5.1 Consider a finite population with five elements labeled A, B, C, D, and E.

Ten

possible simple random samples of size 2 can be selected.

a. List the 10 samples beginning with AB, AC, and so on.

b. Using simple random sampling, what is the probability that each sample of

size 2 is selected?

c. Assume random number 1 corresponds to A, random number 2 corresponds

to B, and so on. List the simple random sample of size 2 that will be selected by

using the random digits 8 0 5 7 5 3 2

Solution:

a) AB, AC,AD,AE,BC,BD,BE,CD,CE,DE

b) When using simple random sample, each sample has an equal chance of being selected:

P(AB) = P(AC) = …= P(DE) = 1/10 = 0.1

c) 8 => ignore (because we don’t have any letter which base on no 8)

0 => ignore

5 => E

7 => ignore

5 => ignore (because just have 5 elements and after take letter E so just have 4 options, if take no 5 means that don’t have anymore
letter E in work to choose)

3 => C

The sample of size 2 is EC

5.2 A population has a mean of 200 and a standard deviation of 50. A simple random

sample of size 100 will be taken and the sample mean will be used to estimate the

population mean.

μ = 200

σ = 50

n = 100
a. What is the expected value of 𝑥̅?

b. What is the standard deviation of 𝑥̅?

c. Show the sampling distribution of 𝑥̅. (normal => sample also have normal distribution, if not

d. What does the sampling distribution of 𝑥̅ show?

Solution

a) The expected value of the sample mean 𝑥̅ is the population mean:

μ(𝑥̅) = μ =200

b)The standard deviation of the sample mean 𝑥̅ is the population standard deviation divided by the square root of the sample size:

σ 50
σ x= = =5
√ n √100
c) The sampling distribution of the sample mean is approximately normal, by the central limit theorem, because the sample size is large

𝑥̅ ~N (200,52)

d) The sampling distribution of the sample mean 𝑥̅ show the probability distribution of 𝑥̅

- the probablility distrubution of all possible 𝑥̅ calculated from sample of size 100

5.3 The mean annual cost of automobile insurance is $939 (CNBC, February 23,

2006). Assume that the standard deviation is σ = $245.

a. What is the probability that a simple random sample of automobile insurance

policies will have a sample mean within $25 of the population mean for each of the

following sample sizes: 30, 50, 100, and 400?

b. What is the advantage of a larger sample size when attempting to estimate the
population mean?

Solution

μ= $ 939 σ = $245 x1 = 939 – 25 = 914 x2 = 936 + 25 = 964

a) P (914<x<964)

n = 30

σ 245
σ x= = =44.73
√ n √30
z1 = 𝑥̅ - Mz-/ σ x = 914 – 939/44.73 = -0.56

z2 = 964 - Mz-/ σ x = 964 - 936 /44.73 = 0.56

=> P(914<x<964) = P(-0.56<z<0.56)

= P(z<0.56) – P(z<-0.56)

= 0.7123 – 0.2877

= 0.4246

n = 50

σ X = σ X /√n = 245/√50 = 34.6482

z1 = 914 – 939/ 34.6482 = -0.72
z2 = 964 – 939/34.6482 = 0.72

=> P(914<x<964) = P(-0.72<z<0.72<)

= P(z<0.72) – P(z<-0.72)

= 0.7642 – 0.2358

= 0.5284

n = 100

σ X = σ X /√n = 245/√100 = 24.5

z1 = -25/24.5 = -1.02

z2 = 25/24.5 = 1.02

=> P(914<x<964) = P(-1.02<z<1.02)

= P(z<1.02) – P(z<-1.02)

= 0.8461 - 0.1539

= 0.6922

n = 400

σ X = σ X /√n = 245/√400 = 12.25

z1 = -25/1225 = -2.04

z2 = 25/12.25 = 2.04

=> P(914<x<964) = P(-2.04<z<2.04)

= P(z<2.04) – P(z<-2.04)

= 0.9793 – 0.0207

= 0.9586

b) A larger sample has a higher probability that the sample mean will be closer to the population mean

5.4 BusinessWeek conducted a survey of graduates from 30 top MBA programs

(BusinessWeek, September 22, 2003). On the basis of the survey, assume that the

mean annual salary for male and female graduates 10 years after graduation is
$168,000 and $117,000, respectively. Assume the standard deviation for the male

graduates is $40,000, and for the female graduates it is $25,000.

a. What is the probability that a simple random sample of 40 male graduates will

provide a sample mean within $10,000 of the population mean, $168,000?

b. What is the probability that a simple random sample of 40 female graduates

will provide a sample mean within $10,000 of the population mean, $117,000?

c. In which of the preceding two cases, part (a) or part (b), do we have a higher

probability of obtaining a sample estimate within $10,000 of the population mean?

Why?

d. What is the probability that a simple random sample of 100 male graduates

will provide a sample mean more than $4000 below the population mean?

Solution

Male μ = $168,000 σ = $40,000 n =40

Female μ = $117,000 σ = $25,000 n = 40

a) x1 =168,000 - 10,000 = 158,000

x2 = 168,000 + 10,000 = 178,000

σ X = σ X /√n = 40,000/√40 = 6324.555

z1 = 158,000 – 168,000/6324.555 = -1.58

z2 = 178,000 – 168,000/6324.555 = 1.58

=>P(158,000<x<178,000) = P(-1.58<z<1.58)

= P(z<1.58) – P(z<-1.58)

= 0.9429 – 0.0571

= 0.8858

b) x1 = 117,000 – 10,000 = 107,000

x2 = 117,000 + 10,000 = 127,000

=> P(107,000<x<127,000)=?

σ X = σ X /√n = 25,000/√40 = 3952.847

z1 = 107,000-117,000/ 3952.847 = -2.53

z2 = 127,000 – 117,000/ 3952.847 = 2.53

=> P(107,000<x<127,000) = P(-2.53<z<2.53)

= P(z<2.53) – P(z<-2.53)

= 0.9943 – 0.0057

= 0.9886

c) The smaller standard deviation, the les distributed the values are meaning it’s more likely to find values within specified
limit. Therefore due to lwer standard deviation it’s more likely to obtain sample of females within $10,00 of the population
mean

d) μ = 168,000 σ = 40,000 n= 100

x = 168,000 – 4,000 = 164,000

σx = σx /√n = 40,000/√100 = 400

z = 164,000 – 168,000/400 = -10

5.5 The Cincinnati Enquirer reported that, in the United States, 66% of adults and

87% of youths ages 12 to 17 use the Internet (The Cincinnati Enquirer, February 7,

2006). Use the reported numbers as the population proportions and assume that

samples of 300 adults and 300 youths will be used to learn about attitudes toward

Internet security.

a. Show the sampling distribution of where is the sample proportion of adults

using the Internet.

b. What is the probability that the sample proportion of adults using the Internet

will be within ± .04 of the population proportion?

c. What is the probability that the sample proportion of youths using the Internet
will be within ± .04 of the population proportion?

d. Is the probability different in parts (b) and (c)? If so, why?

e. Answer part (b) for a sample of size 600. Is the probability smaller? Why?

Solution:

n = 300 p =0.66 of adults

The sample distribution the sample probability is approximately normal (n=300>30)

a) E(p-) p 0.66 = μ

σ p- = √P(1-p)/n = √0.66x(1-0.66)/300 = 0.0273

p- ~N (0.66;0.0272)

b) P1 = 0.66 - 0.04 = 0.62

P2 = 0.66 + 0.04 = 0.70

=> P(0.62<=p-<=0.70)

σ p- = 0.027
z1 = 0.62 - 0.66/0.027 = -1.47

z2 = 0.70-0.66/0.027 = 1.47

=> P(0.62<p-<0.70) = P(-1.47<=z<=1.47)

= P(z<=1.47) - P(z<=-1.47)

= 0.9292 – 0.0708

= 0.8584
c) p1 = 0.87 -0.04 = 0.83 p2 = 0.87 + 0.04 = 0.91

σ p- = √[P(1 – P)/n] = √[0.87 x (1 – 0.87)/300] = 0.0194

z1 = 0.83 – 0.87/ 0.0194 = -2.06

z2 = 0.91 - 0.87/0.0194 = 2.06

P(0.83<=p-<=0.91) = P(-2.06<=z<=2.06)

= P(z<=2.06) – P(z<=-2.06)

= 0.9803 – 0.0197 = 0.9606

d) Part b and c have different probability because the probability of 2 different object is adults and youths which is different
data

e) P(-0.04<p-<0.04)

= P ([-0.04/√0.66x(1-0.66)] < [(p- - P/√P(1-p)] < 0.04/[√0.66x(1-0.66)/600]

= P(-2.07<z<2.07) = P (z<2.07) – P(z<-2.07)

= 0.9808 – 0.0192

= 0.9616

The probability in part b is smaller than that’s in part e because larger samples of size more closely approximate the
population and more accurate result.

5.6 Americans have become increasingly concerned about the rising cost of

Medicare. In 1990, the average annual Medicare spending per enrollee was $3267; in 2003, the average annual Medicare
spending per enrollee was $6883 (Money, Fall 2003). Suppose you hired a consulting firm to take a sample of fifty 2003
Medicare enrollees to further investigate the nature of expenditures. Assume the population standard deviation for 2003 was
$2000.

a. Show the sampling distribution of the mean amount of Medicare spending for

a sample of fifty 2003 enrollees.

b. What is the probability the sample mean will be within ± $300 of the

population mean?

c. What is the probability the sample mean will be greater than $7500? If the
consulting firm tells you the sample mean for the Medicare enrollees they

interviewed was $7500, would you question whether they followed correct simple

random sampling procedures? Why or why not?

Solution:

σ  = 2,000 (1990) μ = 3267 (2003) μ =6883

a) n>=30 =>the central limit theorem: the sampling distribution of the sample mean is approximately normal

=> μ = 6883 σ x = σ/ √ n = 2000/√ 50 =282.843

The sampling distribution of Medicare 𝑥̅ ~N (6883;282.8432)

b) x1 = 6883 - 300 = 6583

x2 = 6883 +300 = 7183

P(6583<x<7183) =?

σ x = 282.843
z1 = 6583-6883/282.843 = -1.06

z2 = 7183-6883/282.843 = 1.06

=> P(6583<x<7183) = P(-1.06<z<1.06)

= P(z<1.06) -P(z<-1.06)

= 0.8554 – 0.1446

= 0.7108

c)P(x->7500)
z = 7500-6883/282.843 = 2.18

=> P(x->7500) = P(z>2.18) = 1- P(z<2.18)

= 1- 0.9854 = 0.0146

You would question if they followed correct simple random because it’s extremely unlikely to obtain this random
sample by chance

5.7 A market research firm conducts telephone surveys with a 40% historical

response rate. What is the probability that in a new sample of 400 telephone

numbers, at least 150 individuals will cooperate and respond to the questions? In other words, what is the probability
that the sample proportion will be at least 150/400 = .375?

Solution:

p = μ = 0.4 n = 400 p- = 150/400 = 0.375

σ p- = √[P(1 – P)/n] = √[0.4x(1-0.4)]/400 = 0.024

z = (0.375-0.4)/0.024 = -1.04

P(p->0.0375) = P(z>-1.04) = P(z<1.04) = 0.8508