PROBLEM SET-4 Continuous Probability - Solutions

# PROBLEM SET-4 Continuous Probability - Solutions -...

This preview shows pages 1–3. Sign up to view the full content.

STATISTICS FOR SOCIAL SCIENCES BUS 152 ___________________________________________________________________________ PROBLEM SET – 4 SOLUTIONS SOME CONTINUOUS PROBABILITY DISTRIBUTIONS Problem 1 (6.1) Given a standardized normal distribution, what is the probability that a. Z is less than 1.57? b. Z is greater than 1.84? c. Z is between 1.57 and 1.84? d. Z is less than 1.57 or greater than 1.84? SOLUTIONS: a) P(Z < 1.57)  = 0.9418 b) P(Z > 1.84)  = 1 – 0.9671  = 0.0329 c) P(1.57  < Z < 1.84)  = 0.9671  – 0.9418  = 0.0253 d) P(Z < 1.57)  + P(Z > 1.84)  = 0.9418  + (1  – 0.9671)  = 0.9747 _________________________________________________________________________________ Problem 2 (6.3) Given a standardized normal distribution, what is the probability that a. Z is less than 1.08? b. Z is greater than -0.21? c. Z is less than -0.21 or greater than the mean? d. Z is less than -0.21 or greater than 1.08? SOLUTIONS: a) P(Z < 1.08)  = 0.8599 b) P(Z > – 0.21)  = 1.0  – 0.4168  = 0.5832   c) P(Z < – 0.21)  + P(Z > 0)  = 0.4168  + 0.5  = 0.9168 d) P(Z < – 0.21)  + P(Z > 1.08)  = 0.4168  + (1  – 0.8599)  = 0.5569 _________________________________________________________________________________ Problem 3 (6.5) Given a standardized normal distribution with 100 = μ and 10 = σ , what is the probability that a. 75 X ? b. 70 < X ? c. 110 X or 80 < X ? d. 80% of the values are between what two X values (symmetrically distributed around the mean)? SOLUTIONS: a) Z = X = 75 –100 10 = – 2.50 P(X > 75)  = P(Z > – 2.50)  = 1 – P(Z< – 2.50)  = 1 – 0.0062  = 0.9938 b) Z = X = 70 – 100 10 = – 3.00 P(X < 70)  = P(Z < – 3.00)  = 0.00135

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
c) Z = X μ σ = 80 – 100 10 = –2.00 Z = X = 110 – 100 10 = 1.00 P(X < 80)  = P(Z < – 2.00)  = 0.0228 P(X > 110)  = P(Z > 1.00)  = 1 – P(Z < 1.00)  = 1.0  – 0.8413  = 0.1587 P(X < 80)  + P(X > 110)  = 0.0228  + 0.1587  = 0.1815 d) P(X lower  < X < X upper ) = 0.80 P(– 1.28 < Z) = 0.10 and P(Z < 1.28) = 0.90 Z = –1.28 = X lower –100 10 Z = + 1.28 = X upper – 100 10 X lower  = 100 – 1.28(10) = 87.20   and   X upper  = 100 + 1.28(10) = 112.80 _________________________________________________________________________________ Problem 4 (6.7) During 2001, 61.3% of U.S. households purchased ground coffee and spent an average of \$36.16 on ground coffee during the year. Consider the annual ground coffee expenditures for households purchasing ground coffee, assuming that these expenditures are approximately distributed as a normal random variable with a mean of \$36.16 and a standard deviation of \$10.00. a.
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/22/2011 for the course BUSINESS bus 152 taught by Professor Çeşmecibaşı during the Spring '11 term at Middle East Technical University.

### Page1 / 8

PROBLEM SET-4 Continuous Probability - Solutions -...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online