Question

Student records suggest that the population of students spends an average of 6.30 hours per week playing organized sports. The population's standard deviation is 2.10 hours per week. Based on a sample of 49 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.
A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?
B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

Answer 1

Answer:

a) 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

b) 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.

Step-by-step explanation:

To solve this question, it is important to know the Normal probability distribution and the Central Limit Theorem

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

In this problem, we have that:

$\mu = 6.3, \sigma = 2.1, n = 49, s = \frac{2.1}{\sqrt{49}} = 0.3$

A) What is the chance HLI will find a sample mean between 5.5 and 7.1 hours?

This is the pvalue of Z when X = 7.1 subtracted by the pvalue of Z when X = 5.5.

By the Central Limit Theorem, the formula for Z is:

$Z = \frac{X - \mu}{s}$

X = 7.1

$Z = \frac{7.1 - 6.3}{0.3}$

$Z = 2.67$

$Z = 2.67$ has a pvalue of 0.9962

X = 5.5

$Z = \frac{5.5 - 6.3}{0.3}$

$Z = -2.67$

$Z = -2.67$ has a pvalue of 0.0038

So there is a 0.9962 - 0.0038 = 0.9924 = 99.24% chance HLI will find a sample mean between 5.5 and 7.1 hours.

B) Calculate the probability that the sample mean will be between 5.9 and 6.7 hours.

This is the pvalue of Z when X = 6.7 subtracted by the pvalue of Z when X = 5.9

X = 6.7

$Z = \frac{6.7 - 6.3}{0.3}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

X = 5.9

$Z = \frac{5.9 - 6.3}{0.3}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918.

So there is a 0.9082 - 0.0918 = 0.8164 = 81.64% probability that the sample mean will be between 5.9 and 6.7 hours.