Question

When a truckload of apples arrives at a packing plant, a random sample of 125 apples are selected and examined for bruises and other defects. In reality, 9% of the apples on a particular truck are bruised or otherwise unsatisfactory. (a) How many standard errors away from 0.09 would you need to go to contain 89% of the sample proportions of bad apples you might expect to find

Answer 1

Answer:

You would need to go 1.6 standard errors away from 0.09.

Step-by-step explanation:

We can solve this using the normal distribution, because, by the Central Limit Theorem, the sampling distribution of the sample proportions is approximately normal.

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.

89% of the sample proportions.

50 - 89/2 = 5.5th percentile

50 + 89/2 = 94.5th percentile

5.5th percentile

X when Z has a pvalue of 0.055. So X when Z = -1.6.

So 1.6 standard error away from 0.09.

94.5th percentile

X when Z has a pvalue of 0.945. So X when Z = 1.6.

So 1.6 standard error away from 0.09.

You would need to go 1.6 standard errors away from 0.09.