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. [tex]\mu=17.4[/tex] and [tex]\sigma=5.2[/tex]
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 :
[tex]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[/tex]
Hence, the probability that the mean wait time is greater than 20 minutes.= 0.0031
Thus , the correct answer is A: 0.0031 .
