A survey reported that 63.63% of moviegoers

Mathematics

1 answer:

jarptica [38.1K]3 years ago

4 0

Answer:

63 63/100

Step-by-step explanation:

You might be interested in

Find the outlier of the data set<br> 3.1, 5.3, 4.3, 0.7, 5.6, 5.2, 5.8, 5.4

Vilka [71]

0.7

Low outlier = Q1 - 1.5(IQR)
High outlier = Q3 + 1.5(IQR)

5 0

3 years ago

According to a study done by the Gallup organization, the proportion of Americans who are satisfied with the way things are goin

UkoKoshka [18]

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

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

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