Question

Some statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.72. Suppose that we randomly pick 25 daytime statistics students.

1. Find the probability that an individual had between $0.68 and $0.92. (Round your answer to four decimal places.)

2.Find the probability that the average of the 25 students was between $0.68 and $0.92. (Round your answer to four decimal places.)

Answer 1

Answer:

1.

Given,

Mean = 1/ = 0.72

P(X <= x) = 1 - e^-x

P(0.68 < X < 0.92) = P(x <= 0.92) - P(x <= 0.68)

= (1 - e^-(0.92/0.72)) - (1 - e^-(0.68/0.72))

= 0.7213 - 0.6111

= 0.1102

2.

sample n = 25

P(0.68 < X < 0.92) = P(z < (0.92 - 0.72)/(0.72/sqrt(25))) - P(z < (0.68 - 0.72)/(0.72/sqrt(25)))

= P(z < 1.39) - P(z < - 0.28)

= 0.9177356 - 0.3897388 [since from z table]

= 0.5280