The time spent waiting in the line is approximately normally distributed. The mean waiting time is 5 minutes and the standard deviation of the waiting time is 2 minutes. Find the probability that a person will wait for more than 8 minutes. Round your answer to four decimal places.

0.0668 = 6.68% probability that a person will wait for more than 8 minutes.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Find the probability that a person will wait for more than 8 minutes.

This is 1 subtracted by the pvalue of Z when X = 8. So

has a pvalue of 0.9332

1 - 0.9332 = 0.0668

0.0668 = 6.68% probability that a person will wait for more than 8 minutes.

4 packs of water and 7 packs of candy bars

step-by-step explanation:

35x4=140

140/20=7

i think asking google will you get the answer to that question