Stats without Tears
Solutions for Chapter 6

1 (a) 0, 1, 2, 3, 4, 5
(b)There are five trials, each die is either a two or not a two, and the dice are independent. This fits the binomial model.

2

x ($)	P(x)
9,999,995	1/10,000,000
95	1/125
5	1/20
−5	.9419999

(a) The probability model is shown at right. (I computed the probability of losing $5 as 1−[1/10000000+1/125+1/20].)

(b) $ in L1, probabilities in L2. 1-VarStats L1,L2 yields μ = −2.70. The expected value of a ticket is −$2.70. This is a bad deal for you. (It’s a very good deal for the lottery company. They’ll make $2.70 per ticket, on average.)

Common mistakes: Students sometimes give hand-waving arguments such as the top prize being very unlikely, or the lottery company always getting to keep the ticket price, but these are not relevant. The only thing that determines whether it’s a good or bad deal for the player is the expected value μ.

3 (a) This is a geometric model: repeated failures until a success, with p = 0.066.

μ = 1/p = 1/.066 ≈ 15.2

Over the course of her undead existence, taking each night’s hunt as a separate experience, the average of all nights has her first getting an O negative drink from her fifteenth victim.

(b) geometcdf(.066,10) = .4947936946 ≈ 0.4948. Velma has almost a 50% chance of getting O negative blood within her first ten victims.

(You could also do this as a binomial, n = 10, p = 0.066, x = 1 to 10.)

(c) This is a binomial model with n = 10, p = 0.066, and x = 2. Use MATH200A part 3 or binompdf(10,.066,2) = .1135207874 ≈ 0.1135. Velma has just over an 11% chance of getting exactly two O negative victims within her first ten.

4 This is a geometric distribution. You’re looking for someone who is opposed to universal background checks, so p = 1−.92 = 0.08.

(a) geometpdf(.08, 3) = .067712 → 0.0677

(b) geometcdf(.08, 3) = .221312 → 0.2213

(You could also do this as a binomial with n = 3, p = 0.08, x = 1 to 3.)

5

(a) This is a binomial distribution: each student passes or not, whether one student passes has nothing to do with whether anyone else passes, and there are a fixed seven trials.

μ = np = 7*0.8 ⇒ μ = 5.6 people

σ = √[npq] = √[7*0.8*(1−0.8)] = 1.058300524 ⇒ σ =  1.1 people

(b) Binomial again, n = 7, p = 0.8, x = 4 to 6. Use binompdf-sum or MATH200A part 3 to find P(4 ≤ x ≤ 6) = 0.7569.

    

(c) Geometric model: p = 0.8, x = 3. geometpdf(.8,3) = 0.0320

(d) geometcdf(.8,2) = 0.9600

Alternative solution: Binomial probability with n = 2, p = 0.8, x = 1 to 2 gives the same answer.

6 This is binomial data, p = .49. For a sample of 40, expected value is μ = np = 40×.49 = 19.6. 13 is less than 19.6, so asking whether 13 is surprising is really asking whether 0 to 13 is surprising; see Surprise!

binomcdf(40,.49,13) or MATH200A Program part 3 with n=40, p=.49, x=0 to 13 gives .0259693307  →  0.0260, less than 5%, so you would be surprised though maybe not flabbergasted. ☺

7 (a) Probability of one equals proportion of all, and therefore a randomly selected 22-year-old male has a 0.1304% chance of dying in the next year. That’s the only “prize”, so multiply it by its probability to find fair price: 100000×0.001304 = $130.40

(b) The company’s gross profit is $180.00−130.40 = $49.60, about 28%. But it could very well cost the company that much to sell the policy, pay the agent’s commission, and enter the policy in the computer. Also, all policies must bear part of the company’s general overhead costs. The price is not necessarily unfair in the plain English sense.

8

(a) x’s in L1, P’s in L2. 1-VarStats L1,L2 yields μ = 2 (exactly) and σ = 1.095353824 or σ ≈ 1.1. Interpretation: In the long run, on average you expect to get two heads per group of five flips. You expect most groups of five flips will yield between μ−σ = 1 head and μ+σ = 3 heads.

(b) (I wouldn’t use this part as a regular quiz question.) The long-term average is 2 heads out of 5 flips, which is p = 2/5 = 40%. Obviously coin flips are independent, so the probability of heads must be the same every time. Therefore you have a binomial model with n = 5 and p = 0.4.

9

(a) Binomial probability with n = 5, p = 0.7, x = 3 to 5. MATH200A part 3 5, .7, 3, 5 yields .83692 or P(x ≥ 3) = 0.8369. Or, binompdf(5,.7)→L6 and then sum(L6,4,6) to get the same answer. Or, use the complement: 1−binomcdf(5,.7,2).

(b) You need the mean of the binomial distribution:

μ = np = 10×0.7 = 7

(c) 5 is less than the expected number, so you compute P(x≤5):

MATH200A part 3 10, .7, 0, 5 yields 0.1503, or

binomcdf(10,.7.5) = 0.1503, not surprising

Common mistake: Don’t just compute P(x=5), which is 0.1029. When you want to know whether a result is unusual or surprising, you have to find the probability of that result or one even further from the expected value.

10 (a) Geometric model, p = 0.34. μ = 1/.34 ≈ 2.94. About three
(b) binompdf(5,.34,0) = .1252332576, about a 12.5% chance