Answer to Question #326330 in Statistics and Probability for VITHU

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

First, we must verify that the following criteria are met:

$np\ge5,\\ n(1-p) \ge5.$

In this case, we have:

$np=90\cdot0.6=54,\\ n(1-p)=90\cdot(1-0.6)=36.$

Both numbers are greater than 5, so we’re safe to use the normal approximation.

We must apply a continuity correction when calculating probabilities. We want to find $P(X\ge60)$. To use the normal distribution to approximate the binomial distribution, we would instead find $P(X\ge59. 5) .$

The mean and standard deviation of the binomial distribution:

$\mu=np=90\cdot0.6=54,\\ \sigma=\sqrt{np(1-p)}=\\ =\sqrt{90\cdot0.6\cdot(1-0.6)}=4.65.$

The z-score:

$z=\cfrac{x- \mu} {\sigma}=\cfrac{59. 5- 54} {4.65}=1.18.$

$P(X>59.5)=P(Z>1.18)=\\ =1-P(Z<1.18)=1-0.8810=0.1190.$