{\$4180}{\sqrt{100}}=\$418$$

Key Takeaways

• The sample mean is a random variable; equally such it is written $\stackrel{-}{\mathit{X}}$ , and $\stackrel{-}{\mathit{ten}}$ stands for individual values information technology takes.
• As a random variable the sample hateful has a probability distribution, a mean ${\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}$ , and a standard divergence ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}.$
• There are formulas that relate the mean and standard deviation of the sample mean to the mean and standard difference of the population from which the sample is drawn.

Exercises

1. Random samples of size 225 are drawn from a population with hateful 100 and standard deviation 20. Discover the mean and standard deviation of the sample mean.

2. Random samples of size 64 are drawn from a population with mean 32 and standard deviation 5. Discover the mean and standard deviation of the sample mean.

3. A population has mean 75 and standard difference 12.

   1. Random samples of size 121 are taken. Find the mean and standard deviation of the sample hateful.
   2. How would the answers to function (a) alter if the size of the samples were 400 instead of 121?

4. A population has mean v.75 and standard deviation 1.02.

   1. Random samples of size 81 are taken. Find the hateful and standard deviation of the sample mean.
   2. How would the answers to part (a) change if the size of the samples were 25 instead of 81?

Answers

1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=100$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\mathrm{ane.33}$

3. 1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{10}}}=75$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\mathrm{ane.09}$
   2. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}$ stays the aforementioned only ${\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}$ decreases to 0.vi

half-dozen.2 The Sampling Distribution of the Sample Mean

Learning Objectives

1. To learn what the sampling distribution of $\stackrel{-}{\mathit{X}}$ is when the sample size is large.
2. To learn what the sampling distribution of $\stackrel{-}{\mathit{X}}$ is when the population is normal.

The Central Limit Theorem

In Note 6.5 "Example 1" in Section half dozen.ane "The Mean and Standard Deviation of the Sample Hateful" we synthetic the probability distribution of the sample mean for samples of size two drawn from the population of four rowers. The probability distribution is:

$$\begin{array}{cccccccc}\stackrel{-}{\mathit{x}}& 152& 154& 156& 158& 160& 162& 164\\ P\left(\stackrel{-}{\mathit{x}}\right)& \frac{1}{16}& \frac{\mathrm{ii}}{16}& \frac{3}{\mathrm{xvi}}& \frac{\mathrm{iv}}{\mathrm{xvi}}& \frac{3}{16}& \frac{2}{16}& \frac{\mathrm{ane}}{16}\end{array}$$

Figure 6.1 "Distribution of a Population and a Sample Mean" shows a side-by-side comparison of a histogram for the original population and a histogram for this distribution. Whereas the distribution of the population is uniform, the sampling distribution of the hateful has a shape budgeted the shape of the familiar bell curve. This phenomenon of the sampling distribution of the mean taking on a bell shape even though the population distribution is not bell-shaped happens in general. Here is a somewhat more realistic example.

Figure half dozen.1 Distribution of a Population and a Sample Mean

Suppose nosotros accept samples of size one, 5, 10, or xx from a population that consists entirely of the numbers 0 and i, half the population 0, half i, so that the population mean is 0.v. The sampling distributions are:

due north = 1:

$$\begin{array}{ccc}\stackrel{-}{\mathit{x}}\hfill & 0& 1\\ P\left(\stackrel{-}{\mathit{x}}\right)\hfill & 0.5& \mathrm{0.v}\end{array}$$

northward = 5:

$$\begin{array}{ccccccc}\stackrel{-}{\mathit{x}}\hfill & 0& 0.2& 0.4& 0.6& 0.8& 1\\ P\left(\stackrel{-}{\mathit{ten}}\right)\hfill & 0.03& 0.16& 0.31& 0.31& 0.16& 0.03\end{array}$$

n = x:

$$\begin{array}{cccccccccccc}\stackrel{-}{\mathit{x}}\hfill & 0& 0.1& 0.2& 0.3& 0.4& 0.5& \mathrm{0.vi}& \mathrm{0.seven}& 0.8& 0.9& 1\\ P\left(\stackrel{-}{\mathit{x}}\right)\hfill & 0.00& 0.01& 0.04& 0.12& 0.21& 0.25& 0.21& 0.12& 0.04& 0.01& 0.00\end{array}$$

north = twenty:

$$\begin{array}{cccccccccccc}\stackrel{-}{\mathit{x}}\hfill & 0& 0.05& 0.10& \mathrm{0.fifteen}& 0.20& 0.25& 0.30& 0.35& \mathrm{0.xl}& 0.45& \mathrm{0.fifty}\\ P\left(\stackrel{-}{\mathit{10}}\right)\hfill & 0.00& 0.00& 0.00& 0.00& 0.00& 0.01& 0.04& 0.07& 0.12& 0.16& \mathrm{0.xviii}\end{array}$$ $$\begin{array}{ccccccccccc}\stackrel{-}{\mathit{ten}}\hfill & 0.55& 0.60& 0.65& 0.70& 0.75& 0.80& 0.85& \mathrm{0.ninety}& 0.95& 1\\ P\left(\stackrel{-}{\mathit{x}}\right)\hfill & 0.16& 0.12& 0.07& 0.04& 0.01& 0.00& 0.00& 0.00& 0.00& 0.00\end{array}$$

Histograms illustrating these distributions are shown in Figure 6.2 "Distributions of the Sample Mean".

Effigy vi.2 Distributions of the Sample Mean

Equally n increases the sampling distribution of $\stackrel{-}{\mathit{X}}$ evolves in an interesting fashion: the probabilities on the lower and the upper ends compress and the probabilities in the eye get larger in relation to them. If we were to keep to increase north then the shape of the sampling distribution would become smoother and more bong-shaped.

What we are seeing in these examples does not depend on the detail population distributions involved. In general, one may beginning with any distribution and the sampling distribution of the sample mean will increasingly resemble the bell-shaped normal curve as the sample size increases. This is the content of the Central Limit Theorem.

The Central Limit Theorem

For samples of size 30 or more, the sample mean is approximately normally distributed, with mean ${\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}=\mathit{\mu }$ and standard departure ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\mathit{\sigma }/\sqrt{n}$ , where n is the sample size. The larger the sample size, the better the approximation.

The Key Limit Theorem is illustrated for several common population distributions in Figure 6.3 "Distribution of Populations and Sample Means".

Figure 6.3 Distribution of Populations and Sample Ways

The dashed vertical lines in the figures locate the population hateful. Regardless of the distribution of the population, as the sample size is increased the shape of the sampling distribution of the sample mean becomes increasingly bell-shaped, centered on the population hateful. Typically past the time the sample size is xxx the distribution of the sample mean is practically the aforementioned as a normal distribution.

The importance of the Cardinal Limit Theorem is that it allows the states to make probability statements nigh the sample mean, specifically in relation to its value in comparison to the population mean, equally we will meet in the examples. But to use the result properly nosotros must showtime realize that there are two divide random variables (and therefore two probability distributions) at play:

1. Ten, the measurement of a unmarried element selected at random from the population; the distribution of X is the distribution of the population, with mean the population mean μ and standard divergence the population standard deviation σ;
2. $\stackrel{-}{\mathit{10}}$ , the mean of the measurements in a sample of size n; the distribution of $\stackrel{-}{\mathit{X}}$ is its sampling distribution, with hateful ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=\mathit{\mu }$ and standard deviation ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\mathit{\sigma }/\sqrt{n}.$

Example 3

Let $\stackrel{-}{\mathit{X}}$ be the mean of a random sample of size 50 drawn from a population with hateful 112 and standard deviation forty.

1. Detect the mean and standard deviation of $\stackrel{-}{\mathit{Ten}}.$
2. Find the probability that $\stackrel{-}{\mathit{X}}$ assumes a value between 110 and 114.
3. Find the probability that $\stackrel{-}{\mathit{Ten}}$ assumes a value greater than 113.

Solution

1. By the formulas in the previous section

   $${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=\mathit{\mu }=112\text{and}{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\frac{\mathit{\sigma }}{\sqrt{n}}=\frac{40}{\sqrt{\mathrm{l}}}=5.65685$$

2. Since the sample size is at least 30, the Central Limit Theorem applies: $\stackrel{-}{\mathit{10}}$ is approximately normally distributed. We compute probabilities using Figure 12.2 "Cumulative Normal Probability" in the usual way, just being careful to utilise ${\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}$ and non σ when we standardize:

   $$\begin{array}{ll}P(110<\stackrel{-}{\mathit{X}}<114)\hfill & =P\left(\frac{110-{\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}}<Z<\frac{114-{\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}}\right)\hfill \\ \hfill & =P\left(\frac{110-112}{\mathrm{v.65685}}<Z<\frac{114-112}{5.65685}\right)\hfill \\ \hfill & =P\left(\text{−}0.35<Z<0.35\right)=0.6368-0.3632=0.2736\hfill \end{array}$$

3. Similarly

   $$\begin{array}{ll}P(\stackrel{-}{\mathit{X}}>113)\hfill & =P\left(Z>\frac{113-{\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}}\right)\hfill \\ \hfill & =P\left(Z>\frac{113-112}{5.65685}\right)\hfill \\ \hfill & =P\left(Z>\mathrm{0.xviii}\right)\hfill \\ \hfill & =\mathrm{ane}-P(Z<0.18)=1-0.5714=0.4286\hfill \end{array}$$

Note that if in Annotation six.11 "Example 3" we had been asked to compute the probability that the value of a unmarried randomly selected element of the population exceeds 113, that is, to compute the number P(Ten > 113), we would not have been able to do so, since we do not know the distribution of X, but only that its mean is 112 and its standard deviation is 40. By contrast we could compute $P\left(\stackrel{-}{\mathit{X}}>113\right)$ fifty-fifty without complete noesis of the distribution of Ten because the Fundamental Limit Theorem guarantees that $\stackrel{-}{\mathit{10}}$ is approximately normal.

Example four

The numerical population of form point averages at a college has mean 2.61 and standard deviation 0.5. If a random sample of size 100 is taken from the population, what is the probability that the sample mean volition be betwixt 2.51 and ii.71?

Solution

The sample hateful $\stackrel{-}{\mathit{10}}$ has hateful ${\mathit{\mu }}_{\stackrel{-}{\mathit{10}}}=\mathit{\mu }=2.61$ and standard deviation ${\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}=\mathit{\sigma }/\sqrt{\mathrm{north}}=0.5/\mathrm{x}=0.05$ , and then

$$\begin{array}{ll}P(2.51<\stackrel{-}{\mathit{Ten}}<2.71)\hfill & =P\left(\frac{\mathrm{two.51}-{\mathit{\mu }}_{\stackrel{-}{\mathit{Ten}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}}<Z<\frac{2.71-{\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}}\right)\hfill \\ \hfill & =P\left(\frac{2.51-2.61}{0.05}<Z<\frac{\mathrm{two.71}-2.61}{0.05}\right)\hfill \\ \hfill & =P\left(\text{−}2<Z<2\right)\hfill \\ \hfill & =P\left(Z<\mathrm{two}\right)-P\left(Z<\text{−}2\right)\hfill \\ \hfill & =0.9772-0.0228=0.9544\hfill \end{array}$$

Normally Distributed Populations

The Central Limit Theorem says that no matter what the distribution of the population is, equally long equally the sample is "large," pregnant of size 30 or more, the sample hateful is approximately usually distributed. If the population is normal to begin with then the sample mean too has a normal distribution, regardless of the sample size.

For samples of whatever size drawn from a normally distributed population, the sample mean is normally distributed, with mean ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=\mathit{\mu }$ and standard difference ${\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}=\mathit{\sigma }/\sqrt{\mathrm{north}}$ , where n is the sample size.

The effect of increasing the sample size is shown in Effigy half-dozen.4 "Distribution of Sample Means for a Normal Population".

Figure 6.4 Distribution of Sample Means for a Normal Population

Instance five

A prototype automotive tire has a design life of 38,500 miles with a standard deviation of ii,500 miles. Five such tires are manufactured and tested. On the assumption that the bodily population mean is 38,500 miles and the actual population standard deviation is 2,500 miles, find the probability that the sample hateful will be less than 36,000 miles. Assume that the distribution of lifetimes of such tires is normal.

Solution

For simplicity we apply units of thousands of miles. Then the sample mean $\stackrel{-}{\mathit{10}}$ has mean ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=\mathit{\mu }=\mathrm{38.v}$ and standard deviation ${\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}=\mathit{\sigma }/\sqrt{n}=\mathrm{2.v}/\sqrt{5}=\mathrm{i.11803}.$ Since the population is normally distributed, so is $\stackrel{-}{\mathit{X}}$ , hence

$$\begin{array}{ll}P(\stackrel{-}{\mathit{X}}<36)\hfill & =P\left(Z<\frac{36-{\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}}\right)\hfill \\ \hfill & =P\left(Z<\frac{36-38.5}{\mathrm{i.11803}}\right)\hfill \\ \hfill & =P\left(Z<\text{−}2.24\right)=0.0125\hfill \end{array}$$

That is, if the tires perform equally designed, in that location is simply about a i.25% chance that the average of a sample of this size would exist so depression.

Example 6

An car battery manufacturer claims that its midgrade battery has a hateful life of fifty months with a standard deviation of half-dozen months. Suppose the distribution of battery lives of this detail brand is approximately normal.

1. On the supposition that the manufacturer'south claims are truthful, notice the probability that a randomly selected battery of this type will last less than 48 months.
2. On the same assumption, find the probability that the mean of a random sample of 36 such batteries will be less than 48 months.

Solution

1. Since the population is known to accept a normal distribution

   $$\begin{array}{ll}P\left(X<48\right)\hfill & =P\left(Z<\frac{48-\mathit{\mu }}{\mathit{\sigma }}\right)=P\left(Z<\frac{48-50}{6}\right)\hfill \\ \hfill & =P\left(Z<\text{−}0.33\right)=0.3707\hfill \end{array}$$

2. The sample mean has hateful ${\mathit{\mu }}_{\stackrel{-}{\mathit{10}}}=\mathit{\mu }=50$ and standard deviation ${\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}=\mathit{\sigma }/\sqrt{\mathrm{northward}}=\mathrm{six}/\sqrt{36}=\mathrm{one}.$ Thus

   $$\begin{array}{ll}P\left(\stackrel{-}{\mathit{X}}<48\right)\hfill & =P\left(Z<\frac{48-{\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}}{{\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}}\right)\hfill \\ \hfill & =P\left(Z<\frac{48-\mathrm{fifty}}{\mathrm{i}}\right)\hfill \\ \hfill & =P\left(Z<\text{−}\mathrm{ii}\right)=0.0228\hfill \end{array}$$

Central Takeaways

• When the sample size is at least 30 the sample mean is normally distributed.
• When the population is normal the sample mean is normally distributed regardless of the sample size.

Exercises

Bones

1. A population has mean 128 and standard divergence 22.

   1. Observe the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 36.
   2. Find the probability that the mean of a sample of size 36 will be within 10 units of the population mean, that is, between 118 and 138.

2. A population has mean 1,542 and standard deviation 246.

   1. Find the hateful and standard deviation of $\stackrel{-}{\mathit{10}}$ for samples of size 100.
   2. Observe the probability that the mean of a sample of size 100 will be within 100 units of the population mean, that is, betwixt i,442 and ane,642.

3. A population has hateful 73.5 and standard departure 2.5.

   1. Find the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 30.
   2. Observe the probability that the mean of a sample of size 30 will be less than 72.

4. A population has hateful 48.4 and standard deviation six.3.

   1. Find the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 64.
   2. Find the probability that the hateful of a sample of size 64 will be less than 46.7.

5. A normally distributed population has hateful 25.6 and standard deviation three.three.

   1. Observe the probability that a single randomly selected element X of the population exceeds xxx.
   2. Find the mean and standard difference of $\stackrel{-}{\mathit{X}}$ for samples of size ix.
   3. Find the probability that the hateful of a sample of size ix fatigued from this population exceeds 30.

6. A normally distributed population has mean 57.7 and standard deviation 12.1.

   1. Find the probability that a single randomly selected element X of the population is less than 45.
   2. Notice the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 16.
   3. Find the probability that the mean of a sample of size 16 drawn from this population is less than 45.

7. A population has mean 557 and standard deviation 35.

   1. Notice the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 50.
   2. Find the probability that the mean of a sample of size 50 will exist more than than 570.

8. A population has mean 16 and standard deviation 1.7.

   1. Find the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 80.
   2. Notice the probability that the mean of a sample of size 80 will be more xvi.4.

9. A unremarkably distributed population has mean 1,214 and standard deviation 122.

   1. Find the probability that a unmarried randomly selected element X of the population is between ane,100 and one,300.
   2. Find the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 25.
   3. Find the probability that the mean of a sample of size 25 drawn from this population is between 1,100 and 1,300.

10. A normally distributed population has mean 57,800 and standard deviation 750.

    1. Discover the probability that a single randomly selected element Ten of the population is between 57,000 and 58,000.
    2. Notice the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 100.
    3. Find the probability that the hateful of a sample of size 100 drawn from this population is between 57,000 and 58,000.

11. A population has mean 72 and standard deviation 6.

    1. Notice the mean and standard deviation of $\stackrel{-}{\mathit{X}}$ for samples of size 45.
    2. Find the probability that the mean of a sample of size 45 will differ from the population mean 72 by at to the lowest degree 2 units, that is, is either less than 70 or more than than 74. (Hint: One way to solve the problem is to kickoff discover the probability of the complementary event.)

12. A population has mean 12 and standard deviation ane.five.

    1. Detect the hateful and standard difference of $\stackrel{-}{\mathit{X}}$ for samples of size xc.
    2. Find the probability that the hateful of a sample of size xc will differ from the population mean 12 by at to the lowest degree 0.3 unit, that is, is either less than 11.7 or more than 12.three. (Hint: One way to solve the problem is to kickoff find the probability of the complementary effect.)

Applications

13. Suppose the mean number of days to germination of a variety of seed is 22, with standard deviation 2.three days. Find the probability that the mean germination time of a sample of 160 seeds will be within 0.5 day of the population hateful.

14. Suppose the hateful length of time that a caller is placed on hold when telephoning a client service center is 23.eight seconds, with standard deviation 4.6 seconds. Observe the probability that the mean length of time on concur in a sample of i,200 calls will be within 0.5 2nd of the population mean.

15. Suppose the mean amount of cholesterol in eggs labeled "large" is 186 milligrams, with standard difference vii milligrams. Find the probability that the mean amount of cholesterol in a sample of 144 eggs will exist within two milligrams of the population mean.

16. Suppose that in i region of the country the hateful amount of credit card debt per household in households having credit card debt is $15,250, with standard departure $7,125. Notice the probability that the hateful amount of credit card debt in a sample of 1,600 such households will be within $300 of the population mean.

17. Suppose speeds of vehicles on a item stretch of roadway are normally distributed with mean 36.6 mph and standard deviation 1.7 mph.

    1. Find the probability that the speed X of a randomly selected vehicle is betwixt 35 and 40 mph.
    2. Detect the probability that the mean speed $\stackrel{-}{\mathit{X}}$ of 20 randomly selected vehicles is between 35 and xl mph.

18. Many sharks enter a state of tonic immobility when inverted. Suppose that in a item species of sharks the time a shark remains in a state of tonic immobility when inverted is normally distributed with mean 11.two minutes and standard divergence ane.1 minutes.

    1. If a biologist induces a state of tonic immobility in such a shark in lodge to study it, notice the probability that the shark will remain in this state for betwixt 10 and 13 minutes.
    2. When a biologist wishes to guess the hateful time that such sharks stay immobile by inducing tonic immobility in each of a sample of 12 sharks, find the probability that mean time of immobility in the sample volition be between x and thirteen minutes.

19. Suppose the hateful cost across the country of a 30-day supply of a generic drug is $46.58, with standard deviation $4.84. Detect the probability that the mean of a sample of 100 prices of 30-day supplies of this drug will be between $45 and $50.

20. Suppose the mean length of fourth dimension between submission of a state tax return requesting a refund and the issuance of the refund is 47 days, with standard deviation 6 days. Notice the probability that in a sample of 50 returns requesting a refund, the hateful such time volition be more than than 50 days.

21. Scores on a common final exam in a big enrollment, multiple-department freshman course are ordinarily distributed with mean 72.vii and standard deviation 13.one.

    1. Find the probability that the score X on a randomly selected exam paper is between 70 and 80.
    2. Find the probability that the hateful score $\stackrel{-}{\mathit{X}}$ of 38 randomly selected exam papers is between 70 and 80.

22. Suppose the mean weight of school children's bookbags is 17.4 pounds, with standard difference ii.2 pounds. Observe the probability that the hateful weight of a sample of 30 bookbags will exceed 17 pounds.

23. Suppose that in a certain region of the land the mean elapsing of first marriages that end in divorce is vii.8 years, standard deviation 1.2 years. Find the probability that in a sample of 75 divorces, the mean historic period of the marriages is at nearly 8 years.

24. Borachio eats at the same fast food restaurant every day. Suppose the fourth dimension X between the moment Borachio enters the restaurant and the moment he is served his food is normally distributed with mean 4.2 minutes and standard deviation one.3 minutes.

    1. Detect the probability that when he enters the eatery today it volition be at least v minutes until he is served.
    2. Find the probability that average time until he is served in eight randomly selected visits to the restaurant will be at to the lowest degree 5 minutes.

Additional Exercises

25. A high-speed packing car can be set to deliver between xi and 13 ounces of a liquid. For any commitment setting in this range the amount delivered is usually distributed with hateful some corporeality μ and with standard deviation 0.08 ounce. To calibrate the machine it is gear up to deliver a particular amount, many containers are filled, and 25 containers are randomly selected and the amount they contain is measured. Detect the probability that the sample mean volition be inside 0.05 ounce of the actual mean amount being delivered to all containers.

26. A tire manufacturer states that a certain blazon of tire has a mean lifetime of sixty,000 miles. Suppose lifetimes are normally distributed with standard difference $\mathit{\sigma }=\text{3,500}$ miles.

    1. Find the probability that if you buy ane such tire, information technology will final just 57,000 or fewer miles. If you had this experience, is it particularly potent bear witness that the tire is non every bit good as claimed?
    2. A consumer grouping buys five such tires and tests them. Find the probability that average lifetime of the five tires will exist 57,000 miles or less. If the mean is so low, is that particularly strong show that the tire is not as good equally claimed?

Answers

1. 1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=128$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=3.67$
   2. 0.9936
3. 1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=73.5$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{10}}}=0.456$
   2. 0.0005
5. 1. 0.0918
   2. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=25.6$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=\mathrm{1.i}$
   3. 0.0000
7. 1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=557$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=4.9497$
   2. 0.0043
9. 1. 0.5818
   2. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=1214$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{X}}}=24.4$
   3. 0.9998
11. 1. ${\mathit{\mu }}_{\stackrel{-}{\mathit{X}}}=72$ , ${\mathit{\sigma }}_{\stackrel{-}{\mathit{Ten}}}=0.8944$
    2. 0.0250

13. 0.9940

15. 0.9994

17. 1. 0.8036
    2. 1.0000

19. 0.9994

21. 1. 0.2955
    2. 0.8977

23. 0.9251

25. 0.9982

6.3 The Sample Proportion

Learning Objectives

1. To recognize that the sample proportion $\widehat{P}$ is a random variable.
2. To understand the meaning of the formulas for the hateful and standard deviation of the sample proportion.
3. To learn what the sampling distribution of $\widehat{P}$ is when the sample size is large.

Often sampling is done in order to estimate the proportion of a population that has a specific feature, such every bit the proportion of all items coming off an assembly line that are defective or the proportion of all people inbound a retail store who brand a purchase before leaving. The population proportion is denoted p and the sample proportion is denoted $\widehat{p}.$ Thus if in reality 43% of people inbound a shop make a purchase before leaving, p = 0.43; if in a sample of 200 people entering the store, 78 make a purchase, $\widehat{p}=78/200=0.39.$

The sample proportion is a random variable: information technology varies from sample to sample in a style that cannot be predicted with certainty. Viewed as a random variable it will be written $\widehat{P}.$ It has a meanThe number about which proportions computed from samples of the same size centre. ${\mathit{\mu }}_{\widehat{P}}$ and a standard divergenceA mensurate of the variability of proportions computed from samples of the same size. ${\mathit{\sigma }}_{\widehat{P}}.$ Here are formulas for their values.

Suppose random samples of size n are drawn from a population in which the proportion with a characteristic of interest is p. The mean ${\mathit{\mu }}_{\widehat{P}}$ and standard deviation ${\mathit{\sigma }}_{\widehat{P}}$ of the sample proportion $\widehat{P}$ satisfy

$${\mathit{\mu }}_{\widehat{P}}=p\text{and}{\mathit{\sigma }}_{\widehat{P}}=\sqrt{\frac{pq}{n}}$$

where $q=1-p.$

The Key Limit Theorem has an analogue for the population proportion $\widehat{P}.$ To run into how, imagine that every element of the population that has the characteristic of interest is labeled with a i, and that every element that does non is labeled with a 0. This gives a numerical population consisting entirely of zeros and ones. Clearly the proportion of the population with the special characteristic is the proportion of the numerical population that are ones; in symbols,

$$p=\frac{\text{number}\text{of}1\text{s}}{N}$$

Merely of course the sum of all the zeros and ones is only the number of ones, then the hateful μ of the numerical population is

$$\mathit{\mu }=\frac{{\displaystyle \mathrm{\Sigma }\mathrm{ten}}}{\mathrm{Due\; north}}=\frac{\text{number}\text{of}\mathrm{one}\text{southward}}{N}$$

Thus the population proportion p is the same every bit the mean μ of the corresponding population of zeros and ones. In the aforementioned style the sample proportion $\widehat{p}$ is the same as the sample hateful $\stackrel{-}{\mathit{x}}.$ Thus the Primal Limit Theorem applies to $\widehat{P}.$ However, the condition that the sample be large is a little more complicated than but being of size at least 30.

The Sampling Distribution of the Sample Proportion

For large samples, the sample proportion is approximately unremarkably distributed, with mean ${\mathit{\mu }}_{\widehat{P}}=p$ and standard deviation ${\mathit{\sigma }}_{\widehat{P}}=\sqrt{pq/\mathrm{due\; north}}.$

A sample is large if the interval $\left[p\text{−}3{\mathit{\sigma }}_{\widehat{P}},p+\mathrm{iii}{\mathit{\sigma }}_{\widehat{P}}\right]$ lies wholly within the interval $\left[\mathrm{0,ane}\right].$

In bodily do p is not known, hence neither is ${\mathit{\sigma }}_{\widehat{P}}.$ In that case in society to check that the sample is sufficiently large we substitute the known quantity $\widehat{p}$ for p. This means checking that the interval

$$\left[\widehat{p}\text{−}3\sqrt{\frac{\widehat{p}(\mathrm{one}-\widehat{p})}{\mathrm{north}}},\widehat{p}+3\sqrt{\frac{\widehat{p}(\mathrm{ane}-\widehat{p})}{n}}\right]$$

lies wholly within the interval $\left[\mathrm{0,1}\right].$ This is illustrated in the examples.

Figure 6.5 "Distribution of Sample Proportions" shows that when p = 0.1 a sample of size fifteen is too pocket-sized but a sample of size 100 is adequate. Figure 6.6 "Distribution of Sample Proportions for " shows that when p = 0.5 a sample of size 15 is acceptable.

Figure vi.5 Distribution of Sample Proportions

Figure half-dozen.6 Distribution of Sample Proportions for p = 0.v and n = fifteen

Example 7

Suppose that in a population of voters in a sure region 38% are in favor of item bond consequence. 9 hundred randomly selected voters are asked if they favor the bail issue.

1. Verify that the sample proportion $\widehat{P}$ computed from samples of size 900 meets the condition that its sampling distribution exist approximately normal.
2. Find the probability that the sample proportion computed from a sample of size 900 volition be within 5 percentage points of the true population proportion.

Solution

1. The data given is that p = 0.38, hence $q=1-p=0.62.$ First we use the formulas to compute the mean and standard deviation of $\widehat{P}$ :

   $${\mathit{\mu }}_{\widehat{P}}=p=0.38\text{and}{\mathit{\sigma }}_{\widehat{P}}=\sqrt{\frac{pq}{\mathrm{north}}}=\sqrt{\frac{\left(0.38\right)\left(0.62\right)}{900}}=0.01618$$

   Then $3{\mathit{\sigma }}_{\widehat{P}}=\mathrm{iii}\left(0.01618\right)=0.04854\approx 0.05$ and then

   $$\left[p\text{−}3{\mathit{\sigma }}_{\widehat{P}},p+3{\mathit{\sigma }}_{\widehat{P}}\right]=\left[0.38-0.\mathrm{05,0}.38+0.05\right]=\left[0.\mathrm{33,0}.43\right]$$

   which lies wholly inside the interval $\left[\mathrm{0,1}\right]$ , so it is safe to assume that $\widehat{P}$ is approximately normally distributed.

2. To be within 5 percentage points of the true population proportion 0.38 means to be between $0.38-0.05=0.33$ and $0.38+0.05=0.43.$ Thus

   $$\begin{array}{ll}P(0.33<\widehat{P}<0.43)\hfill & =P\left(\frac{0.33-{\mathit{\mu }}_{\widehat{P}}}{{\mathit{\sigma }}_{\widehat{P}}}<Z<\frac{0.43-{\mathit{\mu }}_{\widehat{P}}}{{\mathit{\sigma }}_{\widehat{P}}}\right)\hfill \\ \hfill & =P\left(\frac{0.33-0.38}{0.01618}<Z<\frac{0.43-0.38}{0.01618}\right)\hfill \\ \hfill & =P\left(\text{−}3.09<Z<3.09\right)\hfill \\ \hfill & =P\left(3.09\right)-P\left(\text{−}3.09\right)\hfill \\ \hfill & =0.9990-0.0010=0.9980\hfill \end{array}$$

Instance 8

An online retailer claims that xc% of all orders are shipped within 12 hours of being received. A consumer group placed 121 orders of different sizes and at unlike times of mean solar day; 102 orders were shipped inside 12 hours.

1. Compute the sample proportion of items shipped within 12 hours.
2. Confirm that the sample is large enough to assume that the sample proportion is ordinarily distributed. Use p = 0.90, corresponding to the assumption that the retailer'southward claim is valid.
3. Assuming the retailer's claim is true, find the probability that a sample of size 121 would produce a sample proportion and so low as was observed in this sample.
4. Based on the reply to part (c), draw a conclusion most the retailer's claim.

Solution

1. The sample proportion is the number x of orders that are shipped within 12 hours divided by the number n of orders in the sample:

   $$\widehat{p}=\frac{x}{n}=\frac{102}{121}=0.84$$

2. Since p = 0.ninety, $q=1-p=0.10$ , and n = 121,

   $${\mathit{\sigma }}_{\widehat{P}}=\sqrt{\frac{\left(\mathrm{0.xc}\right)\left(0.10\right)}{121}}=0.0\stackrel{-}{27}$$

   hence

   $$\left[p\text{−}3{\mathit{\sigma }}_{\widehat{P}},p+3{\mathit{\sigma }}_{\widehat{P}}\right]=\left[0.90-0.\mathrm{08,0}.\mathrm{ninety}+0.08\right]=\left[0.\mathrm{82,0}.98\right]$$

   Considering $$\left[0.82,0.98\right]\subset \left[\mathrm{0,ane}\right],$$ it is appropriate to use the normal distribution to compute probabilities related to the sample proportion $\widehat{P}.$

3. Using the value of $\widehat{P}$ from part (a) and the computation in part (b),

   $$\begin{array}{ll}P(\widehat{P}\le 0.84)\hfill & =P\left(Z\le \frac{0.84-{\mathit{\mu }}_{\widehat{P}}}{{\mathit{\sigma }}_{\widehat{P}}}\right)\hfill \\ \hfill & =P\left(Z\le \frac{0.84-\mathrm{0.ninety}}{0.0\stackrel{-}{27}}\right)\hfill \\ \hfill & =P\left(Z\le \text{−}2.20\right)=0.0139\hfill \end{array}$$
4. The ciphering shows that a random sample of size 121 has just about a 1.4% adventure of producing a sample proportion as the one that was observed, $\widehat{p}=0.84$ , when taken from a population in which the bodily proportion is 0.90. This is and so unlikely that it is reasonable to conclude that the bodily value of p is less than the 90% claimed.

Primal Takeaways

• The sample proportion is a random variable $\widehat{P}.$
• In that location are formulas for the mean ${\mathit{\mu }}_{\widehat{P}}$ and standard deviation ${\mathit{\sigma }}_{\widehat{P}}$ of the sample proportion.
• When the sample size is large the sample proportion is normally distributed.

Exercises

Basic

1. The proportion of a population with a characteristic of interest is p = 0.37. Find the mean and standard deviation of the sample proportion $\widehat{P}$ obtained from random samples of size 1,600.

2. The proportion of a population with a characteristic of interest is p = 0.82. Find the hateful and standard deviation of the sample proportion $\widehat{P}$ obtained from random samples of size 900.

3. The proportion of a population with a feature of interest is p = 0.76. Notice the mean and standard deviation of the sample proportion $\widehat{P}$ obtained from random samples of size 1,200.

4. The proportion of a population with a characteristic of interest is p = 0.37. Find the hateful and standard deviation of the sample proportion $\widehat{P}$ obtained from random samples of size 125.

5. Random samples of size 225 are drawn from a population in which the proportion with the characteristic of involvement is 0.25. Decide whether or not the sample size is large plenty to assume that the sample proportion $\widehat{P}$ is normally distributed.

6. Random samples of size 1,600 are drawn from a population in which the proportion with the feature of interest is 0.05. Determine whether or not the sample size is big plenty to presume that the sample proportion $\widehat{P}$ is normally distributed.

7. Random samples of size due north produced sample proportions $\widehat{p}$ as shown. In each instance decide whether or non the sample size is big enough to presume that the sample proportion $\widehat{P}$ is usually distributed.

   1. n = 50, $\widehat{p}=0.48$
   2. n = l, $\widehat{p}=0.12$
   3. n = 100, $\widehat{p}=0.12$

8. Samples of size north produced sample proportions $\widehat{p}$ every bit shown. In each case decide whether or not the sample size is large enough to presume that the sample proportion $\widehat{P}$ is normally distributed.

   1. n = 30, $\widehat{p}=0.72$
   2. n = 30, $\widehat{p}=0.84$
   3. north = 75, $\widehat{p}=0.84$

9. A random sample of size 121 is taken from a population in which the proportion with the characteristic of involvement is p = 0.47. Find the indicated probabilities.

   1. $P\left(0.45\le \widehat{P}\le \mathrm{0.l}\right)$
   2. $P\left(\widehat{P}\ge 0.50\right)$

10. A random sample of size 225 is taken from a population in which the proportion with the feature of involvement is p = 0.34. Find the indicated probabilities.

    1. $P\left(0.25\le \widehat{P}\le 0.40\right)$
    2. $P\left(\widehat{P}\le 0.35\right)$

11. A random sample of size 900 is taken from a population in which the proportion with the characteristic of interest is p = 0.62. Find the indicated probabilities.

    1. $P\left(0.60\le \widehat{P}\le 0.64\right)$
    2. $P\left(0.57\le \widehat{P}\le 0.67\right)$

12. A random sample of size 1,100 is taken from a population in which the proportion with the characteristic of involvement is p = 0.28. Find the indicated probabilities.

    1. $P\left(0.27\le \widehat{P}\le 0.29\right)$
    2. $P\left(0.23\le \widehat{P}\le 0.33\right)$

Applications

13. Suppose that 8% of all males endure some grade of color blindness. Notice the probability that in a random sample of 250 men at least 10% will endure some form of colour blindness. Outset verify that the sample is sufficiently large to apply the normal distribution.

14. Suppose that 29% of all residents of a community favor annexation by a nearby municipality. Notice the probability that in a random sample of fifty residents at least 35% will favor looting. First verify that the sample is sufficiently large to apply the normal distribution.

15. Suppose that ii% of all cell phone connections by a certain provider are dropped. Notice the probability that in a random sample of 1,500 calls at nearly 40 will be dropped. First verify that the sample is sufficiently large to employ the normal distribution.

16. Suppose that in xx% of all traffic accidents involving an injury, driver distraction in some form (for example, changing a radio station or texting) is a factor. Find the probability that in a random sample of 275 such accidents between 15% and 25% involve driver distraction in some form. Get-go verify that the sample is sufficiently large to use the normal distribution.

17. An airline claims that 72% of all its flights to a certain region arrive on time. In a random sample of 30 recent arrivals, 19 were on fourth dimension. You may assume that the normal distribution applies.

    1. Compute the sample proportion.
    2. Assuming the airline's claim is true, find the probability of a sample of size thirty producing a sample proportion then depression as was observed in this sample.

18. A humane society reports that 19% of all pet dogs were adopted from an animate being shelter. Assuming the truth of this exclamation, discover the probability that in a random sample of 80 pet dogs, between fifteen% and 20% were adopted from a shelter. You lot may presume that the normal distribution applies.

19. In ane study information technology was plant that 86% of all homes accept a functional smoke detector. Suppose this proportion is valid for all homes. Find the probability that in a random sample of 600 homes, between 80% and ninety% volition have a functional fume detector. You may presume that the normal distribution applies.

20. A state insurance commission estimates that 13% of all motorists in its country are uninsured. Suppose this proportion is valid. Notice the probability that in a random sample of 50 motorists, at to the lowest degree 5 will exist uninsured. You may presume that the normal distribution applies.

21. An outside financial accountant has observed that almost 4% of all documents he examines comprise an error of some sort. Bold this proportion to be accurate, notice the probability that a random sample of 700 documents will incorporate at to the lowest degree thirty with some sort of error. You may presume that the normal distribution applies.

22. Suppose vii% of all households have no abode telephone but depend completely on cell phones. Find the probability that in a random sample of 450 households, between 25 and 35 volition have no domicile phone. You may assume that the normal distribution applies.

Additional Exercises

23. Some countries permit individual packages of prepackaged goods to weigh less than what is stated on the parcel, subject area to sure conditions, such as the average of all packages being the stated weight or greater. Suppose that one requirement is that at virtually 4% of all packages marked 500 grams can counterbalance less than 490 grams. Bold that a product really meets this requirement, find the probability that in a random sample of 150 such packages the proportion weighing less than 490 grams is at least 3%. You may presume that the normal distribution applies.

24. An economist wishes to investigate whether people are keeping cars longer now than in the past. He knows that five years ago, 38% of all passenger vehicles in operation were at least 10 years old. He commissions a report in which 325 automobiles are randomly sampled. Of them, 132 are 10 years sometime or older.

    1. Find the sample proportion.
    2. Find the probability that, when a sample of size 325 is fatigued from a population in which the true proportion is 0.38, the sample proportion volition exist as large every bit the value you computed in part (a). You lot may assume that the normal distribution applies.
    3. Give an estimation of the issue in office (b). Is there stiff prove that people are keeping their cars longer than was the case five years agone?

25. A country public wellness department wishes to investigate the effectiveness of a campaign confronting smoking. Historically 22% of all adults in the state regularly smoked cigars or cigarettes. In a survey commissioned by the public health department, 279 of i,500 randomly selected adults stated that they fume regularly.

    1. Find the sample proportion.
    2. Find the probability that, when a sample of size 1,500 is drawn from a population in which the true proportion is 0.22, the sample proportion will be no larger than the value you computed in role (a). Y'all may assume that the normal distribution applies.
    3. Give an estimation of the outcome in part (b). How strong is the evidence that the campaign to reduce smoking has been constructive?

26. In an effort to reduce the population of unwanted cats and dogs, a group of veterinarians set up a low-cost spay/neuter clinic. At the inception of the clinic a survey of pet owners indicated that 78% of all pet dogs and cats in the community were spayed or neutered. Later on the depression-cost clinic had been in operation for iii years, that figure had risen to 86%.

    1. What information is missing that y'all would need to compute the probability that a sample drawn from a population in which the proportion is 78% (corresponding to the assumption that the depression-price clinic had had no effect) is as high as 86%?
    2. Knowing that the size of the original sample three years ago was 150 and that the size of the recent sample was 125, compute the probability mentioned in role (a). Y'all may assume that the normal distribution applies.
    3. Requite an interpretation of the event in part (b). How strong is the evidence that the presence of the low-price clinic has increased the proportion of pet dogs and cats that have been spayed or neutered?

27. An ordinary die is "fair" or "balanced" if each face has an equal chance of landing on top when the die is rolled. Thus the proportion of times a three is observed in a large number of tosses is expected to be close to one/6 or $0.1\stackrel{-}{\mathrm{half\; dozen}}.$ Suppose a die is rolled 240 times and shows iii on peak 36 times, for a sample proportion of 0.xv.

    1. Find the probability that a fair dice would produce a proportion of 0.fifteen or less. Y'all may assume that the normal distribution applies.
    2. Give an estimation of the result in part (b). How strong is the evidence that the die is not fair?
    3. Suppose the sample proportion 0.15 came from rolling the dice 2,400 times instead of only 240 times. Rework part (a) nether these circumstances.
    4. Give an interpretation of the result in part (c). How strong is the testify that the die is not fair?

Answers

1. ${\mathit{\mu }}_{\widehat{P}}=0.37$ , ${\mathit{\sigma }}_{\widehat{P}}=0.012$

3. ${\mathit{\mu }}_{\widehat{P}}=0.76$ , ${\mathit{\sigma }}_{\widehat{P}}=0.012$

5. $p\pm \mathrm{three}\sqrt{\frac{pq}{n}}=0.25\pm 0.087$ , yes

7. 1. $\widehat{p}\pm \mathrm{iii}\sqrt{\frac{\widehat{p}\widehat{q}}{n}}=0.48\pm 0.21$ , yes
   2. $\widehat{p}\pm 3\sqrt{\frac{\widehat{p}\widehat{q}}{n}}=0.12\pm \mathrm{0.fourteen}$ , no
   3. $\widehat{p}\pm 3\sqrt{\frac{\widehat{p}\widehat{q}}{n}}=0.12\pm 0.10$ , yes
9. 1. 0.4154
   2. 0.2546
11. 1. 0.7850
    2. 0.9980

13. $p\pm 3\sqrt{\frac{pq}{n}}=0.08\pm 0.05$

    and

    $\left[0.\mathrm{03,0}.\mathrm{thirteen}\right]\subset \left[\mathrm{0,1}\right],0.1210$

15. $p\pm \mathrm{three}\sqrt{\frac{pq}{n}}=0.02\pm 0.01$

    and

    $\left[0.\mathrm{01,0}.03\right]\subset \left[\mathrm{0,ane}\right],0.9671$

17. 1. 0.63
    2. 0.1446

19. 0.9977

21. 0.3483

23. 0.7357

25. 1. 0.186
    2. 0.0007
    3. In a population in which the true proportion is 22% the chance that a random sample of size 1500 would produce a sample proportion of 18.6% or less is only 7/100 of ane%. This is strong bear witness that currently a smaller proportion than 22% smoke.
27. 1. 0.2451
    2. Nosotros would expect a sample proportion of 0.15 or less in about 24.v% of all samples of size 240, so this is practically no evidence at all that the die is non off-white.
    3. 0.0139
    4. We would expect a sample proportion of 0.15 or less in only about 1.four% of all samples of size 2400, then this is stiff evidence that the die is not fair.

What Is The Probability That The Sample Mean For A Sample Of Size 34 Will Be More Than 32?,

Source: https://saylordotorg.github.io/text_introductory-statistics/s10-sampling-distributions.html

Posted by: culpepperconifice88.blogspot.com

Share this post