7.4 Using the Central Limit Theorem

A study involving stress is conducted among the students on a college campus. Based on previous work with stress scores, it is known that 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 following:

1. Find the probability that a randomly selected student scores between 2.7 and 3.1.
2. Find the probability that a randomly selected group of 75 students has a mean score between 2.7 and 3.1.

Solution:

Problem 1: It is critical that we identify what random variable is being discussed. Here the question is about a single randomly selected student. The random variable [latex]X[/latex] = stress score and the question is asking about how likely it is for a randomly selected score to be between 2.7 and 3.1. We are asked to find [latex]P(2.7 \lt X \lt 3.1)[/latex], where [latex]X \sim U(1, 5)[/latex].  So we can compute this value as an area or by using technology.  [latex]P(2.7 \lt X \lt 3.1) = 0.1[/latex]. A randomly selected student will have a score between 2.7 and 3.1 with probability 0.1, or about 10% of students will have this range of scores.

Problem 2: Notice the change in wording for this question. We are now being asked about a probability related to the sample mean, which is also a random variable and behaves according to the Central Limit Theorem. The random variable [latex]X[/latex] has a uniform distribution, but because the sample size is large enough, the sample mean has a normal distribution.

We are asked to find [latex]P(2.7 < \bar{x} < 3.1)[/latex], where [latex]X \sim U(1,5)[/latex]. Since [latex]n \geq 30[/latex], we know [latex]\bar{X} \sim N \bigg(\mu_{X}, \frac{\sigma_{X}}{\sqrt{n}}\bigg)[/latex]. For [latex]X \sim U(1, 5)[/latex], we know from our previous study of the uniform distribution that [latex]\mu_{X} = \frac{1+5}{2} = 3[/latex].  Likewise, [latex]\sigma_{X} = \sqrt{ \frac{(5-1)^2}{12} } = 1.1547[/latex] and [latex]\frac{\sigma_{X}}{\sqrt{n}} = \frac{1.1547}{\sqrt{75}} \approx 1.3333[/latex]. Thus, [latex]\bar{X} \sim N(3, 1.3333)[/latex] and we can compute this probability as an area.  We find [latex]P(2.7 \lt \bar{x} \lt 3.1) \approx 0.12[/latex]. See Figure 2.

The probability that a randomly selected group of 75 students has a mean score between 2.7 and 3.1 is approximately 0.12, which is greater than the probability for an individual student.

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. Let the random variable [latex]X[/latex] = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance. Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.

1. Find the probability that the mean excess time used by the 80 customers in the sample is longer than 20 minutes and draw the graph.
2. 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.
3. Explain why the probabilities in parts 1 and 2 are different.
4. Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed their basic contract time allowances.

Solution: Let's get organized and make sure we clearly state the distributions of [latex]X[/latex] and of [latex]\bar{X}[/latex]. We know [latex]X[/latex] refers to the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance and [latex]\bar{X}[/latex] refers to the mean excess time used by a sample of 80 customers who exceed their contracted time allowance. From our previous work with the exponential distribution, we know that the mean and standard deviation are the same, so μ = 22 and σ = 22. We also know [latex]\mu = \frac{1}{\lambda}[/latex], so the parameter [latex]\lambda = \frac{1}{22}[/latex].

[latex]X \sim \text{Exp} \big(\frac{1}{22} \big)[/latex].

[latex]\bar{X} \sim N \big(22,\frac{22}{\sqrt{80}} \big)[/latex] by the Central Limit Theorem.

1. Find: [latex]P( \bar{x} \geq 20 ) \approx 0.7919[/latex]
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.

   

2. Find P(x > 20). Remember to use the exponential distribution for an individual: [latex]X \sim \text{Exp} \big(\frac{1}{22} \big)[/latex].[latex]P(x \gt 20) = e^{-(\frac{20}{22})} \approx 0.4029.[/latex]
3. 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, we use the exponential distribution. When asked to find the probability for a mean, we use the distribution of the mean.
4. Let k = the 95th percentile. Find k where [latex]P(\bar{x} \lt k) = 0.95[/latex]. Using a statistical program, we find k = 26.05. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. The shaded area shows that [latex]P(\bar{x} \lt 26.05) = 0.95[/latex].The 95th percentile for the sample mean excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time. Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.
   

   

Based on data from the National Health Survey, women between the ages of 18 and 24 have a systolic blood pressure (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.

1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
3. If the sample were four women between the ages of 18 to 24, and we did not know the original distribution, could the central limit theorem be used?

Solutions:

1. The random variable [latex]X[/latex] represents the systolic blood pressure for a woman between the ages of 18 and 24. [latex]X \sim N(114.8, 13.1)[/latex]. [latex]P(X \gt 120) \approx 0.3457.[/latex] There is about a 35%, that the randomly selected woman will have a systolic blood pressure greater than 120.
   

   

2. The random variable [latex]\bar{X}[/latex] represents the mean systolic blood pressure of a sample of 40 women. [latex]\bar{X} \sim N \big(114.8, \frac{13.1}{\sqrt{40}} \big)[/latex]. [latex]P( \bar{x} \gt 120) \approx 0.0060[/latex]. There is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.
   

   

3. The Central Limit Theorem cannot be used for a sample size of four when we do not know the original distribution. The sample size would be too small.

A new airplane is being designed and all dimensions such as door height, seat width, etc, must consider human physical dimensions. According to a Vital and Health Statistics report, the mean height of males, age 20 and over in United States, is 69 inches, with a standard deviation (interpreted from the report) of 2.9 inches. Because men have a greater average height, this question focuses on males and heights are normally distributed.

1. What doorway height would allow 95% of men to enter the aircraft without bending?
2. A flight contains 115 men age 20 and over. What doorway height satisfies the condition that there is a 0.95 probability that this doorway height is greater than the mean of the height of 115 men?
3. For engineers designing the 757, which result is more relevant: the height from part 1 or part 2? Why?

Solution:

1. This question directs us to investigate the random variable [latex]X[/latex] representing heights of individual males age 20 and older. We know that  [latex]X \sim N(69, 2.9)[/latex], so using a statistical package, we find the value [latex]k[/latex], such that [latex]P(x \lt k) = 0.95[/latex].  In this case, a door height of 73.77 inches would allow 95% of males to pass through without bending.
   

   

2. Notice how the question has shifted to discussing the mean of 115 men, so we are dealing with the sample mean and have to refer to its sampling distribution. The original population is normal, and we have a large sample size, so we can use the Central Limit Theorem to state [latex]\bar{X} \sim N \big(\mu_{X}, \frac{\sigma_{X}}{\sqrt{n}} \big)[/latex]. Using a statistical package, we find the value [latex]k[/latex], such that [latex]P(\bar{x} \lt k) = 0.95[/latex], with [latex]\bar{X} \sim N (69, 0.27)[/latex].  The doorway height is 69.44 inches.
   

   

3. When designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part 1.

Suppose in a local Kindergarten through 12th 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 between 150 and 175 people favor a charter school. Use the normal approximation to the binomial with and without the continuity correction and use the binomial distribution to calculate the probability directly. Compare the results and provide graphs.

Solution:

The random variable X is defined as the number that favor a charter school for grades K trough 5. Because those surveyed either favor or do not favor a charter school, [latex]X \sim Bin(n, p)[/latex] where [latex]n[/latex] = 300 and [latex]p[/latex] = 0.53. Since [latex]np = (300)(0.53)=159[/latex] and [latex]nq = (300)(0.47)=141[/latex], use the normal approximation to the binomial. The formulas for the mean and standard deviation are [latex]\mu = np[/latex] and [latex]\sigma = \sqrt{npq}[/latex]. The mean is 159 and the standard deviation is 8.6447. Using the normal approximate to the binomial, [latex]X \sim N(159, 8.6447)[/latex].

To incorporate the continuity correction we find [latex]P(149.5 \lt x \lt 175.5) = 0.836[/latex]. See Figure 9. Note, without the continuity correction [latex]P(150 \lt x \lt 175) = 0.819[/latex]. Using the binomial distribution we find [latex]P(150 \leq x \leq 175) = 0.8362[/latex]. See Figure 10. Notice that using the continuity correction provides a much more accurate result.