Question

The mean time from prescription drop-off to medicine pickup for a customer served at Heisenberg's Pharmacy is 64 minutes, with a standard deviation of 18 minutes. Assuming a normal distribution, what is the probability that a randomly chosen customer experiences service done between 64 and 100 minutes? A. 0.6826 B. 0.4772 C. 0.3413 D. 0.5000

Answer 1

Answer: B. 0.4772

Explanation:

Given : The mean time :
`\mu=64\text{ minutes}`

Standard deviation :
`\sigma=18\text{minutes}`

Let X be the service time of a randomly selected customer.

Assuming a normal distribution, the value of z-score is given by :-

`z=(x-\mu)/(\sigma)`

For x = 64

`z=(64-64)/(18)=0`

For x = 100

`z=(100-64)/(18)=2`

The p-value =
`P(64<x<100)=P(0<z<2)`

`P(z<2)-P(z<0)=0.9772498-0.5=0.4772498\approx0.4772`

Hence, the probability that a randomly chosen customer experiences service done between 64 and 100 minutes = 0.4772