IE241 Solutions

1. The binomial probability is

3C3 (1/2)3(1/2)0 = 1/8

Or, since the three tosses are independent, P(H on 1st, H on 2nd, H on 3rd)

= P(H on 1st) x P(H on 2nd) x P(H on 3rd) = ½ x ½ x ½

= 1/8

2. The binomial probability is

100C60 (1/2)60 (1/2)40

3. Total Sample space: 50C10

Sample points in A: 40C10

P(A) = 40C10 / 50C10

4. Sample space: 6 x 6 = 36 points (table next slide)

Ways of getting RR: 4 (X boxes in table)

p(RR) = 4/36 =1/9

Or since roll 1 is independent of roll 2,

P(R roll1 & R roll2) = P(R roll 1) x P(R roll 2) 1/9 = 2/6 x 2/6

4. Second Roll

FirsT

Roll

R1 R2 Y1 Y2 B1

B2

R1 X X

R2 X X

Y1

Y2

B1

B2

5.(a) P(6 boys)= 6C6 (1/2)6 (1/2)0 = 1/64 and P(6 girls) = 1/64

P(6 same sex) = 1/64 + 1/64 = 1/32 (b) P(3 boys) = 6C3 (1/2)3 (1/2)3

= 20 (1/64) = 5/16

6. P(4 w/1 die)) = 1/6 P(sum = 8) = 5/36 (checked boxes)

Roll 2

Roll

1

1 2 3 4 5 6

1

2 X

3 X

4 X

5 X

6 X

7. Binomial: P(3 successes; n = 5)

= 5C3 (1/2)3(1/2)2 = 10 (1/32) = 5/16

8. (a) P(K|face) = P(K and face) = 4/52 = 1 P(face) 12/52 3

(b) P(bK|face) = P(bK and face)= 2/52 = 1 P(face) 12/52 6

9. P(1st = W) = 2/9 P(2nd = B) = 3/8 P(W B) = 2/9 x 3/8 = 6/72 = 1/12

Or, sample space has 72 points, of which 6 are WB, so P(W B) = 6/72 = 1/12

W1B1 W2B1 W1B2 W2B2 W1B3 W2B3

10.

♣ ♦ ♥ ♠ Marginal for face

A 1 1 1 1 4/16=1/4

K 1 1 1 1 4/16=1/4

Q 1 1 1 1 4/16=1/4

J 1 1 1 1 4/16=1/4

Marginal for suit

4/16 =1/4

4/16 =1/4

4/16 =1/4

4/16 =1/4

11. The company can conclude that 2% of its transistors made at plant A during one particular week were defective. They can say nothing about the other plants or other weeks because their sampling frame was rstricted to one week at Plant A.

12. mean = 550 s = 80 n = 100

95% confidence interval:

Because the sample size is so large, t.025 = z.025

so t025 = 1.96 and the confidence interval is

or approximately, 534 ≤ μ ≤ 566

100

80550 t

)8(96.1550

13. The confidence interval based on the t distribution will always be longer than it would be if σ were known, which then would allow use of the normal distribution because a given α value for t is greater than that same α value for z unless the sample size is large.

14. The length of the confidence interval is determined by

and, as the sample size gets larger and larger, the estimate of the standard error gets smaller and smaller. As n → ∞ the standard error → 0.

n

st

15. He can generalize his conclusions only to the 10 male students he has measured because his sample was not random.

16. Mean = 4.429;

Mode = 4; Median = 4.0625. To calculate the median, note that freq < 4 =19and freq > 4 = 21. so the median lies somewhere in the range 3.5 to 4.5.

Divide the frequency at 4 into two proportional distances into the range. You have 9/16 added to the lower limit and 7/16 subtracted from the upper limit.

3.5 + 9/16 = 65/16 = 4.0625 4.5 – 7/16 = 65/16 = 4.0625

histogram for shrub diameter

0

2

4

6

8

10

12

14

16

18

1 2 3 4 5 6 7 8 9 10 11 12

diameter

frequency

17.

Because the distribution is completely

symmetric, mean = median = mode =7.

Roll 2

Roll

1

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10

11

6 7 8 9 10 11

12

0

1

2

3

4

5

6

7

2 3 4 5 6 7 8 9 10 11 12

sum of two faces

frequency

18. Mean = 12 = np variance = 8 = npq

For the binomial: So p = 1/3 variance = npq n = 36 8 = 12q q = 2/3 p = 1/3 mean = np 12 = n(1/3) n = 36

19. binomial cdf point: (3, .17)

P(X ≥ 4) = 1- .17 = .83

20. Point on normal cdf: (1.2, .37)

P(X ≥ 1.2) = 1 - .37 = .63

21. Two groups s1 and n1 s2 and n2

If the two data sets are combined,

1X2X

21

2211

21

1 1

1 2

nn

XnXn

nn

xx

X

n

i

n

jji

combined

22. Based on history, p = 3/30 = 1/10

P(X = 2) = 30C2 (.1)2 (.9)28 = .228

P(X = 1) = 30C1 (.1)1 (.9)29 = .141

P(X = 0) = 30C0 (.1)0 (.9)30 = .042

P(≤ 2) = .411

23. p = 2/3 n = 4

(a) 4C4 (2/3)4 (1/3)0 = .198

(b) 4C4 (1/3)4 (2/3)0 = .012

So .198 +.012 = .210 (c)6C3 (2/3)3 (1/3)3 = .219 = P(tie after 6)

P(A wins 7th) = 2/3 P(A wins in 7) = P(tie @6) x P(A wins 7) = .219 x 2/3 = .146

24. p = .1 n = 20 P(X ≥ 2) = P(at least 2) = 1- P(at most 1)

20C1 (.1)1(.9)19 = .270

20C0 (.1)0(.9)20 = .122

P(X ≤ 1) = .392 = P(at most 1) P(X ≥ 2) = 1 - .392 = .608

25. Var(X +Y)) = Var(X) + Var(Y) + 2Cov(XY)

(a) For the variance of the sum of two random variables to be greater than the sum of their variances, the variables must have a positive relationship so that their covariance > 0.

(b) For the variance of the sum of two random variables to be less than the sum of their variances, the variables must have a negative relationship so that their covariance is negative.

26. Mean = 400 s=15 n =16

Confidence interval

408392

99.740016

15131.2400

025.

n

stX

27. This is a multinomial where x = 1 red ball out of 2 y = 1 green ball out of 3 z = 2 black balls out of 4 n = 4 Because of replacement, the probability

for each ball remains the same.

2857.126

)6)(3)(2(

),(49

241312

C

CCCyxp

28. There are 10 ways of getting a total of 6 on 3 dice:

1 1 4 and there are 216 total 1 2 3 possibilities for 3 dice, 1 3 2 P(dice sum to 6) = 10/216 1 4 1 = .046 2 1 3 2 2 2 2 3 1 3 1 2 3 2 1 4 1 1

29. It would not be surprising to find a relatively high correlation between Wall Street traffic and high tide in Maine because high tide in Maine occurs at 8 am and before that is getting higher and higher and afterward declines gradually. Also traffic in Wall Street rises from about 6 am to its peak at 8 am and then declines gradually.

Of course, the correlation is meaningless.