Question

In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a standard deviation of 5.2 minutes. If 30 days are randomly selected, find the probability that the mean wait time is greater than 20 minutes.
A: 0.0031

B: 0.1023

C: 0.3207

D: 0.9987

Answer 1

Answer: A: 0.0031

Step-by-step explanation:

Given : In a study of wait times at an amusement park, the most popular roller coaster has a mean wait time of 17.4 minutes with a standard deviation of 5.2 minutes.

i.e. $\mu=17.4$ and $\sigma=5.2$

We assume that the wait times are normally distributed.

samples size : n= 30

Let x denotes the sample mean wait time.

Then, the probability that the mean wait time is greater than 20 minutes will be :

$P(x>20)=1-P(x\leq20)\\\\=1-P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{20-17.4}{\dfrac{5.2}{\sqrt{30}}})\\\\=1-P(z\leq2.74)\ \ [\because\ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.9969\ \ [\text{ By z table}]\\\\=0.0031$

Hence, the probability that the mean wait time is greater than 20 minutes.= 0.0031

Thus , the correct answer is A: 0.0031 .