Question

16) Assuming that the number of shoppers who buy a packet of crackers is approximately normal with a mean of 20 and standard deviation of 4, what is the approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

Answer 1

Answer:

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack

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.

In this problem, we have that:

$\mu = 20, \sigma = 4$

What is the approximate probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack?

This probability is 1 subtracted by the pvalue of Z when X = 28. So

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

$Z = \frac{28 - 20}{4}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

1 - 0.9772 = 0.0228

2.28% probability that at least 28 of the next 100 shoppers who sample the crackers will buy a pack