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Chapter 7
Sampling Distributions
U Section 7.
7.1 The probability distribution of the population data is called the population distribution. Table 7.2 in the text provides an example of such a distribution. The probability distribution of a sample statistic is called its sampling distribution. Table 7.5 in the text provides an example of the sampling distribution of the sample mean.
7.2 Sampling error is the difference between the value of the sample statistic and the value of the corresponding population parameter, assuming that the sample is random and no nonsampling error has been made. Example 7–1 in the text displays sampling error. Sampling error occurs only in sample surveys.
7.3 Nonsampling errors are errors that may occur in the collection, recording, and tabulation of data. Example 7–1 in the text exhibits nonsampling error. Nonsampling errors can occur in both a sample survey and a census.
7.4 a. μ = (15 + 13 + 8 + 17 + 9 + 12)/6 = 74/6 = 12.
b. x = (13 + 8 + 9 + 12)/4 = 42/4 = 10. Sampling error = x − μ = 10.50 − 12.33 = −1.
c. Liza’s incorrect x = (13 + 8 + 6 + 12)/4 = 39/4 = 9. x − μ = 9.75 − 12.33 = −2. Sampling error (from part b) = –1. Nonsampling error = −2.58 − (–1.83) = –. d. Sample x x − μ 15, 13, 8, 17 13.25. 15, 13, 8, 9 11.25 –1. 15, 13, 17, 9 13.50 1. 15, 8, 17, 9 12.25 –. 13, 8, 17, 9 11.75 –. 15, 13, 8, 12 12.00 –. 15, 13, 17, 12 14.25 1. 15, 8, 17, 12 13.00. 13, 8, 17, 12 12.50. 15, 13, 9, 12 12.25 –. 15, 8, 9, 12 11.00 –1. 13, 8, 9, 12 10.50 –1. 15, 17, 9, 12 13.25. 13, 17, 9, 12 12.75. 8, 17, 9, 12 11.50 –.
146 Chapter 7 Sampling Distributions
148 Chapter 7 Sampling Distributions
7.15 a. n / N = 300/5000 = .06 >.
x
N n n N
b. Since n / N = 100/5000 = .02 < .05,
σ x = σ n = 25 100 =2.
7.16 a. Since n / N = 2500/100,000 = .025 < .05,
σ x = σ n = 40 2500 =.
b. n / N = 7000/100,000 = .07 >.
x
N n n N
7.17 μ = 125 and σ = 36
a. Since σ x =σ n ,
( ) ( )
n = σ σ x = =
b. ( ) ( ) 2 2
n = σ σ x = 36 2.25 = 256
7.18 μ = 46 and σ = 10
a. Since σ x =σ n ,
( ) ( )
n = σ σ x = =
b. n = (^) ( σ σ (^) x ) 2 = (^) (10 1.6 )^2 ≈ 39
7.19 μ = $25,000, σ = $6280, and n = 400
μ x = μ= $25, 000and
σ x = σ n = 6280 400 =$
7.20 μ = 2300 square feet, σ = 500 square feet, and n = 25
μ x = μ= 2, 300square feet and
σ x = σ n = 500 25 = 100 square feet
7.21 μ = $25,510, σ = $4550, and n = 200
μ x = μ= $25, 510and
σ x = σ n = 4550 200 ≈$321.
7.22 μ = $11,350, σ = $2390, and n = 400
μ x = μ= $11, 350and
σ x = σ n = 2390 400 =$119.
7.23 σ = $2000 and σ x =$
( ) ( ) 2 2
n = σ σ x = 2000 125 = 256 players
7.24 σ = $139.50 million and
σ x = $15.50million
( ) ( )
n = σ σ x = =
7.25 a. x P ( x ) xP x (^^ ) x^2 x^2 P x ( )
76.00 .20 15.200 5776.0000 1155. 76.67 .10 7.667 5878.2889 587. 79.33 .10 7.933 6293.2489 629. 81.00 .10 8.100 6561.0000 656. 81.67 .20 16.334 6669.9889 1333. 84.33 .20 16.866 7111.5489 1422. 85.00 .10 8.500 7225.0000 722. ∑ xP x (^ )= 80.60^ ∑ x^^2 P x (^ )= 6507.
∑ xP x (^ )= 80.60 is the same value found in Exercise 7.6 for^ μ.
b. σ x = (^) ∑ x^2 P x ( ) − μ x^2 = 6507.262 −(80.60)^2 = 3.
c. σ n = 8.09 3 = 4.67is not equal to σ x = 3.30in this case because n / N = 3/5 = .60 > .05.
d.
x 1 3 5 1
N n n N
Section 7.3 Shape of the Sampling Distribution of x 149
U Section 7.
7.26 The population from which the sample is drawn must be normally distributed.
7.27 The central limit theorem states that for a large sample, the sampling distribution of the sample mean is
approximately normal, irrespective of the shape of the population distribution. Furthermore, μ x = μand
σ x = σ n , where μ and σ are the population mean and standard deviation, respectively. A sample size of
30 or more is considered large enough to apply the central limit theorem to x.
7.28 The central limit theorem will apply in cases a and c since n ≥ 30.It will not apply in case b because n <30.
7.29 a. Slightly skewed to the right
b. Approximately normal because n ≥ 30 and the central limit theorem applies c. Close to normal with a slight skew to the right
7.30 a. and b. In both cases the sampling distribution of x would be normal because the population distribution is normal.
7.31 a. and b. In both cases the sampling distribution of x would be normal because the population distribution is normal.
7.32 μ = 7.7 minutes, σ = 2.1 minutes, and n = 16
μ x = μ= 7.7minutes and σ x = σ n = 2.1 16 = .525minute
The sampling distribution of x is normal because the population is normally distributed.
7.33 μ = 20.20 hours, σ = 2.60 hours, and n = 18
μ x = μ= 20.20hours and σ x = σ n = 2.60 18 = .613hour
The sampling distribution of x is approximately normal because the population is approximately normally distributed.
7.34 μ = $140, σ = $30, and n = 25
μ x = μ= $140and σ x = σ n = 30 25 =$
The sampling distribution of x is approximately normal because the population is approximately normally distributed.
7.35 μ = 3.02, σ = .29, N = 5540 and n = 48
μ x = μ=3.
Since n / N = 48/5540 = .009 < .05, σ x = σ n = .29 48 = .042.
The sampling distribution of x is approximately normal because the population is approximately normally distributed.
7.36 μ = 133 pounds, σ = 24 pounds, and n = 45
μ x = μ= 133 pounds and σ x = σ n = 24 45 = 3.578pounds
The sampling distribution of x is approximately normal because the sample size is large ( n ≥ 30).
152 Chapter 7 Sampling Distributions
154 Chapter 7 Sampling Distributions
c. For x = 22: z = ( x − μ ) / σ x = (22 – 20.20)/.61282588 = 2.
P ( x ≥ 22) = P ( z ≥ 2.94) = 1 − P ( z < 2.94) = 1 − .9984 =.
d. For x = 21: z = ( x − μ ) / σ x = (21 – 20.20)/.61282588 = 1.
P ( x ≤ 21) = P ( z ≤ 1.31) =.
7.58 x = 2250 hours, σ = 150 hours, and n = 100
σ x = σ n = 150 100 = 15 hours
We are to find P ( μ − 25 ≤ x ≤ μ + 25)
For x = μ − 25: z = ( x − μ ) / σ x = ( μ − 25 − μ )/15 = −1.
For x = μ + 25: z = ( x − μ ) / σ x = ( μ + 25 − μ )/15 = 1.
P ( μ − 25 ≤ x ≤ μ + 25) = P (−1.67 ≤ z ≤ 1.67) = P ( z ≤ 1.67) − P ( z ≤ −1.67) = .9525 − .0475 =.
7.59 μ = 3 inches, σ = .1 inch, and n = 25
μ x = μ= 3 inches and σ x = σ n = .1 25 = .02inch
For x = 2.95: z = ( x − μ ) / σ x = (2.95 − 3)/.02 = −2.
For x = 3.05: z = ( x − μ ) / σ x = (3.05 − 3)/.02 = 2.
P ( x < 2.95) + P ( x > 3.05) = 1 − [ P (2.95 ≤ x ≤ 3.05)] = 1 − [ P (−2.50 ≤ z ≤ 2.50)] = 1 − [ P ( z ≤ 2.50) − P ( z ≤ −2.50)] = 1 − [.9938 − .0062] = 1 − .9876 =.
U Section7.
7.60 p = 640/1000 = .64 and p ˆ = 24/40 =.
7.61 p = 600/5000 = .12 and p ˆ = 18/120 =.
7.62 Number with characteristic in population = (18,700)(.3) = 5610 Number with characteristic in sample = (250)(.25) = 62.5 =
7.63 Number with characteristic in population = (9500)(.75) = 7125 Number with characteristic in sample = (400)(.78) = 312
7.64 a. μ ˆ p = p
b. σ (^) p ˆ = pq / n
c. The sampling distribution of p ˆ is approximately normal if np > 5 and nq > 5.
7.65 Sampling error = p ˆ – p = .66 – .71 = –.
7.66 Sampling error = p ˆ – p = .33 – .29 =.
7.67 The estimator of p is the sample proportion ˆ p.
The sample proportion ˆ p is an unbiased estimator of p , since the mean of p ˆ is equal to p.
7.68 The sample proportion ˆ p is a consistent estimator of p , since σ (^) p ˆ decreases as the sample size increases.
7.69 σ (^) ˆ p = pq / n , hence σ (^) ˆ p decreases as n increases.
7.70 p = .63 and q = 1 − p = 1 − .63 =. a. n = 100, μ (^) p ˆ = p = .63, and
σ (^) ˆ p = pq / n = (.63)(.37) / 100 =.
b. n = 900, μ (^) p ˆ = p = .63, and
σ (^) ˆ p = pq / n = (.63)(.37) / 900 =.
7.71 p = .21 and q = 1 − p = 1 − .21 =. a. n = 400, μ (^) p ˆ = p = .21, and
ˆ / (.21)(.79) / 400.
σ p = pq n
Section 7.5 Population and Sample Proportion; Mean, Standard Deviation, and 155 Shape of the Sampling Distribution of ˆ p
b. n = 750, μ p ˆ = p = .21, and
σ p = pq n
7.72 p = .12, q = 1 − p = 1 − .12 = .88, and N = 4000 a. n / N = 800/4000 = .20 >.
ˆ 1 (.12)(.88) 4000 800 . 800 4000 1
p
pq N n n N
b. Since n / N = 30/4000 = .0075 < .05,
σ ˆ p = pq / n = (.12)(.88) / 30 =.
7.73 p = .47, q = 1 − p = 1 − .47 = .53, and N = 1400 a. n / N = 90/1400 = .064 >.
ˆ (^1) (.47)(.53) 1400 90 . 90 1400 1
p
pq N n n N
b. Since n / N = 50/1400 = .036 < .05,
σ ˆ p = pq / n = (.47)(.53) / 50 =.
7.74 A sample is considered large enough to apply the central limit theorem if np > 5 and nq > 5.
7.75 a. np = (400)(.28) = 112 and nq = (400)(.72) = 288 Since np > 5 and nq > 5, the central limit theorem applies. b. np = (80)(.05) = 4; since np < 5, the central limit theorem does not apply. c. np = (60)(.12) = 7.2 and nq = (60)(.88) = 52. Since np > 5 and nq > 5, the central limit theorem applies. d. np = (100)(.035) = 3.5; since np < 5, the central limit theorem does not apply.
7.76 a. np = (20)(.45) = 9 and nq = (20)(.55) = 11 Since np > 5 and nq > 5, the central limit theorem applies. b. np = (75)(.22) = 16.5 and nq = (75)(.78) = 58. Since np > 5 and nq > 5, the central limit theorem applies. c. np = (350)(.01) = 3.5; since np < 5, the central limit theorem does not apply. d. np = (200)(.022) = 4.4; since np < 5, the central limit theorem does not apply.
7.77 a. p = 4/6 =.
b. (^) 6 C 5 = 6. c. and d. Let: G = good TV set and D = defective TV set Let the six TV sets be denoted as: 1 = G , 2 = G , 3 = D , 4 = D , 5 = G , and 6 = G. The six possible samples, their sample proportions, and the sampling errors are given in the table below.
p^ ˆ f Relative Frequency ˆ p P p ( ˆ) .60 4 4/6 = .667 .60. .80 2 2/6 = .333 .80. ∑ f^ =^6
Sample TV sets p ˆ^ Sampling error 1, 2, 3, 4, 5 G, G, D, D, G 3/5=.60 .60 – .667 = –. 1, 2, 3, 4, 6 G, G, D, D, G 3/5=.60 .60 – .667 = –. 1, 2, 3, 5, 6 G, G, D, G, G 4/5=.80 .80 – .667 =. 1, 2, 4, 5, 6 G, G, D, G, G 4/5=.80 .80 – .667 =. 1, 3, 4, 5, 6 G, D, D, G, G 3/5=.60 .60 – .667 = –. 2, 3, 4, 5, 6 G, D, D, G, G 3/5=.60 .60 – .667 = –.
158 Chapter 7 Sampling Distributions
b. For p ˆ = .53: z = ( p ˆ − p ) / σ p ˆ= (.53 − .55)/.016583124 = −1.
For p ˆ = .59: z = ( ˆ p − p ) / σ p ˆ= (.59 − .55)/.016583124 = 2.
P (.53 < p ˆ < .59) = P (−1.21 < z < 2.41) = P ( z < 2.41) − P ( z < −1.21) = .9920 − .1131 =.
7.90 p = .85, q = 1 − p = 1 − .85 = .15, and n = 100
σ ˆ p = pq / n = (.85)(.15) / 100 =.
a. For p ˆ = .81: z = ( ˆ p − p ) / σ p ˆ= (.81 − .85)/.03570714 = −1.
For p ˆ = .88: z = ( ˆ p − p ) / σ p ˆ= (.88 − .85)/.03570714 =.
P (.81 < p ˆ < .88) = P (−1.12 < z < .84) = P ( z < .84) − P ( z < −1.12) = .7995 − .1314 =.
b. For p ˆ = .87: z = ( ˆ p − p ) / σ p ˆ= (.87 − .85)/.03570714 =.
P ( ˆ p < .87) = P ( z < .56) =.
7.91 p = .06, q = 1 − p = 1 − .06 = .94, and n = 150
σ ˆ p = pq / n = (.06)(.94) / 150 =.
For ˆ p = .08: z = ( p ˆ − p ) / σ p ˆ= (.08 − .06)/.0193907194 = 1.
P ( ˆ p ≥ .08) = P ( z ≥ 1.03) = 1 − P ( z ≤ 1.03) = 1 − .8485 =.
7.92 p = .80, q = 1 − p = 1 − .80 = .20, and n = 100
σ (^) ˆ p (^) = pq / n = (.80)(.20) / 100 =. a. P ( p ˆ within .05 of p ) = P (.75 ≤ p ˆ ≤ .85)
For p ˆ = .75: z = ( ˆ p − p ) / σ p ˆ= (.75 − .80)/.04 = −1.
For p ˆ = .85: z = ( ˆ p − p ) / σ p ˆ= (.85 − .80)/.04 = 1.
P (.75 ≤ p ˆ ≤ .85) = P (−1.25 ≤ z ≤ 1.25) = P ( z ≤ 1.25) − P ( z ≤ −1.25) = .8944 − .1056 =.
b. P ( p ˆ is less than p by .06 or more) = P ( ˆ p ≤ .74)
For ˆ p = .74: z = ( p ˆ − p ) / σ p ˆ= (.74 − .80)/.04 = −1.
P ( ˆ p ≤ .74) = P ( z ≤ −1.50) =.
c. P ( p ˆ greater than p by .07 or more) = P ( ˆ p ≥ .87)
For ˆ p = .87: z = ( p ˆ − p ) / σ p ˆ= (.87 − .80)/.04 = 1.
P ( ˆ p ≥ .87) = P ( z ≥ 1.75) = 1 − P ( z ≤ 1.75) = 1 − .9599 =.
U Supplementary Exercises
7.93 μ = 8000 hours, σ = 400 hours, and n = 25
μ x = μ= 8000 hours and σ x = σ n = 400 25 = 80 hours
The sampling distribution of x is normal because the population is normally distributed.
7.94 μ = $82.69, σ = $11.17, and n = 23
μ x = μ= $82.69and σ x = σ n = 11.17 23 = $2.329106005 ≈$2.
The sampling distribution of x is approximately normal because the population is normally approximately distributed.
P ( x < 63.75) + P ( x > 64.25) = 1 − [ P (63.75 ≤ x ≤ 64.25)] = 1 − [ P (−2.50 ≤ z ≤ 2.50)]
- 7.5 a. μ = (20 + 25 + 13 + 19 + 9 + 15 + 11 + 7 + 17 + 30)/10 = 166/10 = 16.
- b. x = (20 + 25 + 13 + 9 + 15 + 11 + 7 + 17 + 30)/9 = 147/9 = 16. - Sampling error = x − μ = 16.33 − 16.60 = −.
- c. Rich’s incorrect x = (20 + 25 + 13 + 9 + 15 + 11 + 17 + 17 + 30)/9 = 157/9 = 17. - x − μ = 17.44 − 16.60 =. - Sampling error (from part b) = –. - Nonsampling error = .84 − (–.27) = 1. - 25, 13 19, 9,15,11,7,17,30 16.22 –. d. Sample x x − μ - 20, 13, 19, 9, 15, 11, 7, 17, 30 15.67 –. - 20, 25 19, 9, 15, 11, 7, 17, 30 17.00. - 20, 25, 13, 9, 15, 11, 7, 17, 30 16.33 –. - 20, 25, 13, 19, 15, 11, 7, 17, 30 17.44. - 20, 25, 13, 19, 9, 11, 7, 17, 30 16.78. - 20, 25, 13, 19, 9, 15, 7, 17, 30 17.22. - 20, 25, 13, 19, 9, 15, 11, 17, 30 17.67 1. - 20, 25, 13, 19, 9, 15, 11, 7, 30 16.56 –. - 20, 25, 13, 19, 9, 15, 11, 7, 17 15.11 –1. - 70 .20 14.00 4900 980. 7.6 x P ( x ) xP ( x ) x^2 x^2 P ( x ) - 78 .20 15.60 6084 1216. - 80 .40 32.00 6400 2560. - 95 .20 19.00 9025 1805. - Σ xP ( x ) = 80.60 Σ x^2 P ( x ) = 6561.
- μ = Σ xP ( x ) = 80.
- σ= x P x^2 ( ) − μ^2 = 6561.80 −(80.60) - ∑ = 8. - 15 1/6=. 7.7 a. x P ( x ) - 21 1/6=. - 25 1/6=. - 28 1/6=. - 53 1/6=. - 55 1/6=. - 55, 53, 28, 25 40.25 22.25 1/15 =. b. Sample x x P x ( ) - 55, 53, 28, 21 39.25 28.50 1/15 =. - 55, 53, 28, 15 37.75 29.00 1/15 =. - 55, 53, 25, 21 38.50 29.25 1/15 =. - 55, 53, 25, 15 37.00 29.75 1/15 =. - 55, 53, 21, 15 36.00 30.25 1/15 =. - 55, 28, 25, 21 32.25 30.75 2/15 =. - 55, 28, 25, 15 30.75 31.75 1/15 =. - 55, 28, 21, 15 29.75 32.25 1/15 =. - 55, 25, 21, 15 29.00 36.00 1/15 =. - 53, 28, 25, 21 31.75 37.00 2/15 =. - 53, 28, 25, 15 30.25 37.75 1/15 =. - 53, 28, 21, 15 29.25 38.50 1/15 =. - 53, 25, 21, 15 28.50 39.25 1/15 =. - 28, 25, 21, 15 22.25 40.25 1/15 =. - Section 7.4 Applications of the Sampling of x
- b. For x = 72.4: z = ( x − μ ) / σ x = (72.4 − 75)/3.13049517 = −. - P ( x < 72.4) = P ( z < −.83) =.
- 7.45 μ = 48, σ = 8, and n = - μ x = μ= 48 and σ x = σ n = 8 16 =
- a. For x = 49.6: z = ( x − μ ) / σ x = (49.6 − 48)/2 =. - For x = 52.2: z = ( x − μ ) / σ x = (52.2 − 48)/2 = 2. - P (49.6 < x < 52.2) = P (.80 < z < 2.10) = P ( z < 2.10) − P ( z < .80) = .9821 − .7881 =.
- b. For x = 45.7: z = ( x − μ ) / σ x = (45.7 − 48)/2 = −1. - P ( x > 45.7) = P ( z > −1.15) = 1 − P ( z ≤ −1.15) = 1 − .1251 =.
- 7.46 μ = 60, σ = 10, and n = - μ x = μ= 60 and σ x = σ n = 10 40 =1.
- a. For x = 62.20: z = ( x − μ ) / σ x = (62.20 − 60)/1.58113883 = 1. - P ( x < 62.20) = P ( z < 1.39) =.
- b. For x = 61.4: z = ( x − μ ) / σ x = (61.4 − 60)/1.58113883 =. - For x = 64.2: z = ( x − μ ) / σ x = (64.2 − 60)/1.58113883 = 2. - P (61.4 < x < 64.2) = P (.89 < z < 2.66) = P ( z < 2.66) − P ( z < .89) = .9961 − .8133 =.
- 7.47 μ = 90, σ = 18, and n = - μ x = μ= 90 and σ x = σ n = 18 64 =2.
- a. For x = 82.3: z = ( x − μ ) / σ x = (82.3 − 90)/2.25 = −3. - P ( x < 82.3) = P ( z < −3.42) =.
- b. For x = 86.7: z = ( x − μ ) / σ x = (86.7 − 90)/2.25 = −1. - P ( x > 86.7) = P ( z > −1.47) = 1 − P ( z ≤ −1.47) = 1 − .0708 =.
- 7.48 μ = $190, σ = $20, n = - μ x = μ= $190and σ x = σ n = 20 100 =$
- a. For x = $187: z = ( x − μ ) / σ x = (187 − 190)/2 = −1. - P ( x < $187) = P ( z < −1.5) =.
- b. For x = $193.50: z = ( x − μ ) / σ x = (193.50 – 190)/2 = 1. - P ( x > $193.50) = P ( z > 1.75) = 1 − P ( z ≤ 1.75) = 1 − .9599 =.
- c. For x = $191.70: z = ( x − μ ) / σ x = (191.70 − 190)/2 =. - For x = $194.50: z = ( x − μ ) / σ x = (194.50 − 190)/2 = 2. - P ($191.70 ≤ x ≤ $194.50) = P (.85 ≤ z ≤ 2.25) = P ( z ≤ 2.25) − P ( z ≤ .85) = .9878 − .8023 =.
- 7.49 μ = 3.02, σ = .29, and n = - μ x = μ= 3.02and σ x = σ n = .29 20 =.
- a. For x = 3.10: z = ( x − μ ) / σ x = (3.10 − 3.02)/.06484597 = 1. - P ( x ≥ 3.10) = P ( z ≥ 1.23) = 1 − P ( z ≤ 1.23) = 1 − .8907 =.
- b. For x = 2.90: z = ( x − μ ) / σ x = (2.90 − 3.02)/.06484597 = −1. - P ( x ≤ 2.90) = P ( z ≤ −1.85) =.
- c. For x = 2.95: z = ( x − μ ) / σ x = (2.95 − 3.02)/.06484597 = −1. - For x = 3.11: z = ( x − μ ) / σ x = (3.11 − 3.02)/.06484597 = 1. - P (2.95 ≤ x ≤ 3.11) = P (−1.08 ≤ z ≤ 1.39) = P ( z ≤ 1.39) − P ( z ≤ −1.08) = .9177 − .1401 =.
- 7.50 μ = 7.7 minutes, σ = 2.1 minutes, and n =
- a. For x = 7: z = ( x − μ ) / σ x = (7 − 7.7)/.525 = −1. μ x = μ= 7.7minutes and σ x = σ n = 2.1 16 = .525minute - For x = 8: z = ( x − μ ) / σ x = (8 − 7.7)/.525 =. - P (7 < x < 8) = P (−1.33 < z < .57) = P ( z < .57) − P ( z < −1.33) = .7157 − .0918 =. - For x = 6.7: z = ( x − μ) / σ x = (6.7 − 7.7)/.525 = −1. b. P ( x within 1 minute of μ ) = P (6.7 ≤ x ≤ 8.7) - For x = 8.7: z = ( x − μ) / σ x = (8.7 − 7.7)/.525 = 1. - P (6.7 ≤ x ≤ 8.7) = P (−1.90 ≤ z ≤ 1.90) = P ( z ≤ 1.90) − P ( z ≤ −1.90) = .9713 − .0287 =. - For x = 6.7: z = ( x − μ ) / σ x = (6.7 − 7.7)/.525 = −1. c. P ( x lower than μ by 1 minute or more) = P ( x ≤ 6.7) - P ( x ≤ 6.7) = P ( z ≤ −1.90) =.
- 7.51 μ = $11,350, σ = $2390, and n = - μ x = μ= $11, 350and σ x = σ n = 2390 400 =$119.
- a. For x = $11,540: z = ( x − μ ) / σ x = (11, 540 − 11,350) 119.50 =1. - P ( x > $11,540) = P ( z ≥ 1.59) = 1 − P ( z ≤ 1.59) = 1 − 9441 =.
- b. For x = $11,110: z = ( x − μ ) / σ x = (11,110 − 11,350)/119.50 = −2. - P ( x < $11,110) = P ( z < −2.01) =.
- c. For x = $11,250: z = ( x − μ ) / σ x = (11,250 − 11,350)/119.50 =−. - For x = $11,600: z = ( x − μ ) / σ x = (11,600 − 11,350)/119.50 = 2. - P ($11,250 ≤ x ≤ $11,600) = P (−.84 ≤ z ≤ 2.09) = P ( z ≤ 2.09) − P ( z ≤ −.84) = .9817 − .2005 =.
- 7.52 μ = 8.4 hours, σ = 2.7 hours, and n =
- a. For x = 8: z = ( x − μ ) / σ x = (8 − 8.4)/.40249224 = −. μ x = μ= 8.4hours and σ x = σ n = 2.7 45 = .40249224hour - For x = 9: z = ( x − μ ) / σ x = (9 − 8.4)/.40249224 = 1. - P (8 < x < 9) = P (−.99 < z < 1.49) = P ( z < 1.49) − P ( z < −.99) = .9319 − .1611 =.
- b. For x = 8: z = ( x − μ ) / σ x = (8 − 8.4)/.40249224 = −. - P ( x < 8) = P ( z < −.99) =.
- 7.53 μ = $2840, σ = $672, and n = - μ x = μ= $2840and σ x = σ n = 672 36 =
- a. For x = 2600: z = ( x − μ ) / σ x = (2600 − 2840)/112 = −2. - For x = 2950: z = ( x − μ ) / σ x = (2950 − 2840)/112 =. - P (2600 < x < 2950) = P (−2.14 < z < .98) = P ( z < .98) − P ( z < −2.14) = .8365 − .0162 =.
- b. For x = 3060: z = ( x − μ ) / σ x = (3060 − 2840)/112 = 1. - P ( x < 3060) = P ( z < 1.96) =. - Section 7.4 Applications of the Sampling of x
- 7.54 μ = 151 min, σ = 20 min, and n =
- a. For x = 148.70: z = ( x − μ ) / σ x = (148.70 − 151)/1.414213562 = −1. μ x = μ= 151 min and σ x = σ n = 20 200 = 1.414213562min - For x = 150: z = ( x − μ ) / σ x = (150 − 151)/1.414213562 = −. - P (148.70 < x < 150) = P (−1.63 < z < −.71) = P ( z < −.71) − P ( z < −1.63) = .2389 − .0516 =.
- b. For x = 153: z = ( x − μ ) / σ x = (153 − 151)/1.414213562 = 1. - P ( x > 153) = P ( z ≥ 1.41) = 1 − P ( z ≤ 1.41) = 1 − .9207 =.
- c. For x = 146: z = ( x − μ ) / σ x = (146 − 151)/1.414213562 = −3. - P ( x ≤ 146) = P ( z ≤ −3.54) =.
- 7.55 μ = $140, σ = $30, and n = - μ x = μ= $140and σ x = σ n = 30 75 =$3.
- a. For x = 132: z = ( x − μ ) / σ x = (132 − 140)/3.46410162 = −2. - For x = 136: z = ( x − μ ) / σ x = (136 − 140)/3.46410162 = −1. - P (132 < x < 136) = P (−2.31 < z < −1.15) = P ( z < −1.15) − P ( z < −2.31) = .1251 − .0104 =. - For x = 134: z = ( x − μ ) / σ x = (134 − 140)/3.46410162 = −1. b. P ( x within $6 of μ ) = P (134 ≤ x ≤ 146) - For x = 146: z = ( x − μ ) / σ x = (146 − 140)/3.46410162 = 1. - P (134 ≤ x ≤ 146) = P (−1.73 ≤ z ≤ 1.73) = P ( z ≤ 1.73) − P ( z ≤ −1.73) = .9582 − .0418 =. - For x = 144: z = ( x − μ ) / σ x = (144 − 140)/3.46410162 = 1. c. P ( x greater than μ by at least $4) = P ( x ≥ 144) - P ( x ≥ 144) = P ( z ≥ 1.15) = 1 − P ( z ≤ 1.15) = 1 − .8749 =.
- 7.56 μ = $45,000, σ = $12,720, and n = - μ x = μ= $45, 000and σ x = σ n = 12, 720 144 =$
- a. For x =$42,600: z = ( x − μ ) / σ x = (42,600 – 45,000)/1060 = −2. - P ( x < 42,600) = P ( z < −2.26) =.
- b. For x = $46,240: z = ( x − μ ) / σ x = (46,240 – 45,000)/1060 = 1. - ( x > 46,240) = P ( z > 1.17) = 1 − P ( z ≤ 1.17) = 1 − .8790 =.
- c. For x = $43,190: z = ( x − μ ) / σ x = (43,190 – 45,000)/1060 = −1. - For x = $46,980: z = ( x − μ ) / σ x = (46,980 – 45,000)/1060 = 1. - P (43,190 ≤ x ≤ 46,980) = P (−1.71 ≤ z ≤ 1.87) = P ( z ≤ 1.87) − P ( z ≤ −1.71) = .9693 − .0436 =.
- 7.57 μ = 20.20 hours, σ = 2.60 hours, and n = - For x = 19.20: z = ( x − μ ) / σ x = (19.20 – 20.20)/.61282588 = −1. a. P ( x is not within one hour of μ ) = P ( x < 19.20) + P ( x > 21.20) = 1 − P (19.20 ≤ z ≤ 21.20) - For x = 21.20: z = ( x − μ ) / σ x = (21.20 – 20.20)/.61282588 = 1. - = 1 – [ P ( z ≤ 1.63) − P ( z ≤ −1.63)] = 1 – [.9484 − .0516] =. P ( x < 19.20) + P ( x > 21.20) = 1 − P (19.20 ≤ x ≤ 21.20) = 1 − P (−1.63 ≤ z ≤ 1.63)
- b. For x = 20.0: z = ( x − μ ) / σ x = (20.0 – 20.20)/.61282588 = −. - For x = 20.5: z = ( x − μ ) / σ x = (20.5 – 20.20)/.61282588 =. - P (20.0 < x < 20.5) = P (−.33 < z < .49) = P ( z < .49) − P ( z < −.33) = .6879 − .3707 =. - Section 7.6 Applications of the Sampling Distribution of p ˆ
- 7.85 p = .59, q = 1 − p = 1 − .59 = .41, N = 30,000, and n =
- n / N = 100/30,000 = .003 <.
- σ ˆ p = pq / n = (.59)(.41) / 100 =.
- a. z = ( p ˆ − p ) / σ p ˆ = (.56 − .59)/.04918333 = −.
- b. z = ( p ˆ − p ) / σ p ˆ = (.68 − .59)/.04918333 = 1.
- c. z = ( p ˆ − p ) / σ p ˆ = (.53 − .59)/.04918333 = −1.
- d. z = ( p ˆ − p ) / σ p ˆ = (.65 − .59)/.04918333 = 1.
- 7.86 p = .25, q = 1 − p = 1 − .25 = .75, N = 18,000, and n =
- n / N = 70/18,000 = .004 <.
- σ ˆ p = pq / n = (.25)(.75) / 70 =.
- a. z = ( p ˆ − p ) / σ p ˆ = (.26 − .25)/.05175492 =.
- b. z = ( p ˆ − p ) / σ p ˆ = (.32 − .25)/.05175492 = 1.
- c. z = ( p ˆ − p ) / σ p ˆ = (.17 − .25)/.05175492 = −1.
- d. z = ( p ˆ − p ) / σ p ˆ = (.20 − .25)/.05175492 = −.
- 7.87 p = .86, q = 1 − p = 1 − .86 = .14, and n = - σ ˆ p = pq / n = (.86)(.14) / 400 =.
- a. For ˆ p = .88: z = ( p ˆ − p ) / σ p ˆ = (.88 − .86)/.0173493516 = 1. - P ( ˆ p > .88) = P ( z > 1.15) = 1 − P ( z ≤ 1.15) = 1 − .8749 =.
- b. For ˆ p = .82: z = ( p ˆ − p ) / σ p ˆ= (.82 − .86)/.0173493516 = −2. - For ˆ p = .84: z = ( p ˆ − p ) / σ p ˆ= (.84 − .86)/.0173493516 = −1. - P (.82 < p ˆ < .84) = P (−2.31 < z < −1.15) = P ( z < −1.15) − P ( z < −2.31) = .1251 − .0104 =.
- 7.88 p = .64, q = 1 − p = 1 − .64 = .36, and n = - σ ˆ p = pq / n = (.64)(.36) / 50 =.
- a. For p ˆ = .54: z = ( ˆ p − p ) / σ p ˆ = (.54 − .64)/.06788225 = −1. - For p ˆ = .61: z = ( ˆ p − p ) / σ p ˆ = (.61 − .64)/.06788225 = −. - P (.54 < p ˆ < .61) = P (−1.47 < z < −.44) = P ( z < −.44) − P ( z < −1.47) = .3300 − .0708 =.
- b. For p ˆ = .71: z = ( ˆ p − p ) / σ p ˆ = (.71 − .64)/.06788225 = 1. - P ( ˆ p > .71) = P ( z > 1.03) = 1 − P ( z ≤ 1.03) = 1 − .8485 =.
- 7.89 p = .55, q = 1 − p = 1 − .55 = .45, and n = - σ ˆ p = pq / n = (.55)(.45) / 900 =.
- a. For p ˆ = .58: z = ( ˆ p − p ) / σ p ˆ= (.58 − .55)/.016583124 = 1. - P ( ˆ p < .58) = P ( z < 1.81) =. - Chapter 7 Supplementary Exercises
- 7.95 μ = 8000 hours, σ = 400 hours, and n =
- a. For x = 7890: z = ( x − μ) / σ x = ( 7890 − 8000 ) 80 = −1. - P x ( < 7890 ) = P z ( < −1.38 ) =.
- b. For x = 7850: z = ( x − μ) / σ x = ( 7850 − 8000 ) 80 = −1. - For x = 7910: z = ( x − μ) / σ x = ( 7910 − 8000 ) 80 = −1. - P ( 7850 < x < 7910 ) = P ( −1.88 < z < − 1.13) = P z ( < −1.13 ) − P z ( < −1.88 ) = .1292 − .0301 =. - For x = 7870: z = ( x − μ) / σ x = ( 7870 − 8000 ) 80 = −1. c. P ( x within 130 hours of μ ) = P (7870 ≤ x ≤ 8130) - For x = 8130: z = ( x − μ) / σ x = ( 8130 − 8000 ) 80 =1. - P ( 7870 < x < 8130 ) = P ( −1.63 < z < 1.63) = P z ( < 1.63) − P z ( < − 1.63) = .9484 − .0516 =. - For x = 7850: z = ( x − μ) / σ x = ( 7850 − 8000 ) 80 = −1. d. P ( x is less than μ by 150 hours or more) = P ( x < 7850) - P ( x < 7850) = P ( z < −1.88) =.
- 7.96 μ = $82.69, σ = $11.17, and n = - μ x = μ= $82.69and σ x = σ n = 11.17 23 =$2.
- a. For x = 80: z = ( x − μ ) / σ x = (80 − 82.69)/2.329106005 = −1. - P ( x < 80) = P ( z < −1.15) =.
- b. For x = 75: z = ( x − μ ) / σ x = (75 − 82.69)/2.329106005 = −3. - For x = 85: z = ( x − μ ) / σ x = (85 − 82.69)/2.329106005 =. - P (75 < x < 85) = P (−3.30 < z < .99) = P ( z < .99) − P ( z < −3.30) = .8389 − .0005 =. - For x = 87.69: z = ( x − μ ) / σ x = (87.69 − 82.69)/2.329106005 = 2. c. P ( x within $5 of μ ) = P (77.69 ≤ x ≤ 87.69) - For x = 77.69: z = ( x − μ ) / σ x = (77.69 − 82.69)/2.329106005 = −2. - P (77.69 ≤ x ≤ 87.69) = P (−2.15 ≤ z ≤ 2.15) = P ( z ≤ 2.15) − P ( z ≤ −2.15) = .9842 − .0158 =.
- d. For x = 90: z = ( x − μ ) / σ x = (90 − 82.69)/2.329106005 = 3. - P ( x > 90) = P ( z > 3.14) = 1 − P ( z ≤ 3.14) = 1 − .9992 =.
- 7.97 μ = 50 mpg, σ = 5.9 mpg, and n =
- a. For x = 51.5: z = ( x − μ ) / σ x = (51.5 − 50)/.9571063847 = 1. μ x = μ= 50 mpg and σ x = σ n = 5.9 38 = .9571063847mpg - P ( x > 51.5) = P ( z > 1.57) = 1 − P ( z ≤ 1.57) = 1 − .9418 =.
- b. For x = 48: z = ( x − μ ) / σ x = (48 − 50)/.9571063847 = −2. - For x = 51: z = ( x − μ ) / σ x = (51 − 50)/.9571063847 = 1. - P (48 ≤ x ≤ 51) = P (−2.09 ≤ z ≤ 1.04) = P ( z ≤ 1.04) − P ( z ≤ −2.09) = .8508 − .0183 =.
- c. For x = 53: z = ( x − μ ) / σ x = (53 − 50)/.9571063847 = 3. - P ( x < 53) = P ( z < 3.13) =. - For x = 52.5: z = ( x − μ ) / σ x = (52.5 − 50)/.9571063847 = 2. d. P ( x greater than μ by 2.5 miles or more) = P ( x > 52.5) - P ( x > 52.5) = P ( z > 2.61) = 1 − P ( z ≤ 2.61) = 1 − .9955 =.
- 7.98 μ = 64 ounces, σ = .4 ounce, and n = 160 Chapter 7 Sampling Distributions
- For x = 63.75: z = ( x − μ ) / σ x = (63.75 − 64)/.1 = −2. μ x = μ= 64 ounces and σ x = σ n = .4 16 = .1ounce
- For x = 64.25: z = ( x − μ ) / σ x = (64.25 − 64)/.1 = 2. - = 1 − .9876 =. = 1 − [ P ( z ≤ 2.50) − P ( z ≤ −2.50)] = 1 − [.9938 − .0062]
- 7.99 p = .88, q = 1 − p = 1 − .88 = .12, and n = - μ p ˆ = p = .88, and σ p ˆ = pq / n = (.88)(.12) / 80 =.
- np = (80)(.88) = 70.4 > 5, nq = (80)(.12) = 9.6 >
- 7.100 p = .73, q = 1 − p = 1 − .73 = .27, and n = Since np and nq are both greater than 5, the sampling distribution of p ˆ is approximately normal. - μ p ˆ = p = .73, and σ ˆ p = pq / n = (.73)(.27) / 900 = .0147986486 ≈.
- np = (900)(.73) = 657 > 5, nq = (900)(.27) = 243 >
- 7.101 p = .73, q = 1 − p = 1 − .73 = .27, and n = Since np and nq are both greater than 5, the sampling distribution of p ˆ is approximately normal. - μ p ˆ = p = .73, and σ ˆ p = pq / n = (.73)(.27) / 900 = .0147986486 ≈.
- a. i. For ˆ p = .76: z = ( p ˆ − p ) / σ p ˆ= (.76 − .73) .0147986486 =2. - P ( ˆ p < .76) = P ( z < 2.03) =. - ii. For ˆ p = .70: z = ( p ˆ − p ) / σ p ˆ= (.70 − .73) .0147986486 = −2. - For ˆ p = .75: z = ( p ˆ − p ) / σ p ˆ= (.75 − .73) .0147986486 =1. - P (.70 < p ˆ < .75) = P (−2.03 < z < 1.35) = P ( z < 1.35) − P ( z < −2.03) = .9115 − .0212 =. - For ˆ p = .705: z = ( p ˆ − p ) / σ p ˆ= (.705 − .73) .0147986486 = −1. b. P ( ˆ p within .025 of p ) = P (.705 ≤ p ˆ ≤ .755) - For ˆ p = .755: z = ( p ˆ − p ) / σ p ˆ= (.755 − .73) .0147986486 =1. - P (.705 ≤ ˆ p ≤ .755) = P (−1.69 ≤ z ≤ 1.69) = P ( z ≤ 1.69) − P ( z ≤ −1.69) = .9545 − .0455 =. - For ˆ p = .76: z = ( p ˆ − p ) / σ p ˆ= (.76 − .73) .0147986486 =2. c. P ( ˆ p greater than p by .03 or more) = P ( ˆ p ≥ .76) - P ( ˆ p ≥ .76) = P ( z ≥ 2.03) = 1 − P ( z ≤ 2.03) = 1 − .9788 =.
- 7.102 p = .25, q = 1 − p = 1 − .25 = .75, and n = - σ ˆ p = pq / n = (.25)(.75) / 1000 =. - For ˆ p = .23: z = ( p ˆ − p ) / σ p ˆ= (.23 − .25)/.0136930639 = –1. a. P ( ˆ p within .02 of p ) = P (.23 ≤ p ˆ ≤ .27) - For ˆ p = .27: z = ( p ˆ − p ) / σ p ˆ= (.27 − .25)/.0136930639 = 1. - P (.23 ≤ ˆ p ≤ .27) = P (−1.46 ≤ z ≤ 1.46) = P ( z ≤ 1.46) − P ( z ≤ −1.46) = .9279 − .0721 =.
- b. P ( ˆ p is not within .02 of p ) = 1 – P ( ˆ p is within .02 of p ) = 1 – .8558 =.
Chapter 7 Supplementary Exercises 161
c. P ( ˆ p greater than p by .025 or more) = P ( ˆ p ≥ .275)
For ˆ p = .275: z = ( p ˆ − p ) / σ p ˆ= (.275 − .25)/.0136930639 = 1.
P ( ˆ p ≥ .275) = P ( z ≥ 1.83) = 1 − P ( z ≤ 1.83) = 1 − .9664 =.
d. P ( ˆ p is less than p by .03 or more) = P ( ˆ p ≤ .22)
For ˆ p = .22: z = ( p ˆ − p ) / σ p ˆ= (.22 − .25)/.0136930639 = –2.
P ( ˆ p ≤ .22) = P ( z ≤ −2.19) =.
7.103 σ = $2,845,000, and n = 32
σ x = σ n = 2,845, 000 32 =$502, 929.
The required probability is: P ( μ− 500, 000 ≤ x ≤ μ+500, 000)
For x = μ − 500,000: z = ( x − μ) / σ x = ( μ − 500,000 − μ )/502,929.70 = −.
For x = μ + 500,000: z = ( x − μ) / σ x = ( μ + 500,000 − μ )/502,929.70 =.
P ( μ − 500,000 ≤ x ≤ μ + 500,000) = P (−.99 ≤ z ≤ .99) = P ( z ≤ .99) − P ( z ≤ −.99) = .8389 − .1611 =.
7.104 Given P ( x > 90) = .15 and P ( x < 65) = .30, the corresponding z values are approximately z = 1.04 and z = –.52, respectively. First, we use x = μ + zσ to find σ. We have 90 = μ + 1.04 σ and 65 = μ + (−.52) σ. Subtracting gives 25 = 1.56 σ , so σ = 16.0256. Since 65 = μ + (−.52) σ , we have 65 = μ + (−.52)(16.0256) so μ = 65 + (.52)(16.0256) ≈ 73.33.
7.105 μ = c and σ = .8 ppm
We want P ( μ − .5 ≤ x ≤ μ + .5) = .95. The corresponding z value is 1.96; then 1.96 σ x = .5 and
σ (^) x = .255. Since σ (^) x = σ n , ( ) ( ) 2 2
n = σ σ x = .8 .255 = 9.84.
Thus, 10 measurements are necessary.
7.106 a. p = .60, q = 1 − p = 1 − .60 = .40, and n = 25 Since np = (25)(.60) = 15 > 5, nq = (25)(.40) = 10 > 5, we can use the normal approximation to the
binomial where μ = np = (25)(.60) = 15 and σ = npq = (25)(.60)(.40) =2.44948974.
For x = 12.5: z = ( x - μ )/ σ = (12.5 − 15)/2.44948974 = −1. P ( x > 12.5) = P ( z > −1.02) = 1 − P ( z ≤ −1.02) = 1 − .1539 =. b. For .95 or higher, z = −1.65. Now,
z = ( p ˆ − p ) / σˆ p ,so ˆ
p 1.
p p z
Then, since σ (^) ˆ p (^) = pq / n ,
2 2
n = pq / ( σˆ p ) = .60(.40) / (.0606) =65.35.
The reporter should take a sample of at least 66 voters.
7.107 a. p = .53, q = 1 − p = 1 − .53 = .47, and n = 200
σ ˆ p = pq / n = (.53)(.47) / 200 =.
For ˆ p = .50: z = ( p ˆ − p ) / σ p ˆ= (.50 − .53)/.03529164 = −.
P ( ˆ p > .50) = P ( z > −.85) = 1 − P ( z ≤ −.85) = 1 − .1977 =.
b. For .95 or higher, z = −1.65. Now, z = ( p ˆ − p ) / σˆ p ,so ˆ
p 1.
p p z
. Then,
since σ ˆ p = pq / n , n = pq / ( σ p ˆ)^2 = .53(.47) / (.01818182) 2 =753.53.
The politician should take a sample of at least 754 voters.
Chapter 7 Self-Review Test 163
U Self – Review Test
1. b 2. b 3. a 4. a 5. b 6. b 7. c 8. a 9. a 10. a 11. c 12. a 13. According to the central limit theorem, for a large sample size, the sampling distribution of the sample mean is approximately normal irrespective of the shape of the population distribution. The mean and
standard deviation of the sampling distribution of the sample mean are μ x = μand σ x = σ n ,
respectively. The sample size is usually considered to be large if n ≥ 30. From the same theorem, the sampling distribution of p ˆ is approximately normal for sufficiently large samples. In the case of proportion, the sample is sufficiently large if np > 5 and nq > 5.
14. μ = 145 pounds and σ = 18 pounds
a. μ x = μ= 145 pounds and σ x = σ n = 18 25 = 3.60pounds
b. μ x = μ= 145 pounds and σ x = σ n = 18 100 = 1.80pounds
In both cases the sampling distribution of x is approximately normal because the population has an approximate normal distribution.
15. μ = $650,000 and σ = $140,
a. μ x = μ= $650, 000and σ x = σ n = 140, 000 / 20 =$31, 305
Since the population has an unknown distribution and n < 30, we can draw no conclusion about the shape of the sampling distribution of x.
b. μ x = μ= $650, 000and σ x = σ n = 140, 000 / 100 =$14, 000
Since n ≥ 30, the sampling distribution of x is approximately normal.
c. μ x = μ= $650, 000and σ x = σ n = 140, 000 / 400 =$
Since n ≥ 30, the sampling distribution of x is approximately normal.
16. μ = $650,000 and σ = $140,
μ x = μ= $650, 000and σ x = σ n = 140, 000 / 100 =$14, 000
a. For x = 620,000: z = ( x − μ) / σ x = (620,000 – 650,000)/14,000 = −2.
For x = 635,000: z = ( x − μ) / σ x = (635,000 – 650,000)/14,000 = −1.
P (620,000 < x < 635,000) = P (−2.14 < z < −1.07) = P ( z < −1.07) − P ( z < −2.14) = .1423 − .0162 =. b. P ( x within $24,000 of μ ) = P (626,000 ≤ x ≤ 674,000)
For x = 626,000: z = ( x − μ) / σ x = (626,000 – 650,000)/14,000 = −1.
For x = 674,000: z = ( x − μ) / σ x = (674,000 – 650,000)/14,000 = 1.
P (626,000 ≤ x ≤ 674,000) = P (−1.71 ≤ z ≤ 1.71) = P ( z ≤ 1.71) − P ( z ≤ −1.71) = .9564 − .0436 =.
c. For x = 630,000: z = ( x − μ) / σ x = (630,000 – 650,000)/14,000 = −1.
P ( x ≥ 630,000) = P ( z ≥ −1.43) = 1 − P ( z ≤ −1.43) = 1 − .0764 =.
164 Chapter 7 Sampling Distributions
d. P ( x not within $20,000 of μ ) = 1 − P ( x ≤ 630,000) + P ( x ≥ 670,000)
For x = 630,000: z = ( x − μ) / σ x = (630,000 – 650,000)/14,000 = −1.
For x = 670,000: z = ( x − μ) / σ x = (670,000 – 650,000)/14,000 = 1.
P ( x ≤ 630,000) + P ( x ≥ 670,000) = 1 − P (630,000 < x < 670,000) = 1 − P (−1.43 ≤ z ≤ 1.43) = 1 – [ P ( z ≤ 1.43) − P ( z ≤ −1.43)] = 1 – [.9236 − .0764] = 1 − .8472 =.
e. For x = 640,000: z = ( x − μ) / σ x = (640,000 – 650,000)/14,000 = −.
P ( x < 640,000) = P ( z < −.71) =.
f. For x = 660,000: z = ( x − μ) / σ x = (660,000 – 650,000)/14,000 =.
P ( x < 660,000) = P ( z < .71) =.
g. For x = 670,000: z = ( x − μ) / σ x = (670,000 – 650,000)/14,000 = 1.
P ( x ≥ 670,000) = P ( z ≥ 1.43) = 1 − P ( z ≤ 1.43) = 1 − .9236 =.
h. For x = 640,000: z = ( x − μ) / σ x = (640,000 – 650,000)/14,000 = −.
For x = 665,000: z = ( x − μ) / σ x = (665,000 – 650,000)/14,000 = 1.
P (640,000 < x < 665,000) = P (−.71 < z < 1.07) = P ( z < 1.07) − P ( z < −.71) = .8577 − .2389 =.
17. μ = 16 ounces, σ = .18 ounce, and n = 16
μ x = μ= 16 ounces and σ x = σ n = .18 16 = .045ounce
a. i. For x = 15.90: z = ( x − μ ) / σ x = (15.90 − 16)/.045 = −2.
For x = 15.95: z = ( x − μ ) / σ x = (15.95− 16)/.045 = −1.
P (15.90 < x < 15.95) = P (−2.22 < z < −1.11) = P ( z < −1.11) − P ( z < −2.22) = .1335 − .0132 =.
ii. For x = 15.95: z = ( x − μ ) / σ x = (15.95− 16)/.045 = −1.
P ( x < 15.95) = P ( z < −1.11) =.
iii. For x = 15.97: z = ( x − μ ) / σ x = (15.97 − 16)/.045 = −.
P ( x > 15.97) = P ( z > −.67) = 1 − P ( z ≤ −.67) = 1 − .2514 =. b. P ( x within .10 ounce of μ ) = P (15.90 ≤ x ≤ 16.10)
For x = 15.90: z = ( x − μ ) / σ x = (15.90 − 16)/.045 = −2.
For x = 16.10: z = ( x − μ ) / σ x = (16.10 − 16)/.045 = 2.
P (15.90 ≤ x ≤ 16.10) = P (−2.22 ≤ z ≤ 2.22) = P ( z ≤ 2.22) − P ( z ≤ −2.22) = .9868 − .0132 =.
c. P ( x is less than μ by .135 ounce or more) = P ( x < 15.865)
For x = 15.865: z = ( x − μ ) / σ x = (15.865 − 16)/.045 = −3.
P ( x < 15.865) = P ( z < −3.00) =.
18. p = .15, and q = 1 − p = 1 − .15 =.
a. n = 30, μ ˆ p = p = .15, and σ p ˆ = pq / n = (.15)(.85) / 30 =.
np = (30)(.15) = 4.5 and nq = (30)(.85) = 25. Since np < 5, we can draw no conclusion about the shape of the sampling distribution of p ˆ.
b. n = 300, μ (^) ˆ p = p = .15, and σ (^) p ˆ = pq / n = (.15)(.85) / 300 =. np = (300)(.15) = 45 and nq = (300)(.85) = 255 Since np > 5 and nq > 5, the sampling distribution of ˆ p is approximately normal.
