_newbold_ism_07.pdf

Chapter 7: Sampling and Sampling Distributions

7.1

a. Probability distribution for one die: Die outcome 1 2 3 4 5 6

Probability 1/6 1/6 1/6 1/6 1/6 1/6

b. Sampling distribution of the sample means from rolling a pair of dice: x Total Sample Prob. of x 2 11 1 1/36 3 1 2, 2 1 1.5 2/36 4 1 3, 3 1, 2 2 2 3/36 5 1 4, 4 1, 2 3, 3 2 2.5 4/36 6 1 5, 5 1, 2 4, 4 2, 3 3 3 5/36 7 1 6, 6 1, 2 5, 5 2, 3 4, 4 3 3.5 6/36 8 2 6, 6 2, 3 5, 5 3, 4 4 4 5/36 9 3 6, 6 3, 4 5, 5 4 4.5 4/36 10 4 6, 6 4, 5 5 5 3/36 11 5 6, 6 5 5.5 2/36 12 66 6 1/36 7.2

a. Binomial random variable with n = 2, p = .5 Probability Density Function Binomial with n = 2 and p = 0.5 x P( X = x ) 0 0.25 1 0.50 2 0.25

b. Binomial random variable with n = 4, p = .5 Probability Density Function Binomial with n = 4 and p = 0.5 x P( X = x ) 0 0.0625 1 0.2500 2 0.3750 3 0.2500 4 0.0625

c. Binomial random variable with n = 10, p = .5


Probability Density Function Binomial with n = 10 and p = 0.5 x P( X = x ) 0 0.000977 1 0.009766 2 0.043945 3 0.117188 4 0.205078 5 0.246094 6 0.205078 7 0.117188 8 0.043945 9 0.009766 10 0.000977

7.3 The sampling distribution of the sample mean can be generated by listing out all possible samples of size n, calculate each possible x , determine the probability of each possible x and generate the sampling distribution. Alternatively, the probabilities of each x can be generated by use of the binomial formula. a. When n = 5: Use the binomial formula for x = 0, x = 1, etc.: x X P(X) 0 .07776 0 1 .25920 .2 2 .34560 .4 3 .23040 .6 4 .07680 .8 5 .01024 1.0 p(1 − p ) .4(.6) 2 E ( px ) = np = (5)(.4) = 2.0, σ p = = = = .048, σ p = .2191 n 5

b. Using the result from part a: 2

E ( px ) = np = (100)(.4) = 40, σ p =

σ p = .04899

p(1 − p ) .4(.6) = = = .0024, n 100

7.4 The response should note that there will be errors in taking a census of the entire population as well as errors in taking a sample. Improved accuracy can be achieved via sampling methods versus taking a complete census (see reference to Hogan, 90). By using sample information, we can make valid inferences about the entire population without the time and expense involved in taking a census. 7.5

a. mean and variance of the sampling distribution for the sample mean μ x = μ = 100

σ 2 x = σ n = 81 25 = 3.24 σ x = σ x 2 = 3.24 2

145

146

th

Statistics for Business & Economics, 6 edition

102 − 100 = 1.11 1 – Fz(1.11) = .1335 3.24 98 − 100 = -1.11 Fz = 1 – Fz(1.11) c. Probability that 98 ≤ x ≤ 101 z x = 3.24 = .1335 101 − 100 zx = = .56 Fz(.56) – [1-Fx(1.11)] = .7123 - .1335 = .5788 3.24 101.5 − 100 = .83 Fz = .7967 d. Probability that x ≤ 101.5 z x = 3.24 b. Probability that x > 102 z x =

7.6


σ 2 x = σ n = 900 30 = 30 σ x = σ x 2 = 30 2

109 − 100 = 1.64 1 – Fz(1.64) = .0505 30 96 − 100 = -.73 1 – Fz(.73) = .2327 c. Probability that 96 ≤ x ≤ 110 z x = 30 110 − 100 zx = = 1.83 Fz = .9664. .9664 - .2327 = .7337 30 107 − 100 = 1.28 Fz = .8997 d. Probability that x ≤ 107 z x = 30

b. Probability that x > 109 z x =

7.7


σ 2 x = σ n = 625 25 = 25 σ x = σ x 2 = 25 209 − 200 = 1.80 1 – Fz(1.80) = .0359 b. Probability that x > 209 z x = 25 198 − 200 = -.40 1 – Fz(.40) = c. Probability that 198 ≤ x ≤ 211 z x = 25 .3446 211 − 200 zx = = 2.20 Fz(2.20) = .9861. .9861 - .3446 = .6415 25 202 − 200 = .40 Fz = .6554 d. Probability that x ≤ 202 z x = 25 2

7.8


σ 2 x = σ n = 1600 35 = 45.7143 σ x = σ x 2 = 45.7143 2


412 − 400 = 1.77 1 – Fz(1.77) = .0384 45.7143 407 − 400 = 1.04 Fz(1.04) = c. Probability that 393 ≤ x ≤ 407 z x = 45.7143 .8508 393 − 400 zx = = -1.04 1 – Fz(1.04) = .1492. .8508 - .1492 = 45.7143 .7016 389 − 400 = -1.63 1-Fz(1.63) = 1-.9484 d. Probability that x ≤ 389 z x = 45.7143 = .0516 b. Probability that x > 412 z x =

7.9

a. E( X ) = μ x = 92. 2

b. σ x =

σ

2

n

=

( 3.6 )

c. σ x =

σ n

2

= 3.24

4

=

3.6 = 1.8 2

93 − 92 ) = P(Z > .56) = 1.8

d. P(Z >

.2877 7.10

a. E( X ) = μ x = 1,200 2

b. σ x =

σ2 n

=

( 400 ) 9

c. σ x =

2

= 17,778

d. P(Z<

σ n

=

400 = 133.33 3

1, 050 − 1, 200 ) = P(Z<-1.13) 133.33

=.1292 24 − 25 24 − 25 ) = P(Z < -.5) = .3085. ii) P(Z < ) = P(Z < -1) 2 2 4 =.1587 24 − 25 ) = P(Z < -2) = .0228 iii) P(Z < 2 16 b. As the sample size increases, the standard error of the sampling distribution will decrease. That is, as the sample size increases, the sampling distribution of the sample means will clump up tighter around the true population mean. The graph would show a tighter distribution with less area in the tails.

7.11 a. i) P(Z >

7.12

110 − 115 ) = P(Z < -2) = .9772 25 100 113 − 115 117 − 115 b. P(
a. P(Z >

147

148

th


114 − 115 116 − 115
c. P(

7.13

7.14

60 = 20 9 270 − 280 ) = P(Z < -.5) = 1 - .6915 = .3085 b. P(Z < 20 250 − 280 ) = P(Z > -1.5) = .9332 c. P(Z > 20 d. If the population standard deviation is smaller, then the standard error of the sampling distribution of the means will also be smaller. Since Z is higher, tail areas are smaller and the probabilities calculated for parts a and b will both be smaller. a. σ x =

b. c. d. e.

7.15

22 = 5.5 16 100 − 87 ) = P(Z < 2.36) = .9909 P(Z < 5.5 80 − 87 ) = P(Z > -1.27) = .8980 P(Z > 5.5 85 − 87 95 − 87 >Z> ) = P(-.36 > Z > 1.45) = .4329 P( 5.5 5.5 Higher, higher, lower. The graph will show that the standard error of the sample means will decrease with an increased sample size.

a. σ x =

.6 = .3 4 19.7 − 20 ) = P(Z < -1) = .1587 b. P(Z < .3 20.6 − 20 ) = P(Z > 2) = .0228 c. P(Z > .3 19.5 − 20 20.5 − 20

7.16

7.17

7.18

40 =4 100 b. P(Z > 5/4) = P(Z > 1.25) = .1056 c. P(Z < -4/4) = P(Z < -1) = .1587 d. P(-3/4 > Z > 3/4) = P(-.75 > Z > .75) = .4532 a. σ x =

8 = 4. P(Z > 2/4) = P(Z > .5) = .3085 4 b. P(Z < -3/4) = P(Z < -.75) = .2266 c. P(-4/4 > Z > 4/4) = P(-1 > Z > 1) = .3174 d. Lower, lower, lower a. σ x =

a. σ x =

Difference 1.6 , Difference = = .16, P(Z>1.645) =.05, 1.645 = .16 100

±.2632 Difference , Difference = -.2048 .16 Difference , Difference = .2304 c. P(Z > 1.44) = .075, 1.44 = .16

b. P(Z < -1.28) = .1, -1.28 =

1 , n = 39.075, take n = 40 3.8 n c. larger

7.19 a. P(Z > 1.645) = .10, 1.645 = b. larger 7.20

a. P(Z > 1.96) = .025, 1.96 =

2 8.4

n

, n = 67.766, take n = 68

b. smaller c. larger 7.21

2

a. σ x =

1 N

∑ ( X i − X )2 =

∑x

i

2

− Nx 2

N

=

205 − (6)(5.5) 2 23.5 47 = = 6 6 12

b. (4.5 − 5.5) 2 2(4.75 − 5.5) 2 2(5 − 5.5) 2 + + + 15 15 15 2(5.25 − 5.5)2 1(5.5 − 5.5) 2 3(5.75 − 5.5) 2 (6 − 5.5) 2 + + + + + 15 15 15 15 2(6.25 − 5.5) 2 (6.75 − 5.5) 2 47 + + = 15 15 120 2 σ N − n 47 12 6 − 4 47 = = c. σ x 2 = n N −1 4 6 − 1 120

σ x 2 = ∑ ( X i − μ ) 2 Px ( x ) =

149

150

th


0 19 20 N = 40, correction factor = 39 80 N = 100, correction factor = 99 980 N = 1,000, correction factor = 999 9,980 N = 10,000, correction factor = 9,999 b. When the population size (N) equals the sample size (n), then there is no variation away from the population mean and the standard error will be zero. As the sample size becomes relatively small compared to the population size, the correction factor tends towards 1 and the correction factor becomes less significant in the calculation of the standard error c. The correction factor tends toward a value of 1 and becomes progressively less important as a modifying factor when the sample size decreases relative to the population size

7.22 a. N = 20, correction factor =

7.23

7.24

300, 000 400 = 26,859,689 499 100 825, 000 − 800, 000 )= P(Z > .93) = .1762 b. P(Z > 26,859.689 780, 000 − 800, 000 )= P(Z > -.74) = .7704 c. P(Z > 26,859.689 790, 000 − 800, 000 820, 000 − 800, 000
200 = 3.8023 249 2.5 )= P(Z > .66) = .2546 a. P(Z > 3.8023 −5 )= P(Z < -1.31) = .0951 b. P(Z < 3.8023 −10 10
σx =

30 50


7.25

a. b. c. d.

7.26

b.

c. d.

7.28

(.4)(.6) = .04899 100 Probability that the sample proportion is greater than .45 .45 − .4 z= = P(Z >1.02) = .1539 .04899 Probability that the sample proportion is less than .29 .29 − .4 z= = P(Z< -2.25) = .0122 .04899 Probability that the sample proportion is between .35 and .51 .35 − .4 .51 − .4 P(
E ( pˆ ) = .4 σ pˆ = a.

7.27

10 450 = .7077 150 599 31 − 32 )= P(Z > -1.41) = .9207 P(Z > .7077 33 − 32 )= P(Z < 1.41) = .9207 P(Z < .7077 Normal probability graph. Due to the property of symmetry, the area in the tails of the normal probability distribution are the same. 31 − 32 33 − 32 ) or P(Z > ) = P(Z < -1.41) or P(Z > 1.41) = P(Z < .7077 .7077 .1586

σx =

(.25)(.75) = .0306186 200 a. Probability that the sample proportion is greater than .31 .31 − .25 z= = P(Z > 1.96) = .0250 .0306186 b. Probability that the sample proportion is less than .14 .14 − .25 z= = P(Z < -3.59) = .0002 .0306186 c. Probability that the sample proportion is between .24 and .40 .24 − .25 .4 − .25 P(
E ( pˆ ) = .25 σ pˆ =

(.6)(.4) = .04899 100 a. Probability that the sample proportion is greater than .66 .66 − .6 = P(Z > 1.22) = .1112 z= .04899 b. Probability that the sample proportion is less than .48

E ( pˆ ) = .60 σ pˆ =

151

152

th


.48 − .6 = P(Z < -2.45) = .0071 .04899 c. Probability that the sample proportion is between .52 and .66 .52 − .6 .66 − .6 P( z =
7.29

7.30

(.5)(.5) = .01667 900 a. Probability that the sample proportion is greater than .52 .52 − .5 = P(Z > 1.20) = .1152 z= .01667 b. Probability that the sample proportion is less than .46 .46 − .5 = P(Z < -2.40) = .0082 z= .01667 c. Probability that the sample proportion is between .47 and .53 .47 − .5 .53 − .5 P( z =
E ( pˆ ) = .50 σ pˆ =

a. E ( pˆ ) = .424 (.424)(.576) b. σ pˆ 2 = = .00244 100 c. σ pˆ = .0494 d. P(Z >

.5 − .424 )= P(Z > 1.54) = .0618 .0494

7.31 a. E ( pˆ ) = .75 (.75)(.25) b. σ pˆ 2 = = .001875 100 c. σ pˆ = .0433 d. P(Z > 7.32

7.33

.8 − .75 )= P(Z > 1.15) = .1251 .0433

a. E ( pˆ ) = .20 (.2)(.8) b. σ pˆ 2 = = .000889 180 c. σ pˆ = .0298 d. P(Z <

.15 − .2 )= P(Z < -1.68) = .0465 .0298

a. σ pˆ =

(.3)(.7) = .0324 200


.25 − .3 )= P(Z < -1.54) = .0618 .0324 .33 − .3 )= P(Z > .93) = .1762 c. P(Z > .0324 .27 − .3 .33 − .3
7.34

7.35

7.36

7.37

(.4)(.6) = .0447 120 .35 − .4 .45 − .4
σ pˆ =

(.42)(.58) = .0285 300 .5 − .42 )= P(Z > 2.81) = .0025 b. P(Z > .0285 .4 − .42 .45 − .42
a. σ pˆ =

(.2)(.8) = .0351 130 .15 − .2 ) = P(Z > -1.42) = .9222 b. P(Z > .0351 .18 − .2 .22 − .2
a. σ pˆ =

(.3)(.7) = .02739 280 .32 − .3 )= P(Z < .73) = .7673 b. P(Z < .02739 c. .29 - .31

a. σ pˆ =

153

154

th


The largest value for σ pˆ is when p = .5. In this case, σ pˆ =

7.38

(.5)(.5) = 100

.05 7.39

P(Z > 1.96) = .025, .03 = 1.96

(.5)(.5) , solving for n = 1067.11. Take a n

sample of size 1,068. 7.40

(.25)(.75) = .0395 120 Difference , Difference = .0506 b. P(Z > 1.28), 1.28 = .0395 Difference , Difference = .065 c. P(Z < -1.645), -1.645 = .0395 Difference , Difference = .0409 d. P(Z > 1.036), 1.036 = .0395

a. σ pˆ =

7.41

σ pˆ =

.56 − .5 (.5)(.5) = .04082, P(Z > )= P(Z > 1.47) = .0708 .04082 150

7.42

σ pˆ =

.58 − .5 (.5)(.5) = .03162, P(Z > )= P(Z > 2.53) = .0057 .03162 250

7.43

7.44

(.55)(.45) 419 = .05065 81 499 .5 − .55 )= P(Z < -.99) = .1611 P(Z < .05065

σ pˆ =

211 = .3996 528 (.3996)(.6004) 408 σ pˆ = = .03934 120 527 .33 − .3996 )= P(Z < -1.77) = .0384 b. P(Z < .03934 .5 − .3996 .6 − .3996

239 (.5457)(.4543) 358 = .5457, σ pˆ = = .05038 438 80 437 .5 − .5457 )= P(Z < -.91) = .1814 b. P(Z < .05038 .5 − .5457 .6 − .5457
7.45 a. pˆ =

7.46

.1 − .122 (.2709)(.7291) = P(Z < -.61) = .2709, σ pˆ = = .04969 .036 81 .5 − .2709 ) = P(Z > 4.61) ≈ .0000 P(Z > .04969 P(Z <

7.47

a. Find the probability that the sample mean is > 101. 101 − 100 Probability that x > 101 z x = = .80 1 – Fz(.80) = .2119 5 16 b. Find the probability that the sample variance is > 45 (n − 1) s 2 15(45) P( ) = P ( χ 2 (15) > 27) = between .05 and .025 > 25 σ2 c. Find the probability that the sample variance is > 60 (n − 1) s 2 15(60) P( ) = P ( χ 2 (15) > 36) = Less than .005 > 2 25 σ

7.48

a. Probability that the sample mean is > 200. 200 − 198 Probability that x > 200 z x = = 1.00 1 – Fz(1.00) = .1587 10 25 b. 5% of the sample variances would be less than this value ⎡ (n − 1) s 2 ⎤ 24s 2 2 χ = 13.85 < 13.85 s 2 < 57.702 P(s 2 > k ) = P ⎢ 24,.95 ⎥ 2 100 ⎣ σ ⎦ c. 5% of the samples variances would be greater than this value ⎡ (n − 1) s 2 ⎤ 24s 2 2 > 36.42 s 2 > 151.879 χ = 36.42 P(s 2 > k ) = P ⎢ 24,.05 ⎥ 2 100 ⎣ σ ⎦

7.49 a. Probability that the sample mean is > 50 50 − 46 Probability that x > 50 z x = = 2.40 1 – Fz(2.40) = .0082 7.07107 18 b. 5% of the sample variances would be less than this value ⎡ (n − 1) s 2 ⎤ 17 s 2 2 2 P(s > k ) = P ⎢ < 8.67 s 2 < 25.50 χ 17,.95 = 8.67 ⎥ 2 50 ⎣ σ ⎦

155

156

th


c. 5% of the samples variances would be greater than this value ⎡ (n − 1) s 2 ⎤ 17 s 2 2 > 27.59 s 2 > 81.147 P(s 2 > k ) = P ⎢ χ 17,.05 = 27.59 ⎥ 2 50 ⎣ σ ⎦ 7.50 P (

(n − 1) s 2

σ

2

>

19(3.1) ) = P ( χ 2 (19) > 33.66) = between .01 and .025 (.0201 1.75

exactly) 11(2.5) 2 ) = P ( χ 2 (11) < 23.79) = between .975 and .99 2 2 (1.7) σ (.9864 exactly) (n − 1) s 2 11(1)1 > ) = P ( χ 2 (11) > 3.81) = between .975 and .99 (.9751 b. P ( 2 2 (1.7) σ exactly)

7.51.1 a. P (

7.52 a. P (

(n − 1) s 2

(n − 1) s 2

σ2

<

>

15(3, 000) 2 ) = P ( χ 2 (15) > 21.6) = greater than .1 (.1187 2 (2,500)

exactly) (n − 1) s 2 15(1,500) 2 < ) = P ( χ 2 (15) < 5.4) = between .01 and .025 b. P ( 2 2 (2,500) σ (.0118 exactly) 7.53

7.54

7.55

(n − 1) s 2

19(100) ) = P( χ 2 (19) < 7.6) = about .01 ( .0097 exactly) 250 σ 2 (n − 1) s 19(500) > ) = P( χ 2 (19) > 38) = between .005 and .01 (.0059 b. P ( 2 250 σ exactly) a. P (

2

a. P (

(n − 1) s 2

a. P (

(n − 1) s 2

<

24(75) 2 ) = P ( χ 2 (24) < 13.5) = between .025 and .05 (100) 2 σ2 (.0428 exactly) (n − 1) s 2 24(150) 2 > ) = P ( χ 2 (24) > 54) = less than .005 (.0004 b. P ( (100) 2 σ2 exactly) <

29(3.5) 2 ) = P ( χ 2 (29) > 17.54) = between .95 and .975 (4.5) 2 σ2 (.9531 exactly) – yes >


b. P (

(n − 1) s 2

29(6) 2 ) = P ( χ 2 (29) < 51.56) = between .99 and .995 2 2 (4.5) σ (.9939 exactly) - yes <

7.56 Descriptive Statistics: C20, C21, C22, C23, C24, C25, C26, C27, ... Variable C20 C21 C22 C23 C24 C25 C26 C27 C28 C29 C30 C31 C32 C33 C34

Mean 3.00 4.00 4.00 4.50 5.00 5.00 5.00 6.00 6.0000 6.500 7.00 6.500 7.00 7.500 5.50

Variance 2.00 8.00 8.00 12.50 18.00 2.00 2.00 8.00 0.000000000 0.500 2.00 0.500 2.00 0.500 4.50

Descriptive Statistics: Variance Variable Variance

x=

Mean 4.72

StDev 5.26

Variance 27.62

Sum 70.80

70.8 = 4.72 E ( s 2 ) = 15(3.91667) = 4.1964 (14) 15

7.57

a. P( χ 2 (5) > 11.07) = .05, 11.07 = 5(Difference), Difference = 2.214 (221.4%) b. P ( χ 2 (5) < 1.61) = .1, 1.61 = 5(Difference), Difference = .322 (32.2%)

7.58

a. P ( χ 2 (9) > 14.68) = .10, 14.68 = 9(Difference), Difference = 1.6311 (163.11%) b. P ( χ 2 (9) < 2.7) = .025, P ( χ 2 (9) > 19.02) = .025, 2.7 = 9a, a = .3, 19.02 = 9b, b = 2.1133 The probability is .95 that the sample variance is between 30% and 211.33% of the population variance c. The interval in part b. will be smaller

7.59

a. P( χ

2 (14)

14 s 2 x > 29.14) = .01, 29.14 = , sx = 2.597 (1.8) 2 14 s 2 x , sx = 1.141 (1.8) 2 < 6.57) = .05, P ( χ 2 (14) > 23.68) = .05

b. P ( χ 2 (14) > 5.63) = .025, 5.63 = c. P( χ 2 (14 )

between 6.57 =

14 s 2 a 14 s 2b , s = 1.233, and 23.68 = , sb = 2.341 a (1.8) 2 (1.8) 2

157

158

th


7.60

a. P( χ 2 (11) > 4.57) = .95, 4.57 = 11Difference, Difference = .4155 (41.55%) b. P ( χ 2 (11) > 5.58) = .90, 5.58 = 11Difference, Difference = .5073 (50.73%) c. P ( χ 2 (11) < 3.82) = .025, P( χ 2 (11) > 21.92) = .025, 3.82 = 11a, a = .34727, 21.92 = 11b, b = 1.9927 The probability is .95 that the sample variance is between 34.727% and 199.27% of the population variance

7.61

P(

(n − 1) s 2

σ

2

>

19(2.05) ) = P ( χ 2 (19) > 25.97) = more than 10% (.1310 1.5

<

24(12.2) ) = P( χ 2 (24) < 19.01) = less than .90 (.5438 exactly) 15.4

exactly) 7.62

P(

(n − 1) s 2

σ

2

7.63 A sample mean can be thought of as a random variable since there are a very large number of possible samples, sample size n, that can be drawn from a population. Each of those samples is likely to have a different sample mean. Therefore, all possible sample means, sample size n, will have its own probability distribution. This probability distribution is made up of all possible sample means calculated from all possible samples of a certain size drawn from a specific population 7.64

6! = 15 possible samples 2!4! b. (41, 39), (41, 35), (41, 35), (41, 33), (41, 38), (39, 35), (39, 35), (39, 33), (39, 38), (35, 35), (35, 33), (35, 38), (35, 33), (35, 38), (33, 38) 2 c. 34 PX (34) = 34 = 4.5333 15 35 = 2.3333 35PX (35) = 15 35.5 35.5PX (35.5) = = 2.3667 15 36 = 2.4 36 PX (36) = 15 2 36.5PX (36.5) = 36.5 = 4.8667 15 3 37 PX (37) = 37 = 7.4 15 a. C26 =


2 = 5.0667 15 38.5 38.5PX (38.5) = = 2.5667 15 39.5 = 2.6333 39.5PX (39.5) = 15 40 = 2.6667 40 PX (40) = 15 d. The mean of the sampling distribution of the sample mean is ∑ xPx ( x ) = 36.8333 which is exactly equal to the population mean: 38PX (38) = 38

1 ∑ xi = 36.8333 . This is the result expected from the Central Limit N Theorem. 7.65 The central limit theorem states that as the sample size increases, the sampling distribution of the sample mean tends toward the normal probability distribution, allowing use of the normal probability distribution for estimating population means. 450 − 420 ) = P(Z > 1.5) = .0668 100 25 400 − 420 450 − 420 b. P( 1.28) = .1, 1.28 = , x = 445.6 100 25 x − 420 d. P(Z < -1.28) = .1, -1.28 = , x = 394.4 100 25

7.66 a. P ( Z >

e. P( χ 2 (24) > 36.42) = .05, 36.42 =

24 s 2 , s = 123.1868 (100) 2

24 s 2 , s = 75.966 (100) 2 g. Smaller. A larger sample size would lead to a smaller standard error and the graph of the normal distribution would be tighter with less area in the tails.

f. P( χ 2 (24) < 13.85) = .05, 13.85 =

65 − 60 ) = P(Z > 1) = .1587 10 4 Xi − 60 b. P(Z < -1.28) = .1, -1.28 = , Xi = 53.6 10 4

7.67 a. P(Z >

159

160

th


c. P( χ 2 (3) > 6.25) = .1, 6.25 =

3s 2 , s = 14.4337 (10) 2

d. P( χ 2 (3) < .584) = .1, .584 =

3s 2 x , s = 4.4121 (10) 2

65 − 60 ) = P(Z > 1.0) = .1587 10 4 Use the binomial formula: P(X >2) = P(X = 3) + P(X = 4) C34 (.8413)3 (.1587)1 + C44 (.8413) 4 (.1587)0 = .87896

e. P(Z >

19 − 14.8 ) = P(Z > 2) = .0228 6.3 9 10.6 − 14.8 19 − 14.8 b. P(
7.68 a. P(Z >

d. P( χ 2 (8) > 13.36) = .1, 13.36 =

8s 2 , s = 8.1414 (6.3) 2

e. Smaller 1,500 − 1, 600 ) = P(Z > -1.00) = .8413 400 16 X − 1, 600 b. P(Z > 1.04) = .15, 1.04 = i , Xi = 1,704 400 16

7.69 a. P(Z >

c. P ( χ 2 (15) > 22.31) = .1, 22.31 =

15s 2 , s = 487.825 (400) 2

7.70 Let n = N, then X = μ x : E[∑ i =1 ( X i − X ) 2 ] = nσ 2 x − n N

σ

σ 2x N − n n N −1

= nσ 2 x −

N −n 2 σ x= N −1

Nσ x (n − 1) N −1 N −1 1 1 Nσ 2 x 2 2 Therefore, E[ ∑ ( X i − X ) ] = n − 1 E[∑ ( X i − X ) ] = N − 1 n −1 2

x

2

(nN − n − N + n) =

120 − 100 ) = P(Z > 2) = .0228 30 9 X − 100 , Xi = 91.57 b. P(Z < -.843) = .20, -.843 = i 30 9

7.71 a. P(Z >


c. P ( χ 2 (8) < 2.73) = .05, 2.73 =

8s 2 , s = 17.525 (30) 2

.7 − .8 ) = P(Z < -1.94) = .0262 (.8)(.2) / 60 b. Use the binomial formula: 6(.7) = 4.2, P(X ≤ 4) = 1 – P(X > 4) = 16(.8)5(.2)-(.8)6 = .3446 30, 000 − 29, 000 ) = P(Z > .61) = .2709 c. P(Z > 4, 000 6 30, 000 − 29, 000 d. (.8)P(Z > ) = (.8)P(Z > .25) = (.8)(.4013) = .3210 4, 000

7.72 a. P(Z <

15s 2 , s = 2.5701 (1.8) 2 Difference b. P(Z > 1.04) = .15, 1.04 = , Difference = .468 1.8 16 Difference c. P(Z > 1.96) = .025, 1.96 = , Difference = ±.882 1.8 16

7.73 a. P( χ 2 (15) > 30.58) = .01, 30.58 =

7.74 P(

(n − 1) s 2

σ2

> 20(2)) = P ( χ 2 ( 20 ) > 40) = .005

.6 − .5 ) = P(Z > 2) = .0228 (.5)(.5) /100 .4 − .5 .55 − .5 b. P(
7.75 a. P(Z >

7.76 10 < μ x < X + 10, -10 < X - μ x < 10 −10 10 P(
P(Z <

29.5 − 30 ) = P(Z < -1.54) = .0618 1.3 16

161

162

7.78

th


(.4)(.6) = .03098 250

a. σ x =

p − .4 , p = .3739 .03098 p − .4 , p = .4397 b. P(Z < 1.28) = .9, 1.28 = .03098 Difference , Difference = ±.0322 c. P(Z > 1.04) = .35, 1.04 = .03098

P(Z > -.843) = .8, -.843 =

7.79

7.80

.28 − .2 ) = P(Z < 3.46) = .9997 (.2)(.8) / 300 .28 − .4 b. P(Z < ) = P(Z < -4.24) ≈ .0000 (.4)(.6) / 300

a. P(Z >

a. P (

(n − 1) s 2

σ2

>

24(4, 000) 2 ) = P( χ 2 (24) > 8.82) = more than .99 (.9979 2 (6, 600)

exactly) (n − 1) s 2 24(8, 000) 2 b. P ( < ) = P ( χ 2 (24) < 35.62) = between .9 and .95 2 2 (6, 600) σ (.9354 exactly) 7.81

P(

(n − 1) s 2

σ2 exactly)

>

19(2.5) 2 ) = P( χ 2 (19) > 29.69) = between .05 and .1 (.0559 2 (2)

_newbold_ism_07.pdf

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