Example 7.8
A study involving stress is conducted among the students on a college campus. The stress scores follow a uniform distribution with the lowest stress score equal to one and the highest equal to five. Using a sample of 75 students, find:
 the probability that the mean stress score for the 75 students is less than 2
 the 90^{th} percentile for the mean stress score for the 75 students
 the probability that the total of the 75 stress scores is less than 200
 the 90^{th} percentile for the total stress score for the 75 students
Let X = one stress score.
Problems (a) and (b) ask you to find a probability or a percentile for a mean. Problems (c) and (d) ask you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75.
Because the individual stress scores follow a uniform distribution, X ~ U(1, 5) where a = 1 and b = 5 (see Continuous Random Variables for an explanation of a uniform distribution),
$${\mu}_{X}\text{=}\frac{a+b}{2}=\frac{1+5}{2}=3$$
$${\sigma}_{X}\text{=}\sqrt{\frac{{(b\u2013a)}^{2}}{12}}=\sqrt{\frac{{(5\u20131)}^{2}}{12}}\text{=1}\text{.15}$$
In the formula above, the denominator is understood to be 12, regardless of the endpoints of the uniform distribution.
For problems (a) and (b), let $\overline{X}$ = the mean stress score for the 75 students. Then,
$$\overline{X}\text{~}N\text{}\left(3,\frac{1.15}{\sqrt{75}}\right)\text{where}n\text{=75}\text{.}$$
a. Find P($\overline{x}$ < 2). Draw the graph.
Solution 7.8
a. P($\overline{x}$ < 2) = 0
The probability that the mean stress score is less than 2 is about zero.
normalcdf
$\left(\text{1,2,3,}\frac{\text{1}\text{.15}}{\sqrt{\text{75}}}\right)$ = 0
Reminder
The smallest stress score is one.
b. Find the 90^{th} percentile for the mean of 75 stress scores. Draw a graph.
Solution 7.8
b. Let k = the 90^{th} precentile.
Find k, where P($\overline{x}$ < k) = 0.90.
$$k\text{=3}\text{.2}$$
The 90^{th} percentile for the mean of 75 scores is about 3.2. This tells us that 90 percent of all the means of 75 stress scores are at most 3.2, and that 10 percent are at least 3.2.
invNorm
$\left(\text{0}\text{.90,3,}\frac{1.15}{\sqrt{75}}\right)$ = 3.2
For problems (c) and (d), let ΣX = the sum of the 75 stress scores. Then, ΣX ~ N[(75)(3),$(\sqrt{75})$(1.15)].
c. Find P(Σx < 200). Draw the graph.
Solution 7.8
c. The mean of the sum of 75 stress scores is (75)(3) = 225.
The standard deviation of the sum of 75 stress scores is $(\sqrt{75})$(1.15) = 9.96.
$$P\text{(}\Sigma x\text{}\text{200)=0}$$
The probability that the total of 75 scores is less than 200 is about zero.
normalcdf
(75,200,(75)(3),$(\sqrt{75})$(1.15)).
Reminder
The smallest total of 75 stress scores is 75, because the smallest single score is one.
d. Find the 90^{th} percentile for the total of 75 stress scores. Draw a graph.
Solution 7.8
d. Let k = the 90^{th} percentile.
Find k where P(Σx < k) = 0.90.
$$k\text{}=\text{}237.8$$
The 90^{th} percentile for the sum of 75 scores is about 237.8. This tells us that 90 percent of all the sums of 75 scores are no more than 237.8 and 10 percent are no less than 237.8.
invNorm
(0.90,(75)(3),$(\sqrt{75})$(1.15)) = 237.8
Example 7.9
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract. The analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
X ∼ Exp$\left(\frac{1}{22}\right)$. From previous chapters, we know that μ = 22 and σ = 22.
Let $\overline{X}$ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance.
$\overline{X}$ ~ N$\left(\text{22,}\frac{\text{22}}{\sqrt{\text{80}}}\right)$ by the central limit theorem for sample means.
Using the clt to find probability
 Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes. This is asking us to find P($\overline{x}$ > 20). Draw the graph.
 Suppose that one customer who exceeds the time limit for his cell phone contract is randomly selected. Find the probability that this individual customer's excess time is longer than 20 minutes. This is asking us to find P(x > 20).
 Explain why the probabilities in parts (a) and (b) are different.
Solution 7.9

Find: P($\overline{x}$ > 20)
P($\overline{x}$ > 20) = 0.79199 using normalcdf
$\left(\text{20,1E99,22,}\frac{\text{22}}{\sqrt{\text{80}}}\right)$
The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
Reminder
1E99 = 10^{99} and –1E99 = –10^{99}. Press the EE
key for E. Or just use 10^{99} instead of 1E99.
 Find P(x > 20). Remember to use the exponential distribution for an individual. $X~Exp\left(\frac{1}{22}\right)$.
$$P(x>20)={e}^{\left(\left(\frac{1}{22}\right)(20)\right)}\text{or}{e}^{\text{(}\u2013\text{0}\text{.04545(20))}}\text{=0}\text{.4029}$$

 P(x > 20) = 0.4029, but P($\overline{x}$ > 20) = 0.7919
 The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means.
 When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. Use the clt with the normal distribution when you are being asked to find the probability for a mean.
Using the clt to find percentiles
Find the 95
^{th} percentile for the
sample mean excess time for a sample of 80 customers who exceed their basic contract time allowances. Draw a graph.
Solution 7.9
Let k = the 95^{th} percentile. Find k where P($\overline{x}$ < k) = 0.95.
k = 26.0 using invNorm
$\left(\text{0}\text{.95,22},\frac{22}{\sqrt{80}}\right)$ = 26.0
The 95^{th} percentile for the sample mean excess time used is about 26.0 minutes for a random samples of 80 customers who exceed their contractual allowed time.
95 percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.
Example 7.12
Suppose in a local kindergarten through 12^{th} grade (K–12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.
 Find the probability that at least 150 favor a charter school.
 Find the probability that at most 160 favor a charter school.
 Find the probability that more than 155 favor a charter school.
 Find the probability that fewer than 147 favor a charter school.
 Find the probability that exactly 175 favor a charter school.
Let X = the number that favor a charter school for grades K through 5. X ~ B(n, p) where n = 300 and p = 0.53. Because np > 5 and nq > 5, use the normal approximation to the binomial. The formulas for the mean and standard deviation are μ = np and σ = $\sqrt{npq}$. The mean is 159, and the standard deviation is 8.6447. The random variable for the normal distribution is Y. Y ~ N(159, 8.6447). See The Normal Distribution for help with calculator instructions.
For Part (a), you include 150 so P(X ≥ 150) has a normal approximation P(Y ≥ 149.5) = 0.8641.
normalcdf
(149.5,10^99,159,8.6447) = 0.8641.
For Part (b), you include 160 so P(X ≤ 160) has a normal approximation P(Y ≤ 160.5) = 0.5689.
normalcdf
(0,160.5,159,8.6447) = 0.5689
For Part (c), you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.
normalcdf
(155.5,10^99,159,8.6447) = 0.6572.
For Part (d), you exclude 147 so P(X < 147) has normal approximation P(Y < 146.5) = 0.0741.
normalcdf
(0,146.5,159,8.6447) = 0.0741
For Part (e), P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083.
normalcdf
(174.5,175.5,159,8.6447) = 0.0083
Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have computer software that calculates binomial probabilities. Many students have access to calculators that calculate probabilities for binomial distribution. If you type in binomial probability distribution calculation in an internet browser, you can find at least one online calculator for the binomial.
For Example 7.10, the probabilities are calculated using the following binomial distribution: (n = 300 and p = 0.53). Compare the binomial and normal distribution answers. See Discrete Random Variables for help with calculator instructions for the binomial.
P(X ≥ 150) :1  binomialcdf
(300,0.53,149) = 0.8641
P(X ≤ 160) :binomialcdf
(300,0.53,160) = 0.5684
P(X > 155) :1  binomialcdf
(300,0.53,155) = 0.6576
P(X < 147) :binomialcdf
(300,0.53,146) = 0.0742
P(X = 175) :(You use the binomial pdf.)binomialpdf
(300,0.53,175) = 0.0083