Question

Of the 1,300 children participating in a town's parks and recreations programs, 765 are under the age of 8. To encourage participation across all ages, city officials are studying which programs are the most popular in each age group. In the sampling distribution of sample proportions of size 100, above what proportion will 35% of all sample proportions be?

Answer 1

Answer:

35% of all sample proportions be will be above 0.6074.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

Of the 1,300 children participating in a town's parks and recreations programs, 765 are under the age of 8. This means that $\mu = \frac{765}{1300} = 0.5885$

In the sampling distribution of sample proportions of size 100, above what proportion will 35% of all sample proportions be?

We are using sample proportions of size 100, so

$\sigma = sqrt{\frac{0.5885*0.4115}{100}} = 0.0492$.

In the sampling distribution of sample proportions of size 100, above what proportion will 35% of all sample proportions be?

This proportion is the value of X in the 65th percentile, that is, when Z has a pvalue of 0.65. This is $Z = 0.385$. So

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

$0.385 = \frac{X - 0.5885}{0.0492}$

$X - 0.5885 = 0.385*0.0492$

$X = 0.6074$

35% of all sample proportions be will be above 0.6074.