Question

A restaurant has found that the probability that a customer will complete an online survey is 0.12. In a group of 300 customers, find the mean and standard deviation of the number who will complete the survey.

Answer 1

We know that the population proportion for this study is p=0.12.

If the sample has a size of n=300, we can estimate the mean as the proportion times the sample size:

$M=n\cdot p=300\cdot0.12=36$

The stantard deviation can be calculated as:

$s=np(1-p)=300\cdot0.12\cdot(1-0.12)=300\cdot0.12\cdot0.88=31.68$

Answer:

Mean = 36

Standard deviation = 31.68