Section 7.5: Summary of Applications of the Normal Distribution

This section is a summary of the applications of the normal distribution. As you read through this chapter, it is easy to become confused. Understanding which distribution is appropriate to the situation is essential to solving problems correctly.

In this chapter we have studied the following applications of the normal distribution:

•Finding probability using a normal distribution.

•Finding percentiles using a normal distribution.

•The distribution of sample means.

•The distribution of sample proportions.

This section is designed to assist you in determining which distribution and method is appropriate. See figure 7.33 for a flowchart summarizing the ways we have used the normal distribution in this chapter. The binomial and uniform distributions also appeared in this chapter but are not summarized here.

Figure 7.33: Flowchart of uses of the normal distribution in chapter 7.

Solution:This problem gives us information about American spending on impulse purchases. There are no samples being taken, so we are not dealing with sampling distributions. What is given is a normal distribution with mean $\mu =23$ and standard deviation $\sigma =9$. So, this is the left branch of the flowchart and is a section 7.2 problem. We will solve it using the normal calculator.

Using the Normal Calculator in STATBELT with mean 23 and standard deviation 9, $P\left(X<10\right)\approx 0.0743$. The interpretation is that about 7.4% of individual impulse purchases were for items that cost $10 or less.

Using the Normal Calculator in STATBELT, $P\left(X\ge 50\right)\approx 0.0013$. The interpretation is that about 0.1% of individual impulse purchases were for items that cost $50 or more. This suggests that it is rare to spend larger sums on impulse. When most consumers spend greater amounts, they tend to think more carefully about those purchases.

Once again, we use the normal calculator, this time for a percentile. Since we want to find $P_{90}$, we want the area on the left, or the probability on the left, to be 0.90. In the calculator, choose “P(X < _) = a” under “Find a value from an area” and set the area to 0.90.

$$\begin{array}{cc}& P\left(X\le \right)=0.90\mathrm{.}\end{array}$$

STATBELT computes the 90^th percentile to be $34.53. The interpretation is that 90% of impulse purchases are for items that cost less than $34.53, and any impulse purchase for more than $34.53 is in the top 10% of impulse purchase costs.

Solution:In this problem, a description of consumer spending is given with a mean and standard deviation, but no description of the shape of the distribution. The question states that samples of size n = 60 are randomly selected, meaning this is a sampling distribution question. Following the flowchart, we must determine whether this question is about the distribution of sample means (section 7.3) or the distribution of sample proportions (section 7.4). But part (a) clearly asks us to describe the distribution of sample means of 60 consumers, so this is a 7.3 question, and we will follow the second branch of the flowchart.

The parent distribution for X = amount spent by consumers on impulse purchases has mean $\mu =276$ dollars and standard deviation $\sigma =108$ dollars. The shape of the distribution is not stated.

The distribution of sample means, $\bar{X}$, of samples of size n = 60 has the same mean, ${\mu }_{\bar{X}}=\mu =276.$ The standard deviation is

$$\begin{array}{cc}{\sigma }_{\bar{X}}& =\frac{\sigma }{\sqrt{n}}=\frac{\text{original standard deviation}}{\sqrt{\text{sample size}}}\end{array}$$

$$\begin{array}{cc}& =\frac{108}{\sqrt{60}}\end{array}$$

$$\begin{array}{cc}& \approx \frac{108}{7.7459667}\end{array}$$

$$\begin{array}{cc}& \approx 13.94.\end{array}$$

We don’t know whether the original distribution is normally distributed, but $n=60\ge 30$, so $\bar{X}$ is automatically normally distributed. The distribution for $\bar{X}$ = the mean amount spent on impulse purchases in samples of 60 consumers:

•$\bar{X}$ has mean ${\mu }_{\bar{X}}=276$

•$\bar{X}$ has standard deviation ${\sigma }_{\bar{X}}\approx 13.94$

•$\bar{X}$ is normally distributed

Now, since $\bar{X}$ is normally distributed, we use the normal calculator with mean ${\mu }_{\bar{X}}=276$ and standard deviation ${\sigma }_{\bar{X}}\approx 13.94$. The probability that a sample would have mean spending of at least $300 is $P\left(\bar{X}\ge 300\right)\approx 0.0426$. This is less than 5%, so it would be unusual to have a sample of 60 consumers who spent more than $300 per month on impulse purchases.

The probability that a sample would have mean spending of less than $183 is $P\left(\bar{X}\le 183\right)\approx 1.27\times {10}^{-11}\approx 0.00000000001$.

There is about a one in 100 billion chance this sample would occur randomly in 2021. It is far more likely that the spending habits of consumers have changed since 2020. In fact, for much of 2020, consumers were stuck at home without access to impulse purchases, except online. In 2021, consumers had more access to traditional in-person shopping as well as online, which can explain some of the additional impulse spending. There is also the phenomenon called COVID fatigue that may have fueled “retail therapy” spending. In any case, we are using data from the two years to infer that there is a difference in spending patterns. More inference is coming up in Unit 3!

Solution:In this problem, we are given a percentage of a population, which is a proportion, but no mean, no standard deviation, and no description of the shape of a distribution. The question does state that samples of size n = 300 are randomly selected, meaning this is a sampling distribution question. The percentage that is given suggests it could be a proportions problem. In fact, part (a) asks us to describe the distribution of sample proportions, so this is a 7.4 question, and we will follow the right branch of the flowchart. Further, there is no parent distribution described so it is not a question about sample means.

The distribution of sample proportions has n = 300 and p = 0.66, which is 66% written in decimal form. The variable is  $\widehat{p}$, which in can be interpreted as a proportion or percent.

•The mean is ${\mu }_{\widehat{p}}=p=0.66$.

•The standard deviation needs to be calculated:

${\sigma }_{\widehat{p}}$ $=\sqrt{\left(\frac{p(1-p)}{n}\right)}=\sqrt{\left(\frac{0.66(1-0.66)}{300}\right)}$

$$\begin{array}{cc}& =\sqrt{\left(\frac{\left(0.66\right)\left(0.34\right)}{300}\right)}\end{array}$$

$$\begin{array}{cc}& =\sqrt{0.000748}\end{array}$$

$$\begin{array}{cc}& \approx 0.027350\end{array}$$

•To determine the shape, we check if $n\cdot p\cdot \left(1-p\right)\ge 10$.

$$\begin{array}{cc}n\cdot p\cdot \left(1-p\right)& =300\times \left(0.66\right)\times \left(1-0.66\right)\\ & =300\times \left(0.66\right)\times \left(0.34\right)\\ & =67.32\ge 10\mathrm{.}\end{array}$$

Since $n\cdot p\cdot \left(1-p\right)\ge 10$, the distribution of sampling proportions will have an approximately normal distribution.

So, the distribution of sample proportions is described as follows:

•$\widehat{p}$ has mean ${\mu }_{\widehat{p}}=0.66$

•$\widehat{p}$ has standard deviation ${\sigma }_{\widehat{p}}\approx 0.027350$

•$\widehat{p}$ is normally distributed

First, 70% = 0.70 = 0.7 as a decimal. So, we wish to find the probability that a sample of 300 students would have 70% or more attend college. Symbolically, that means we want $P(\widehat{p}\ge 0.70)$. Since $\widehat{p}$ is normally distributed, we use the Normal Calculator in STATBELT, using the mean 0.66 and standard deviation 0.027350.

Choosing “P(X > a)” and typing 0.7 for 70%, we get the probability $P\left(\widehat{p}\ge 0.70\right)\approx 0.0718$. There is about a 7.2% chance of obtaining a sample of 300 students who attended college immediately.

In this case, we will use the normal calculator to calculate the probability that a sample of 300 students would have 61% or less attend college immediately. Here, 61% is written in decimal form as 0.61. Symbolically, we want to find $P(\widehat{p}\le 0.61)$.

Choosing “P(X < a)” and typing 0.61, we get the probability $P\left(\widehat{p}\le 0.61\right)\approx 0.0338$. There is about a 3.4% chance that such a sample would have occurred by random chance in 2019. Since 3.4% is less than 5%, this would be considered unusual when compared to 2019 data. It appears that the number of graduates who enrolled in college immediately in 2020 was unusually low compared to pre-pandemic rates. Again, we are using inference to reach the conclusion that fewer high school graduates attended college immediately in 2020 than in the prior year.

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