Submitted by: Submitted by joplinz
Views: 329
Words: 298
Pages: 2
Category: Science and Technology
Date Submitted: 08/12/2012 09:42 PM
Practice Exam 1 Solutions
1. P(Bill applies) = 1
P(Tina applies) = .6
P(Bill awarded grant | Tina applies) = .3
P(Bill awarded grant | Tina NOT applies) = .65
Find P(Tina NOT applies | Bill awarded grant):
P(Tina NOT applies | Bill awarded grant)
= P(Tina NOT applies ∩ Bill awarded grant) / P(Bill awarded grant)
= P(Bill awarded grant | Tina NOT applies) * P(Tina NOT applies) / P(Bill awarded grant)
= .65*.4 / (P(Bill awarded grant | Tina not applies)*P(Tina NOT applies)
+ P(Bill grant|Tina applies)*P(Tina applies))
= .65*.4 / (.65*.4 + .3*.6)
= .591
2. a. Use formulas:
Mean=7.73, median=7.75, variance=1.83, sd=1.35
b. Easy.
c. n = 23, p = .4 np = 9.2 10th number is 40th percentile = 7
(Wrong! n should be 22, so, np=8.8 -> 9th number)
d. Boxplot (all you need is max, min, Q1, Q2, Q3)
e. 7 classes. Values 5 to 11. Distribution looks like this:
Class Frequency
[5, 6) 2
[6,7) 4
[7,8) 6
[8,9) 7
[9,10) 2
[10,11] 2
Now draw histogram…
3. sample of television sets from 10 with 3 defective.
Let X = # of defectives in sample.
a. sample of 4 sets
b.
4. f(x) = kx^2, x = 0, 2, 4, 6, 8
f(0) = k*0 = 0
f(2) = k*2^2 = 4k
f(4) = k*4^2 = 16k
f(6) = k*6^2 = 36k
f(8) = k*8^2 = 64k
To be a proper probability distribution, the probabilities must be > 0 and all add to 1.
Therefore, k > 0, and 0 + 4k + 16k + 36k + 64k = 120k = 1 so k = 1/120
As a result, f(4) = P(X = 4) = 1/120 * 4^2 = 16/120 = .133
Also, F(2) = P(X