Question

The management of a restaurant has been studying whether or not new customers return within a month. The collected data reveal that 60% of the new customers have returned. If 90 new customers dine at the restaurant this month, what is the probability that at least 60 will return next month? Use Normal approximation to Binomial distribution.

Answer 1

Using the normal approximation to the binomial, it is found that there is a 0.119 = 11.9% probability that at least 60 will return next month.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

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

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

• The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$.

In this problem, we have that:

• 60% of the new customers have returned, hence p = 0.6.

• 90 new customers dine at the restaurant this month, hence n = 90.

Thus, the mean and the standard error for the approximation are given as follows:

• $\mu = np = 90(0.6) = 54$

• $\sigma = \sqrt{np(1-p)} = \sqrt{90(0.6)(0.4)} = 4.6476$

Using continuity correction, the probability that at least 60 will return next month is P(X > 60 - 0.5) = P(X > 59.5), which is 1 subtracted by the p-value of Z when X = 59.5, hence:

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

$Z = \frac{59.5 - 54}{4.6476}$

Z = 1.18

Z = 1.18 has a p-value of 0.881.

1 - 0.881 = 0.119.

0.119 = 11.9% probability that at least 60 will return next month.

More can be learned about the normal distribution at https://brainly.com/question/24663213