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?

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

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 and standard deviation

In this question, we have that:

So

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

By the Central Limit Theorem

has a pvalue of 0.0418

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

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1