The time spent waiting in the line is approximately normally distributed. The mean waiting time is 55 minutes and the standard deviation of the waiting time is 33 minutes. Find the probability that a person will wait for more than 88 minutes. Round your answer to four decimal places.

Zero.

Step-by-step explanation:

The variable here is t where t is = time spent waiting in the line

The values for this variable and in this study, are normally distributed.

The mean waiting time = 55 minutes

The standard deviation = 33 minutes

What is the probability that t is greater than 88 minutes?

Now, the confidence interval about the mean for this study is:

[55-33], [55+33]  = [22, 88]

Therefore, all values of t fall within this lower and upper limit or interval.

If that is the case, and recalling also that the distribution of values is normal, then it is impossible for t to be greater than 88 minutes.

Hence, P[t>88] = 0

To 4 decimal places, that would be 0.0000

14.3

step-by-step explanation:

add all the numbers together.

9 + 12 + 15 + 16 + 17 + 17 = 86

now, divide 86 by how many numbers we have: 6.

86 / 6 = 14.3