Question

According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things are going in their lives is 0.82. In a sample of 100 Americans, what is the probability that the proportion who are satisfied with the way that things are going in their life exceeds 0.85

Answer 1

Answer:

21.77% probability that the proportion who are satisfied with the way that things are going in their life exceeds 0.85

Step-by-step explanation:

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.

For proportions p in a sample of size n, we have that $\mu = p, \sigma = \sqrt{\frac{p(1 - p)}{n}}$

In this problem:

$\mu = 0.82, \sigma = \sqrt{\frac{0.82*0.18}{100}} = 0.0394$

In a sample of 100 Americans, what is the probability that the proportion who are satisfied with the way that things are going in their life exceeds 0.85

This is 1 subtracted by the pvalue of Z when X = 0.85. So

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

$Z = \frac{0.85 - 0.82}{0.0384}$

$Z = 0.78$

$Z = 0.78$ has a pvalue of 0.7823

1 - 0.7823 = 0.2177

21.77% probability that the proportion who are satisfied with the way that things are going in their life exceeds 0.85