MAT 107 Test #2 NAME:
Directions: Read and answer each question carefully. Do NOT remove the staple! Use backs of pages if necessary. Show all work! Probabilities should be carried out to 3 decimal places or left as fractions. Circle or box your answers!
a. Is male.
b. Is both female and had no hangovers
c. Had a hangover twice or more, given she is female.
d. Is male, given he had only one hangover.
e. Are being Male and having one hangover independent?
|
|
Number |
Of |
Hangovers |
|
Gender |
None |
One |
Two or More |
|
Male |
61 |
23 |
40 |
|
Female |
66 |
25 |
36 |
|
X |
-5 |
0 |
5 |
10 |
|
p(x) |
0.40 |
0.25 |
? |
0.10 |
Answers:
1. P( X > 15) = 0.050, P( X < 5) = 0.000, P(5 <= X <=15) = 0.949
μ = 20 * .6 = 12, σ = sqrt( 20*
.6 * .4) = 2.191
2. P(male) = 124/251, P(female ∩
None) = 66/251
P(Twice
| Female) = 36/127, P(Male | One) = 23/48
Male and One are not independent,
P(male) * P(one) not = P(male ∩one)
P(one)
= 48/251, P(male) * P(one) = .094, P(male ∩one) = .092
3. P( X
< 137) = P( Z < -1.27) = .1020
P( X
> 215) = P( Z > 1.33) = .0918
P(137
< X < 200) = P(Z < 0.83) – P(Z < -1.27) = .7967 - .1020 = .6947
P( Z
> 1.28) = .10, w = 1.28*30 + 175 = 213.4
4. P(X = 5) = .25
P((X =
10)’ ) = 1 - .10 = .90
P(X >= 0) = 1 - .40 = .60
E(X) = -5*.40 + 0*.25 + 5*.25 +
10*.10 = .25