Section 4.3 Solutions

 

4.42

Although a picture is not included with this solution, I strongly encourage you to draw one.

 

Y is N(266, 16)

 

Then,

, from Table A.

 

 

4.47

P(message passed) = 0.37 and the steps are independent

 

Let T be the total length of the chain. Then,

P(T = 2) = P(your friend does not pass on the message) = 1 – 0.37 = 0.63

 

P(T = 3) = P(your friend passes the message and your friend’s friend does not pass the message) = P(your friend passes the message)P(your friend’s friend does not pass the message) = (0.37)(0.63) = 0.2331, by independence

 

 is the probability that the message does not make it to the fifth person (i.e., the probability that one of the first three people to receive the message does not pass it on).

 

 

 

4.49

S = {(1,0), (1,0), (1,0), (1,6), (1,6), (1,6),

    (2,0), (2,0), (2,0), (2,6), (2,6), (2,6),

    (3,0), (3,0), (3,0), (3,6), (3,6), (3,6),

    (4,0), (4,0), (4,0), (4,6), (4,6), (4,6),

    (5,0), (5,0), (5,0), (5,6), (5,6), (5,6),

    (6,0), (6,0), (6,0), (6,6), (6,6), (6,6)}

 

All these possible outcomes are equally likely. (Note this sample space takes into consideration that a 1 on the standard die and a 0 on the non-standard die can occur in three ways.)

 

Then, P(Y = 1) = P[(1,0) or (1,0) or (1,0)] = 3/36 = 1/12

In fact, all the possible values of Y have probability 1/12.

 

Then the probability distribution of Y, the sum of the spots, is

 

Value of Y

1

2

3

4

5

6

7

8

9

10

11

12

Probability

 

 

Note that these probabilities could have been determined without writing out the sample space. For example,

 

P(Y = 1) = P(1 on standard die and 0 on non-standard die) = P(1 on standard die)P(0 on non-standard die) = (1/6)(3/6) = 1/12

 

 

4.52

a.      P[(A supports) and (B supports) and (C opposes)] = P(A supports)P(B supports)P(C opposes) = (0.6)(0.6)(0.4) = 0.144, since the opinions of the students are independent.

 

b.      Let a subscript of s mean the student supports and a subscript of o mean the student opposes. Then, S = {AsBsCs, AsBsCo, AsBoCs, AsBoCo, AoBsCs, AoBsCo, AoBoCs, AoBoCo}, and the respective probabilities are {0.216, 0.144, 0.144, 0.096, 0.144, 0.096, 0.096, 0.064}.

 

c.       The probability distribution of X is:

 

Value of X

0

1

2

3

Probability

0.216

0.432

0.288

0.064

 

d.      The event that “a majority of the advisory board opposes funding” occurs if 2 or 3 of the members oppose funding. The probability of this event is P(X  2) = 0.288 + 0.064 = 0.352.

 

 

4.54

a.      P(X  0.27) = 1 – 0.27 = 0.73

 

b.      P(X = 0.27) = 0 (for any continuous random variable, the probability is 0 that it exactly equals a certain value)

 

c.       P(0.27 < X < 1.27) = P(0.27 < X < 1) = 1 – 0.27 = 0.73 (because the density curve stops at 1)

 

d.      P(0.1  X  0.2 or 0.8  X  0.9) = 0.1 + 0.1 = 0.2

 

e.      P( X is not in the interval 0.3 to 0.8) = P(0  X  0.3 or 0.8  X  1) = 0.3 + 0.2 = 0.5; or, using the complement rule, P(X is not in the interval 0.3 to 0.8) = 1 – P(X is in the interval 0.3 to 0.8) = 1 – 0.5 = 0.5

 

4.55

a.      Because the area under the density curve must be 1, the height of the density curve must be 0.5 (then the area under the curve is 2(0.5) = 1).

 

b.     

 

c.      

 

d.