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

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

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 = (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{(p(1 - p))/(n)}`

In this problem:

`\mu = 0.82, \sigma = \sqrt{(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 = (X - \mu)/(\sigma)`

`Z = (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