Question

The waiting times at a certain bank are normally distributed with a mean waiting time of 3.7 minutes and a standard deviation of 1.4 minutes. Find the waiting time at the 97th percentile. That is, find the wait time that separates the bottom 97% from the rest.
Select one:

a. .02

b. .97

c. 1.9

d. 6.3

Answer 1

Answer:

From the given options we can say the correct answer is d.) 6.3

Step-by-step explanation:

The waiting times at the bank are given to be normally distributed.

The mean of the waiting time is given as  = 3.7 minutes

The given standard deviation is = 1.4 minutes

It is required to find the waiting time at the 97th percentile.

So we are required to find the time which can be said to be the waiting time which occurs with 97 % probability.

Let the time required be designated as y.

Therefore we can write p(X < y) = 0.97 (where 0.97 means the 97th percentile)

This is a left tailed test.

Using the formula from MSEXCEL to find y we get

y = NORMINV(0.97, 3.7, 1.4) = 6.333.

The other approach would to find the Z value corresponding to the probability value from the Z Table

The Z value for a probability of 0.97 gives us Z  = 1.881  .

Therefore $\frac{y - 3.7}{1.4} = 1.881$. Therefore y = (1.881 × 1.4) + 3.7  = 6.33

Therefore from the given options we can say the correct answer is d.) 6.3