Question

What are the 10th and 90th percentiles of the distribution of sample proportion for the population proportion below?

Answer 1

Answer:

The 10th percentile is 0.0784.

The 90th percentile is 0.1616.

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

In this problem, we have that:

$\mu = 0.12, \sigma = \sqrt{\frac{0.12*0.88}{100}} = 0.0325$

10th percentile:

X when Z has a pvalue of 0.1. So X when Z = -1.28.

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

$-1.28 = \frac{X - 0.12}{0.0325}$

$X - 0.12 = -1.28*0.0325$

$X = 0.0784$

The 10th percentile is 0.0784.

90th percentile:

X when Z has a pvalue of 0.9. So X when Z = 1.28.

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

$1.28 = \frac{X - 0.12}{0.0325}$

$X - 0.12 = 1.28*0.0325$

$X = 0.1616$

The 90th percentile is 0.1616.