Ams 310: Probability and Statistics Example

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