Question

The wait times in line at a grocery store are roughly distributed normally with an average wait time of 7.6 minutes and a standard deviation of 1 minute 45 seconds. What is the probability that the wait time is less than 7.9 minutes

Answer 1

Answer:

$0.56749$.

Step-by-step explanation:

We have been given that the wait times in line at a grocery store are roughly distributed normally with an average wait time of 7.6 minutes and a standard deviation of 1 minute 45 seconds. We are asked to find the probability that the wait time is less than 7.9 minutes.

First of all, we will convert 45 seconds into minutes by dividing by 60 as:

$45\text{ Seconds}=\frac{45}{60}\text{ minutes}=0.75\text{ minutes}$

So 1 minute 45 seconds will be equal to 1.75 minutes.

Now, we will find z-score corresponding to 7.9 minutes using z-score formula.

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

z = z-score,

x = Random sample score,

$\mu$ = Mean,

$\sigma$ = Standard deviation.

$z=\frac{7.9-7.6}{1.75}$

$z=\frac{0.3}{1.75}$

$z=0.17142\approx 0.17$

Now we will use normal distribution to find area under normal curve that corresponds to a z-score of 0.17 that is $P(z<0.17)$.

$P(z<0.17)=0.56749$

Therefore, the probability that wait time is less than 7.9 minutes would be 0.56749.