Question

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

People end up tossing 12% of what they buy at the grocery store (Reader’s Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 100 grocery shoppers to further investigate their behavior.

Answer 1

The 10th percentile is 0.0784.

The 90th percentile is 0.1616.

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

In this problem, we have that:

`\mu = 0.12, \sigma = \sqrt{(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 = (X - \mu)/(\sigma)`

`-1.28 = (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 = (X - \mu)/(\sigma)`

`1.28 = (X - 0.12)/(0.0325)`

`X - 0.12 = 1.28*0.0325`

`X = 0.1616`

The 90th percentile is 0.1616.