Math 207 – Solutions to Assignment 4

 

4.62

Important Note: It’s appropriate, and perhaps easier, to solve this problem with a tree diagram—as long as it’s correct, thorough, and carefully labeled. (My solutions do not include a tree simply because Word doesn’t draw them.) If you use a tree, the first set of tree branches is smoking status, and the conditional set of tree branches is lung-cancer-death status.  In this situation, you need to find a probability within the branches of the tree.

P(smoker) = 0.20

P(lung cancer death| smoker) 10P(lung cancer death| not smoker)

P(lung cancer death) = 0.006

 

The event (lung cancer death) can be written as (lung cancer death  smoker) (lung cancer death  not smoker).

 

Using the addition rule for mutually exclusive events and the general multiplication rule,

0.006 = P(lung cancer death) = P[(lung cancer death  smoker) (lung cancer death  not smoker)] =

P(lung cancer death  smoker) + P(lung cancer death  not smoker) =

P(smoker)P(lung cancer death| smoker)  + P(not smoker)P(lung cancer death| not smoker).

(Note: This is simply the Law of Total Probability.)

 

Then by substitution,

0.006 = (0.20)(10)P(lung cancer death| not smoker) + (0.80)P(lung cancer death| not smoker)

and therefore, P(lung cancer death| not smoker) = 0.00214.

Then the final answer is P(lung cancer death| smoker) = 10(0.00214) = 0.0214.

 

 

4.99

Let X be the net profit per policy for the insurance company, and let C be the premium charged per policy. Then, based on the information given, the probability distribution of X is

 

x

C

C – 250,000

C – 800,000

P(X=x)

0.94

0.05

0.01

 

and E(X) = C(0.94) + (C – 250,000) (0.05) + (C – 800,000)(0.01) = C – 20,050. We want this expected value to be $0 per policy, so   That is, the company should charge a premium of $20,050 for the policy in order to break even, on average. (Although it seems odd the company wants to break even rather than make money.)

 

Alternative (and perhaps more intuitive) Solution

You can also define your variable as the net profit if the company doesn’t charge at all. Then the expected profit is 0(0.4) + (-250,000)(0.05) + (-800,000)(0.01) = - $20,050. That is, if the company doesn’t charge, they pay out $20,050 on average per policy. Hence, to break even, the company should charge $20,050 for each policy.

 

 

4.114

Important Note: It’s appropriate, and perhaps easier, to solve this problem with a tree diagram—as long as it’s correct, thorough, and carefully labeled. (My solutions do not include a tree simply because Word doesn’t draw them.)

P(Bus) = 0.3

P(Subway) = 0.7

P(Late | Bus) = 0.3

P(Late | Subway) = 0.2

 

Using the definition of conditional probability, the law of total probability, and the general multiplication rule:

 

 

 

4.135

The two orchestra representatives will be randomly selected from a list of six volunteers, consisting of four men and two women.

a.      Let X be the number of women chosen as orchestra representatives. Then X can take on the values 0, 1, or 2, and the probability distribution of X is shown below. (Note the probabilities sum to 1.)

P(0 women) =

 

P(1 women) =

 

P(2 women) =

 

b.      Looking ahead to the next chapter, we know X follows a hypergeometric distribution with M = 2, N = 6, and n = 2. So we know the mean of X is   and the variance of X is . We can also derive these values directly from the distribution above (using the definition of mean and variance):

 

 

 

 

c.       The probability that both representatives are women can be taken directly from the probability distribution above:

 

 

5.22

a.      Check BINS: Binary outcomes—took SAT or didn’t take SAT; approximate Independence—population size is over 20 times the sample size; fixed Number of observations—n=100 students sampled; Same probability of success on each trial—p=0.45 probability that a student took the SAT. This is an approximate binomial setting and the number of students (of the 100) that took the SAT is an approximate binomial random variable with n = 100 and p = 0.45.

 

b.      Check BINS: Binary outcomes—the outcomes are not binary, as there are more than two possible scores. This is not a binomial setting.

 

c.       Check BINS: Binary outcomes—scored above 1518 or scored at or below 1518; approximate Independence—population size is over 20 times the sample size; fixed Number of observations—n=100 students sampled; Same probability of success on each trial—fixed, yet unknown probability (p) of scoring above 1518. This is an approximate binomial setting and the number of students (of the 100) that scored above 1518 is an approximate binomial random variable with n = 100 and p unknown (if the distribution of scores is symmetric, then we know p = 0.5).

 

d.      Check BINS: Binary outcome—the outcomes are not binary, as there are more than two possible time amounts. This is not a binomial setting.

 

 

5.27

Let X be the number of patient bills that will have to be forgiven. This is an approximate binomial setting (Binary outcomes—forgiven or not forgiven; approximate Independence—population is 20 times the sample size; fixed Number, n=4, of patients; same Probability, p=0.3, of each patient failing to pay a bill). Then X has an approximate binomial distribution with n = 4 and p = 0.3.

 

a.      P(X = 4) =  (or, using Table 1, P(X = 4) = 1.000 – 0.992 = 0.008)

 

b.      P(X = 1) =  (or, using Table 1, P(X = 4) = 0.652 – 0.240 = 0.412)

 

c.       P(X = 0) =  (or, using Table 1, P(X = 0) = 0.240)

 

 

5.45

Let X be the number of injuries per year for a school-age child. Note that the number of injuries is not fixed (i.e., this is not a binomial experiment). Then X can be modeled as Poisson random variable with .

a.      P(X = 2) =  (or, using Table 2, P(X = 2) = 0.677 – 0.406 = 0.271)

 

b.      P(X  2) = 1 – P(X  1) = 1 – 0.406 = 0.594 (using Table 2)

 

c.       P(X  1) = P(X = 0) + P(X = 1) =  (or, using Table 2, P(X  1) = 0.406)

 

 

5.47

Let X be the number of bacteria per water specimen. Then X has a Poisson distribution with , and P(X > 5) = 1 – P(X  5) = 1 – 0.983 = 0.017 (using Table 2). This is very unlikely (i.e., the probability is very small).

 

 

5.56

There are 10 seeds: 5 treated and 5 untreated. From these seeds, 4 plants emerged. Let X be the number of emerged plants from treated seeds (note that X can take on the values 0, 1, 2, 3, 4). Note the binary-outcomes, fixed-sample-size, and same-probability-of-emergence conditions hold in this setting. But approximate independence does not hold (since the population size, 10, is not more than 20 times the sample size, n=4). Then X has a hypergeometric distribution (not a binomial distribution). To calculate the probabilities you can either use the hypergeometric probability distribution or simply use counting rules.

 

a.       

 

 

b.      Use the complement rule: P(X  3) = 1 – P(X = 4) = 1 – 0.024 = 0.976

 

 

c.       P(2  X  3) = P(X = 2) + P(X = 3) =