This document contains instructions, some answers, comments, tips and tricks, and further explanations for the exercises.

Exercise 6.57; explanation

X1 a random variable with \(\mu_{X_1}\) = 303.35 USD and \(\sigma_{X_1}\) = 9,707.57 USD.
X2 a random variable with \(\mu_{X_2}\) = 303.35 USD and \(\sigma_{X_2}\) = 9,707.57 USD.

(or shorter, for i = 1, 2: Xi a random variable with \(\mu_{X_i}\) = 303.35 USD and \(\sigma_{X_i}\) = 9,707.57 USD.)

Note:

  1. the expected value of Xi is positive, so if the insurance company has a lot of clients with this insurance, the company will make profit of on average 303.35 USD per client
  2. X1 has a large standard deviation; the standard deviation measures the ‘typical’ distance of the observations to the mean; if this company has only one client, the probability that the company makes a loss is high; to mitigate this risk, a company should have a many clients; this exercise shows why more clients helps to decrease the risk

W = \(\frac{X_1+X_2}{2} = \frac{1}{2}(X_1+X_2\))
Combine rule (4) and rule (2) (see handout week 49): \(\mu_W = \frac{1}{2}(\mu_{X_1}+\mu_{X_2}) = \frac{1}{2}(303.35 + 303.35) = 303.35\)

Define first S = X1 + X2
According to rule (5):
\(VAR_S~=~VAR_{X_1}~+VAR_{X_2}~=~9,707.57^2~+9,707.57^2 = 2\times9,707.57^2\)

\(\sigma_S=\sqrt{2\times 9,707.57^2}=\sqrt2 \times 9,707.57\)

And according to rule (2) \(\sigma_W~=~\frac{1}{2}\times \sigma_S~=~\frac{\sqrt2}{2} \times 9,707.57\) \(= \frac{9,707.57}{\sqrt{2}}\)

Conclusion the standard deviation of W is a factor \(\sqrt2\) smaller than the standard deviation of X. So if two customers have such a life insurance, the risk for making a loss is lower.

Exercise 6.58 remark

V = \(\frac{(X_1+X_2+X_3+X_4)}{4}\) = \(\frac{1}{4}\times(X_1+X_2+X_3+X_4)\)

Following the reasoning in exercise 6.57 above:

\(\mu_V = .....\) and \(\sigma_W = \frac{\sigma_{X}}{\sqrt{4}} = \frac{\sigma_{X}}{2}\)

Exercise 6.60

X1: resistance 100-ohm resistor
X2: resistance 250-ohm resistor

X1 ~ N(\(\mu\) = 100; \(\sigma\) = 2.5) ohm
X2 ~ N(\(\mu\) = 250; \(\sigma\) = 2.8) ohm

X: resistance of a 100-ohm and a 250-ohm resistor in series
X = X1 + X2 (a) X ~ N(\(\mu\) = 100 + 250 = 350; \(\sigma\) = $ = 3.754)
(b) P(345 < X < 355) = normalcdf(345, 355, 350, 3.754) = 0.817

Exercise 6.61

T, total time, is the sum of the four times; T is the sum of four independent normally distributed random variables. T ~ N(\(\mu\) = 55.2 + 58.0 + 56.3 + 54.7 = 224.2; \(\sigma\) = \(\sqrt{2.8^2 + 3.0^2 + 2.6^2 + 2.7^2}\) = \(\sqrt{30.89}\) = 5.558)

P(T < 220) = normcdf(-inf, 220, 224.2, 5.558) = 0.225

Exercise 6.63

X: NOX present in randomly chosen car
X ~ N(1.4; 0.3) g/mi
D: Difference NOX pesent between two randomly selected cars
D = X1 - X2; X1 and X2 have both the same distribution as X above
D ~ N(\(\mu\) = 1.4 -/- 1.4 = 0; \(\sigma\) = $ = 0.4243)
P(D < -0.8 or D > 0.8) = 1 = normalcdf(-0.8, 0.8, 0, 0.4243) = 1 - 0.9406 = 0.059

Exercise 6.69

X has a binomial distribution; X ~ bin(n = 20; p = 0.85) Justification: the random sample is a repetition of 20 independent repeats of a success/failure experiment with P(success) = 0.85 for each repeat.

Exercise 6.70

Y does not have a binomial distribution. The names are drawn without replacement, the sample size 4 is more than 10% of the population size. The outcomes of the different draws are not independent.

Exercise 6.71

V does not have a binomial distribution. This is not an experiment in which the number of ‘successes’ is counted.
In fact this is, if the total population is large, a geometric experiment.

Exercise 6.72

W ~ bin(n = 6; p = 0.10)
Justification: the days are chosen at random, so whether the train is late is independent from wether the train is late on an other day in the sample; the probability of being late is assumed to be 10%.

Exercise 6.73

  1. Binomial setting. The numbers are chosen at random; the different repeats are independent, for every call in the sample the probability that a person is reached is 20%. So X ~ bin(n = 15, p = 0.20)
  2. Not a binomial setting. Not the number of successes (reaching a person) is counted, but the number of trials needed to reach one person. This is a geometric setting.

Exercise 6.74

  1. Not a binomial setting. Although the repeats are independent, the probability of success is not the same for each repeat
  2. Binomial setting. Independent repeats, probability success can be assumed to be approximately the same for each repeat

Exercise 6.80

W: number of trains that arrive late in 6 randomly chosen days
W ~ bin(n = 6, p = .10)

  1. P(W = 2) = \(\binom{6}{2}\times 0.10^2 \times 0.90^4\) = …
    or use TI84 P(W = 2) = binompdf(6, 0.10, 2) = 0.098
  2. P(W >= 2) = 1 - P(W < 2) = 1 - P(W <= 1) = 1 - binomcdf(6, 0.10, 1) = 1 - 0.885735 = 0.114

Exercise 6.95

  1. Not a geometric setting. After you have one diver’s card, the probability of getting another card (a success) is 0.8. After you have two different driver’s cards, the probability of a success is 0.6.
  2. A geometric setting. Independent trials, success chance unchanged, waiting till the first success. K: the number of games until the first success; K ~ geometric(p = 0.259)

Exercise 6.96

  1. Not geometric. The trials are not independent. The probability of turning an ace changes.
  2. Geometric. It says ‘on any shot, he has about 10% chance of hitting …’. This means, the shots are independent. Shooting until hitting the bull’s eye is then a geometric setting, and X: the trial in which the bull’s eye is hit has a geometric distribution with p = 0.1.

Exercise 6.99

Let X denote the number of invoices chosen until the first invoice that has a first digit 8 or 9 because these are independent trials of the same binary process and the success probbability stays 0.097.
X ~ geometric(p = 0.097)

  1. E(X) = 1/0.097 = 10.3
  2. P(X >= 40) = 1 - P(X <= 39) = 1 - geomcdf(0.097, 39) = 1 - 0.98130 = 0.019; because this is a very small probability, it is reasonable to doubt the genuineness of the invoices.