← Exercises for Ch 6
(b) $ in L1, probabilities in 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
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.
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.
.4947936946 ≈ 0.4948. Velma has almost a
50% chance of getting O negative blood within her first ten
(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.
(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.)
(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
(c) Geometric model: p = 0.8, x = 3.
geometpdf(.8,3) = 0.0320
geometcdf(.8,2) = 0.9600
Alternative solution: Binomial probability with n = 2, p = 0.8, x = 1 to 2 gives the same answer.
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. ☺
(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.
(a) x’s in L1, P’s in 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
(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.
(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:
(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,
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.
binompdf(5,.34,0)= .1252332576, about a 12.5% chance