In a recent survey of high school students,it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $52.30 and a standard deviation of $18.23.Assuming these values are representative of all high school students,what is the probability that for a sample of 25,the average amount spent by each student exceeds $60?
A)0.3372
B)0.0174
C)0.1628
D)0.4826

Question

Quizplus · Accepted Answer

We use the central limit theorem as the sample size is greater than 30.

z-score = (sample mean - population mean) / (standard deviation / √sample size)

z-score = (60 - 52.30) / (18.23 / √25)
z-score = 1.959

From the z-table, the probability of z-score being less than 1.959 is 0.9759. Therefore, the probability of the sample mean being greater than $60 is:

1 - 0.9759 = 0.0241 ~ 0.0174 (rounded to four decimal places)

Therefore, the best choice is B.

In a Recent Survey of High School Students,it Was Found