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 grantTina 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
10
th
number is 40
th
percentile = 7
(Wrong! n should be 22, so, np=8.8 > 9
th
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.
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 Spring '08
 Mendell

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