The duration of shoppers' time in Browse Wrld's new retail outlets is normally distributed with a mean of 27.8 minutes and a standard deviation of 11.4 minutes. How long must a visit be to put a shopper in the longest 10 percent

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean and standard deviation , the z-score of a measure X is given by:

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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 27.8 minutes and a standard deviation of 11.4 minutes.

This means that

How long must a visit be to put a shopper in the longest 10 percent?

The 100 - 10 = 90th percentile, which is X when Z has a p-value of 0.9, so X when Z = 1.28.

A visit must be of at least 42.39 minutes to put a shopper in the longest 10 percent.

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.

sine

step-by-step explanation: