# Stats without Tears

Solutions for Chapter 6

Updated 13 Mar 2015
(What’s New?)

Copyright © 2013–2023 by Stan Brown, BrownMath.com

Solutions for Chapter 6

Updated 13 Mar 2015
(What’s New?)

Copyright © 2013–2023 by Stan Brown, BrownMath.com

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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.

(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

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

x ($) | P(x) |
---|---|

9,999,995 | 1/10,000,000 |

95 | 1/125 |

5 | 1/20 |

−5 | .9419999 |

(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 = √[*n**p**q*] =
√[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 = *n**p* = 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)

(b)

`binompdf(5,.34,0)`

= .1252332576,
about a 12.5% chance
11
Your words will vary, but you should have the idea that the binomial
model is a fixed number of trials with varying number of successes,
whereas the geometric model is a varying number of trials that ends
with the first success.

12
Your words will vary, but you should have the idea that a pdf
is the probability of a specific outcome, and the cdf is the
cumulative probability of all outcomes 0 through a specified number.
I’m not so concerned that you know what pdf and cdf stand for,
as long as you understand what they mean and when to use each.

**13 Mar 2015**: Simplify the solution to part (a) of the fair-price problem.- (intervening changes suppressed)
**16 Apr 2013**: New document.

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