Question

A recent survey found that 86% of employees plan to devote at least some work time to follow games during the NCAA Men's Basketball Tournament. A random sample of 100 employees was selected. What is the probability that less than 80% of this sample will devote work time to follow games?

Answer 1

Answer:

4.18% probability that less than 80% of this sample will devote work time to follow games

Step-by-step explanation:

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question, we have that:

$p = 0.86, n = 100$

So

$\mu = 0.86, s = \sqrt{\frac{0.86*0.14}{100}} = 0.0347$

What is the probability that less than 80% of this sample will devote work time to follow games?

This is the pvalue of Z when X = 0.8. So

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

By the Central Limit Theorem

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

$Z = \frac{0.8 - 0.86}{0.0347}$

$Z = -1.73$

$Z = -1.73$ has a pvalue of 0.0418

4.18% probability that less than 80% of this sample will devote work time to follow games