Section 5.1 Solutions

 

5.1

  1. B – binary outcomes: succeed or don’t succeed; I – independent outcomes: since the population of phone numbers in New York City is larger than 20(500) = 10,000, we have approximate independence; N – number of observations fixed: n = 500 numbers called; S – same probability of success for each observation: p = (1/12) of successfully talking with someone for each phone number

 

This is an approximate binomial setting. Hence, X (the number of calls that reach a live person) has a binomial distribution with n = 500 and p = (1/12).

 

  1. This is not a binomial setting, because the number of observations is not fixed (Darci continues to log in until she succeeds).

 

  1. This is not a binomial setting, because the number of observations is not fixed (e.g., we aren’t looking at a fixed number of ducks), and also there are really no well-defined observations (e.g., we don’t look at a duck and record whether it “fouls” the pond or not).

 

 

5.3

B – binary outcomes: catch word error or don’t catch word error

I – independent outcomes: no information is given to us about independence; to answer this question we must assume that whether the student catches one word error is independent of whether he/she catches another word error

N – number of observations fixed: n = 20 word errors in the essay

S – same probability of success for each observation: p = 0.70 of successfully catching a word error

 

a.      Assuming independence, the distribution of the number of errors caught is binomial with n = 20 and p = 0.7. Assuming independence, the distribution of the number of errors missed is binomial with n = 20 and p = 0.3.

 

b.      Let X be the number of word errors caught, and let Y be the number of word errors missed. In terms of X, the probability we’re interested in is . But we can’t use Table C to find the distribution of X, as Table C only goes up to p = 0.5. Hence, we can rephrase the probability in terms of Y and then use Table C: 0.0654 + 0.0308 + 0.0120 + 0.0039 + 0.0010 + 0.0002 = 0.1133

 

 

5.8

a.      We are sampling without replacement and the population size, 75, is not 20 times as large as the sample size, 25. Hence, the independence assumption is violated.

 

b.      This is an approximate binomial setting with n = 500 and p = 0.002. Since np = 500(0.002) = 1 < 10, we should not use the normal approximation (it may be very inaccurate).

 

 

5.9

B – binary outcomes: 0 or not 0

I – independent outcomes: digits independent of each other

N – number of observations fixed: n = 5 digits for part a and n = 40 for part b

S – same probability of success for each observation: p = 0.1 of any digit being a 0

 

a.      The number of 0s in a group of 5 random digits (call this X) has a binomial distribution with n = 5 and p = 0.1. Then (using the complement rule and Table C),

 

b.      The number of 0s in a group of 40 random digits has a binomial distribution with n = 40 and p = 0.1. The mean for this binomial distribution is np = 40(0.1) = 4 digits.

 

 

5.11

Although normal curve pictures aren’t included with this solution, I encourage you to draw them.

 

Let X be the number of home runs hit my McGwire. As given in the problem X has an approximate binomial distribution with n = 509 and p = 0.116.

 

a.      The mean number of home runs is

 

b.      Since n > 20, we can’t use Table C. Will the normal approximation be accurate? Since np = 509(0.116) = 59.044 > 10 and n(1-p) = 509(1-0.116) = 449.956 > 10, the normal approximation will be accurate.

 

Then the distribution of X is approximately normal with  and

 

Thus,

 

 

c.       Let X be the number of home runs hit my Bonds. As given in the problem, X has an approximate binomial distribution with n = 476 and p = 0.0865. Since n > 20, we can’t use Table C. Will the normal approximation be accurate? Since np = 476(0.0865) = 41.174 > 10 and n(1-p) = 476(1-0.0865) = 434.826 > 10, the normal approximation will be accurate.

 

Then the distribution of X is approximately normal with  and

 

Thus,

 

Note:  The standard normal table only extends to a z-value of 3.49. Technically, all we can say about  is that it is less than (1 – 0.9998) = 0.0002 (since this is the last probability listed in the table). But since 5.19 is so far beyond 3.49, the area to right of 5.19 is probably much, much less than 0.0002. Practically speaking, the area (and, hence, the probability) is about 0.

 

 

5.12

a.      X is binomial with n = 6 and p = 0.65

 

b.      Because p = 0.65, we can’t use Table C directly to get the probabilities (since the table only goes up to p = 0.5). Let Y be the number of years among the next 6 years in which the index falls. Then Y is binomial with n = 6 and p = 0.35. We can use the probability distribution of Y (from Table C) to find the probability distribution of X:

 

P(X = 0) = P(Y = 6) = 0.0018

(important note: Table C incorrectly lists this probability as 0.0083; if you use the incorrect value, the probabilities sum to more than 1)

P(X = 1) = P(Y = 5) = 0.0205

P(X = 2) = P(Y = 4) = 0.0951

P(X = 3) = P(Y = 3) = 0.2355

P(X = 4) = P(Y = 2) = 0.3280

P(X = 5) = P(Y = 1) = 0.2437

P(X = 6) = P(Y = 0) = 0.0754

 

In table form, the probability distribution of X is

 

Value of X

0

1

2

3

4

5

6

Probability

0.0018

0.0205

0.0951

0.2355

0.3280

0.2437

0.0754

 

 

 

 

 

            In graphical form, the distribution is shown below.

 

 

c.        The mean number of years in which the stock price index rises is .

 

d.      The standard deviation of X is  Then the probability that X takes a value within one standard deviation of its mean is

 

 

5.19

a.      The probability of a successful guess on a single card is

b.      Let X be the number of correct guesses. Then X is B(20, 0.25). From Table C, = 0.0099 + 0.0030 + 0.0008 + 0.0002 = 0.0139.

 

c.       The mean and standard deviation of X are  and

 

d.      In this case, the draws of the cards are not independent – the population size is 25 (all the cards) and all 25 cards are included in the sample (n = 25). Hence, knowing what happened on the previous cards changes the probability of happens on the latter cards. Because the independence assumption is violated, a binomial model is no longer appropriate.

 

 

 

5.20

Although normal curve pictures aren’t included with this solution, I encourage you to draw them.

 

The number of students who accept (denoted X) has a binomial distribution with n = 1500 and p = 0.7.

 

a.      The mean and standard deviation of X are  and

 

b.      Since np = 1500(0.7) = 1050 > 10 and n(1 – p) = 1500(0.3) = 450 > 10, the normal approximation should be good. Then,

 

c.       , so the college needn’t be worried

 

Note:  The standard normal table only extends to a z-value of 3.49. Technically, all we can say about  is that it is less than (1 – 0.9998) = 0.0002 (since this is the last probability listed in the table). But since 8.45 is so far beyond 3.49, the area to right of 8.45 is probably much, much less than 0.0002. Practically speaking, the area (and, hence, the probability) is about 0.

 

d.      Now X is B(1700, 0.7), so  and  Then,

 

 

5.21

Although normal curve pictures aren’t included with this solution, I encourage you to draw them.

 

This is a binomial setting with n = 1000 and p = (1/5) = 0.2.

 

a.      Let X be the number of correct answers. Then  and

 

b.      Let be the proportion of correct answers. Then  and

 

 

c.       Since n > 20, we can’t use Table C. Will the normal approximation be accurate? Since np = 1000(0.2) = 200 > 10 and n(1-p) = 1000(1-0.2) = 800 > 10, the normal approximation will be accurate.

 

Thus,

 

d.      We want to find the value x such that We know the 99th percentile of the standard normal distribution is 2.33 (from reverse look-up in Table A). Thus,  Thus, a person would need to get at least 23% correct answers to be judged to have ESP.