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

Step-by-step 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 :-

$z=\dfrac{x-\mu}{\sigma}$

For x = 64

$z=\dfrac{64-64}{18}=0$

For x = 100

$z=\dfrac{100-64}{18}=2$

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