Download Darwin`s height Men have heights with a mean of 1

BASKIN SCHOOL OF ENGINEERING
Department of Applied Mathematics
and Statistics
AMS 7 Spring 2012
Name:
Quiz 2 Solutions
Please show your work in all the problems.
Problem 1: Darwin’s height
Men have heights with a mean of 176 cm and a standard deviation of 7 cm. Charles
Darwin had a height of 182 cm.
(1) What is the difference between Darwin’s heights and the mean?
182 − 176 = 6 cm
(2) How many standard deviations is that [the difference found in part (1)]?
6/7 = 0.857 standard deviations
(3) Convert Darwin’s height to a z score.
z = (182 − 176)/7 = 0.857
(4) If we consider usual heights to be those that convert to z scores between -2 and 2, is
Darwin’s height usual or unusual?
USUAL
(4 pts (1 each))
Problem 2: Consider a couple that has three children:
(1) Find all the possible outcomes of the sample space corresponding to the three children
gender.
Sample space= {BBG,BBB,BGB,BGG,GBG,GBB,GGB,GGG} (8 possibilities= 2 ×
2 × 2)
(2) Find the probability that among the three children, there is exactly one girl.
There are 3 possibilities among 8 (BBG,BGB,GBB). Probability = 3/8
(3) Find the probability that among the three children, there is at least one girl.
There are 7 possibilities among 8 (BBG,BGB,BGG,GBG,GBB,GGB,GGG). Probability = 7/8. Also Prob. of at least 1 girl= 1- Prob(No Girls)=1- P(BBB)=1-1/8=7/8
(6 pts (2 each))
Problem 3: Consider the following table on a pregnancy test results:
Positive test result Negative test result Total
Subject is pregnant
80
5
85
Subject is not pregnant
3
11
14
Total
83
16
99
(1) What is the probability of getting someone who tested positive, given that the selected
person was pregnant.
P r(T ested pos.|P regnant) = 80
= 0.94.
85
Also P r(T ested pos.|P regnant) = P r(T estedpos. and P regnant)/P r(P regnant) =
80/99
= 0.94.
85/99
(2) What is the probability of getting someone who resulted pregnant, given that the
selected person tested positive.
P r(P regnant|T ested pos.) = 80
= 0.96.
83
Also P r(P regnant|T ested pos.) = P r(P regnant and T ested pos.)/P r(T ested pos.) =
80/99
= 0.96
83/99
(4 pts (2 each) )
Problem 4: Use the following distribution for the next three questions.
x P(x)
0 0.03
1 0.19
2 .25
3 .28
4 .37
5 .39
A statistics student surveyed the probabilities for x, the number of classmates who were
not paying attention to the lecture.
(1) Identify the random variable for the statistics class distribution.
The random variable is: x= Number of classmates who were not paying attention in
the lecture.
(2) Identify the random variable as either discrete or continuous.
DISCRETE
(3) Determine whether a probability distribution is given in the statistics class distribution. If not, identify the requirements that are not satisfied
∑
The given distribution is not a probability distribution because
P (x) = 0.03 +
0.19 + . . . + 0.39 = 1.51. This value is greater than 1 and it should be = 1.
(6 pts (2 each))