Question

A restaurant supply store surveys all chefs in the area and finds that 64 of the 85 head chefs in the area prefer their stores to the other restaurant supply store in town. To increase their sales, they are doing a statistical study of all chefs in the area. Use a calculator to find above what proportion will 80% of all sample proportions be in the sampling distribution of sample proportions of size 45.

Answer 1

Answer:

80% of all sample proportions in the sampling distribution of sample proportions of size 45 will be above 0.6988.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = \frac{64}{85} = 0.7529, n = 45, \mu = 0.7529, s = \sqrt{\frac{0.7529*0.2471}{45}} = 0.0643$

Above what proportion will 80% of all sample proportions be in the sampling distribution of sample proportions of size 45.

Above the 100 - 80 = 20th percentile, which is X when Z has a pvalue of 0.2. So X when Z = -0.842.

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

By the Central Limit Theorem

$-0.842 = \frac{X - 0.7529}{0.0643}$

$X - 0.7529 = -0.842*0.0643$

$X = 0.6988$

80% of all sample proportions in the sampling distribution of sample proportions of size 45 will be above 0.6988.