Question

Suppose that the weight students gain in their first year of college is Normally distributed with a mean of 10 pounds and a standard deviation of 2 pounds. (a) What is the probability that a randomly selected student will gain between 6 and 8 pounds? (b) Suppose we pick 5 students at random. What is the probability at least one will gain between 6 and 8 pounds? (c) If 30 students are randomly selected, what is the probability that the average weight gain of the students is between 6 and 8 pounds? Check the appropriate conditions and state the corresponding sampling distribution as part of your work. (a) The average weight gain of the 30 students in the sample was actually 6.5 pounds with a standard deviation of 3.5 lbs. Find a 90% confidence interval for the average weight gain based on the mean and standard deviation of this sample.

Answer 1

Answer:

Given,

Mean = 10

Standard deviation = 2

a)

P(6 < X < 8) = P((6-10)/2 < z < (8-10)/2)

= P(-2 < z < -1)

= P(z < -1) - P(z < -2)

= 0.1586553 - 0.0227501 [since from z table]

= 0.1359

b)

P(6 < X < 8) = P((6-10)/(2/sqrt(5)) < z < (8-10)/(2/sqrt(5)))

= P(- 4.47 < z < -2.24)

= P(z < - 2.24) - P(z < - 4.47)

= 0.0125455 - 0.0000039 [since from z table]

= 0.0125

c)

P(6 < X < 8) = P((6-10)/(2/sqrt(30)) < z < (8-10)/(2/sqrt(30)))

= P(-10.95 < z < -5.48)

= P(z < - 5.48) - P(z < - 10.95)

= 0

d)

degree of freedom = n - 1 = 30 - 1 = 29

alpha = 0.1

t = 1.699127 = 1.699

90% CI = xbar +/- t*s/sqrt(n)

substitute values

= 6.5 +/- 1.699*3.5/sqrt(30)

= 6.5 +/- 1.09

= (5.41 , 7.59)