Question

THE NEXT QUESTIONS ARE BASED ON THE FOLLOWING INFORMATION:

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. Assume that these values are representative of all high school students.

Showing the work on all questions is required.

1) What is the probability that for a sample of 25 students, the average amount spent exceeds $60?

2) What is the likelihood of finding a sample of 20 students spent an average between $50 and $60?

3) Find the average expenditure on entertainment for the 95th percentile for a group of 10 students.

Answer 1

#1.
P(X > 60)
= P(t > (60 - 52.3)/(18.23/sqrt(25)))
= P(t > 2.1119)
= 0.0226

#2.
P(50 < X < 60)
= P((50 - 52.3)/(18.23/sqrt(20)) < t < (60 - 52.3)/(18.23/sqrt(20)))
= P(-0.5642 < t < 1.8889)
= 0.6733

#3.
t-value = 1.8331
x = 52.3 + 1.8331*(18.23/sqrt(10))
x = 62.8675