Question

The number of times that students go to the movies per year has mean is a normal distribution with a mean of 17 with standard deviation of 8. What is the probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times?

Answer 1

Answer:

53.84% probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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 sample of size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 17, \sigma = 8, n = 10, s = \frac{8}{\sqrt{10}} = 2.53$

What is the probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times?

This probability is the pvalue of Z when X = 18 subtracted by the pvalue of Z when X = 14. So

X = 18

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

By the Central Limit Theorem

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

$Z = \frac{18 - 17}{2.53}$

$Z = 0.4$

$Z = 0.4$ has a pvalue of 0.6554

X = 14

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

$Z = \frac{14 - 17}{2.53}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170

0.6554 - 0.1170 = 0.5384

53.84% probability that for a group of 10 students, the mean number of times they go to the movies each year is between 14 and 18 times.