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Chapter 6
Continuous Random Variables and the
Normal Distribution
Section 6.
6.1 The probability distribution of a discrete random variable assigns probabilities to points while that of a continuous random variable assigns probabilities to intervals.
6.2 The probability that a continuous random variable x assumes a single value is always zero, that is, P ( x = a ) = 0.
6.3 Since P ( a ) = 0 and P ( b ) = 0 for a continuous random variable, P ( a < X < b ) = P ( a < X ≤ b ) = P ( a ≤ X < b ) = P ( a ≤ X ≤ b ). This means that the probability that X assumes a value in the interval a to b is the same whether or not the values a and b are included in the interval. This is true because P ( X = a ) = 0 and P ( X = b ) = 0.
6.4 The following are the three main characteristics of a normal distribution.
- The total area under a normal curve is 1.0.
- A normal curve is symmetric about the mean. Consequently, 50% of the total area under a normal distribution curve lies on the left side of the mean, and 50% lies on the right side of the mean.
- The tails of a normal distribution curve extend indefinitely in both directions without touching or crossing the horizontal axis. Although a normal curve never meets the horizontal axis, beyond
the points represented by μ − 3 σ and
μ + 3 σ it becomes so close to this axis
that the area under the curve beyond these points in both directions is very close to zero.
These three characteristics are described with graphs on page 271 of the text.
6.5 The standard normal distribution is a special case of the normal distribution. For the standard normal distribution, the value of the mean is equal to zero and the value of the standard deviation is 1. The units of the standard normal distribution curve are denoted by z and are called the z -values or z - scores. The z -values on the right side of the mean (which is zero) are positive and those on the left side are negative. A specific value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.
6.6 The parameters of the normal distribution are the mean μ and the standard deviation σ.
6.7 As its standard deviation decreases, the width of a normal distribution curve decreases and its height increases.
6.8 The width and height of a normal distribution do not change when its standard deviation remains the same but its mean increases.
6.9 For the standard normal distribution, the z value gives the distance between the mean and the point represented by z in terms of the standard deviation. The z values on the right side of the mean are positive and those on the left side are negative.
128 Chapter 6 Continuous Random Variables and the Normal Distribution
6.10 Area between μ − σ and μ + σis the area
between z = −1 and z = 1. Then P (−1 < z < 1) = P ( z < 1) − P ( z < −1) = .8413 − .1587 =.
6.11 Area between μ − 1.5σand μ + 1.5σ is the
area between z = −1.5 and z = 1.5. Then, P (−1.5 < z < 1.5) = P ( z < 1.5) − P ( z < −1.5) = .9332 − .0668 =.
6.12 Area within two standard deviations of the mean is: P (−2 < z < 2) = P ( z < 2) − P ( z < −2) = .9772 − .0228 =.
6.13 Area within 2.5 standard deviations of the mean is: P (−2.5 < z < 2.5) = P ( z < 2.5) − P ( z < −2.5) = .9938 − .0062 =.
6.14 Area within three standard deviations of the mean is: P (−3 < z < 3) = P ( z < 3) − P ( z < −3) = .9987 − .0013 =.
6.15 a. P (0 < z < 1.95) = P ( z < 1.95) − P ( z < 0) = .9744 − .5000 =. b. P (−2.05 < z < 0) = P ( z < 0) − P ( z < −2.05) = .5000 − .0202 =. c. P (1.15 < z < 2.37) = P ( z < 2.37) − P ( z < 1.15) = .9911 − .8749 =. d. P (−2.88 ≤ z ≤ −1.53) = P ( z ≤ −1.53) − P ( z ≤ −2.88) = .0630 − .0020 =. e. P (−1.67 ≤ z ≤ 2.24) = P ( z ≤ 2.24) − P ( z ≤ −1.67) = .9875 − .0475 =.
6.16 a. P (0 ≤ z ≤ 2.34) = P ( z ≤ 2.34) − P ( z ≤ 0) = .9904 − .5000 =. b. P (−2.58 < z < 0) = P ( z < 0) − P ( z < −2.58) = .5000 − .0049 =.
c. P (.84 ≤ z ≤ 1.95) = P ( z ≤ 1.95) − P ( z ≤ .84) = .9744 − .7995 =. d. P (−2.49 < z < −.57) = P ( z < −.57) − P ( z < −2.49) = .2843 − .0064 =. e. P (−2.15 < z < 1.87) = P ( z < 1.87) − P ( z < −2.15) = .9693 − .0158 =.
6.17 a. P ( z > 1.36) = 1 − P ( z ≤ 1.36) = 1 − .9131 =. b. P ( z < −1.97) =. c. P ( z > −2.05) = 1 − P ( z ≤ −2.05) = 1 − .0202 =. d. P ( z < 1.76) =.
6.18 a. P ( z > 1.43) = 1 − P ( z ≤ 1.43) = 1 − .9236 =. b. P ( z < −1.65) =. c. P ( z > −.65) = 1 − P ( z ≤ −.65) = 1 − .2578 =. d. P ( z < .89) =.
6.19 a. P (0 < z < 4.28) = P ( z < 4.28) − P ( z < 0) = 1 −. = .5 approximately b. P (−3.75 ≤ z ≤ 0) = P ( z ≤ 0) − P ( z ≤ −3.75) = .5 − 0 = .5 approximately c. P ( z > 7.43) = 1 − P ( z ≤ 7.43) = 1 − 1 = 0 approximately d. P ( z < −4.69) = 0 approximately
6.20 a. P (0 ≤ z ≤ 3.94) = P ( z ≤ 3.94) − P ( z < 0) = 1 −. = .5 approximately b. P (−5.16 < z < 0) = P ( z < 0) − P ( z < −5.16) = .5 − 0 = .5 approximately c. P ( z > 5.42) = 1 − P ( z ≤ 5.42) = 1 − 1 = 0 approximately d. P ( z < −3.68) = 0 approximately
6.21 a. P (−1.83 ≤ z ≤ 2.57) = P ( z ≤ 2.57) − P ( z ≤ −1.83) = .9949 − .0336 =.
130 Chapter 6 Continuous Random Variables and the Normal Distribution
For x = 16.54: z = ( x − μ )/ σ = (16.54 − 12)/2 = 2.
Section 6.4 Determining the z and x Values When an Area Under the Normal Distribution is Known 133
6.48 μ = 7.59 and σ =. a. For x = 6: z = ( x − μ )/ σ = (6 − 7.59)/.73 = −2. P ( x < 6) =. P ( z < −2.18) = .0146 or 1.46% b. For x = 7: z = ( x − μ )/ σ = (7 − 7.59)/.73 = −. For x = 8: z = ( x − μ )/ σ = (8 − 7.59)/.73 =. P (7 < x < 8) = P (−.81 < z < .56) = P ( z < .56) − P ( z < −.81) = .7123 − .2090 =. c. For x = 9: z = ( x − μ )/ σ = (9 − 7.59)/.73 = 1. P ( x ≥ 9) = P ( z ≥ 1.93) = 1 − P ( z < 1.93) = 1 − .9732 = .0268 or 2.68%
6.49 μ = 54.3 pounds and σ = 14.5 pounds
a. For x = 10: z = ( x − μ )/ σ = (10 − 54.3)/14.5 = −3. P ( x < 10) = P ( z < −3.06) = .0011 or .11% b. For x = 40: z = ( x − μ )/ σ = (40 − 54.3)/14.5 = −. For x = 60: z = ( x − μ )/ σ = (60 − 54.3)/14.5 =. P (40 < x < 60) = P (−.99 < z < .39) = P ( z < .39) − P ( z < −.99) = .6517 − .1611 = .4906 or 49.06% c. For x = 90: z = ( x − μ )/ σ = (90 − 54.3)/14.5 = 2. P ( x > 90) = P ( z > 2.46) = 1 − P ( z ≤ 2.46) = 1 − .9931 = .0069 or .69% d. For x = 50: z = ( x − μ )/ σ = (50 − 54.3)/14.5 = −. For x = 70: z = ( x − μ )/ σ = (70 − 54.3)/14.5 = 1. P (50 < x < 70) = P (−.30 < z < 1.08) = P ( z < 1.08) − P ( z < −.30) = .8599 − .3821 = .4778 or 47.78%
6.50 μ = 15 minutes and σ = 2.4 minutes a. For x = 20: z = ( x − μ )/ σ = (20 − 15)/2.4 = 2. P ( x > 20) = P ( z > 2.08) = 1 − P ( z ≤ 2.08) = 1 − .9812 = .0188 or 1.88% b. For x = 25: z = ( x − μ )/ σ = (25 − 15)/2.4 = 4. P ( x > 25) = P ( z > 4.17) = 1 − P ( z ≤ 4.17) = 1 − 1 = 0 approximately Although it is possible that a given car may take more than 25 minutes for oil and lube service; however, this probability is very close to zero.
6.51 μ = 3.0 inches and σ = .009 inch For x = 2.98: z = ( x − μ )/ σ = (2.98 − 3.0)/.009 = −2. For x = 3.02: z = ( x − μ )/ σ = (3.02 − 3.0)/.009 = 2. P ( x < 2.98) + P ( x > 3.02) = 1 − [ P (2.98 ≤ x ≤ 3.02)] = 1 − [ P (−2.22 ≤ z ≤ 2.22)] = 1 − [ P ( z ≤ 2.22) − P ( z ≤ −2.22)] = 1 − [.9868 − .0132] = 1 − .9736 = .0264 or 2.64%
6.52 μ = 5.75 ounces and σ = .11 ounce For x = 5.5: z = ( x − μ )/ σ = (5.5 − 5.75)/.11 = −2. For x = 6.0: z = ( x − μ )/ σ = (6.0 − 5.75)/.11 = 2. P ( x < 5.5) + P ( x > 6.0) = 1 − [ P (5.5 ≤ x ≤ 6.0)] = 1 − [ P (−2.27 ≤ z ≤ 2.27)] = 1 − [ P ( z ≤ 2.27) − P ( z ≤ −2.27)] = 1 − [.9884 − .0116] = 1 − .9768 = .0232 or 2.32%
Section 6.
6.53 a. z = 2.
b. z = −2.02 approximately c. z = −.37 approximately d. z = 1.02 approximately
6.54 a. z = .51 approximately b. z = −.75 approximately c. z = −.82 approximately d. z = 1.25 approximately
134 Chapter 6 Continuous Random Variables and the Normal Distribution
6.55 a. z = 1.65 approximately
b. z = −1. c. z = −2.33 approximately d. z = 2.58 approximately
6.56 a. z = 1.
b. z = −1.65 approximately c. z = −3.08 approximately d. z = 2.33 approximately
6.57 μ = 200 and σ = 25 a. z =. x = μ + zσ = 200 + (.34)(25) = 208. b. z = 1. x = μ + zσ = 200 + (1.65)(25) = 241. c. z = −. x = μ + zσ = 200 + (−.86)(25) = 178. d. z = −2. x = μ + zσ = 200 + (−2.17)(25) = 145. e. z = −1. x = μ + zσ = 200 + (−1.67)(25) = 158. f. z = 2. x = μ + zσ = 200 + (2.05)(25) = 251.
6.58 μ = 550 and σ = 75 a. z = −1. x = μ + zσ = 550 + (−1.96)(75) = 403 b. z = −1. x = μ + zσ = 550 + (−1.51)(75) = 436. c. z = 1. x = μ + zσ = 550 + (1.92)(75) = 694 d. z = 1. x = μ + zσ = 550 + (1.75)(75) = 681. e. z = −1. x = μ + zσ = 550 + (−1.88)(75) = 409 f. z = 1. x = μ + zσ = 550 + (1.34)(75) = 650.
6.59 μ = 15 minutes and σ = 2.4 minutes Let x denote the time to service a randomly chosen car. We are to find x so that the area in the right tail of the normal distribution curve is .05. Thus, z = 1.65 and x = μ + zσ = 15 + (1.65)(2.4) = 18.96 minutes.
The maximum guaranteed waiting time should be approximately 19 minutes.
6.60 μ = $95 and σ = $ Let x denote the amount spent by a randomly chosen customer on a visit to this store. We are to find x such that the area in the right tail of the normal distribution curve is .10. Thus, z = 1.28 and x = μ + zσ = 95 + (1.28)(20) = $120.60. So, a required minimum purchase of $121 would meet the condition.
6.61 μ = 1650 kwh and σ = 320 kwh Let x denote the amount of electric consumption during the winter by a randomly selected customer. We are to find x such that the area to the left of x in the normal distribution curve is .90. Thus, z = 1.28 and x = μ + zσ = 1650 + (1.28)(320) = 2059.6 kwh. Bill Johnson’s monthly electric consumption is approximately 2060 kwh.
6.62 μ = 70 months and σ = 8 months Let x denote the warranty period. a. We are to find x such that the area in the left tail of the normal distribution curve is .01. Thus, z = −2.33 and x = μ + zσ = 70 + (−2.33)(8) = 51.36 months. The warranty period should be about 51 months. b. We are to find x such that the area in the left tail of the normal distribution curve is .05. Thus, z = −1.65 and x = μ + zσ = 70 + (−1.65)(8) = 56.8 months. The warranty period should be about 56 months.
6.63 σ = $36.35 and P ( x ≥ 184.52) =. The area to the left of x = 184.52 is 1 – .20 = .80 and z = .84 approximately. Then, we have x = μ + zσ ⇒ μ = x − zσ = 184.52 − (.84)(36.35) = $153.99. The mean price of all college textbooks is approximately $153.99.
6.64 σ = .35 and P ( x ≥ 64) =. The area to the left of x = 64 is 1 – .95 =. and z = −1.65 approximately. Then, μ = x − zσ = 64 − (−1.65)(.35) = 64. ounces. The mean amount of detergent poured by this machine into jugs should be approximately 65 ounces.
For x = 330.5: z = (330.5 − 312)/13.76565149 = 1.
Supplementary Exercises
- b. For x = 23: z = ( x − μ )/ σ = (23 − 20)/4 =.
- For x = 26: z = ( x − μ )/ σ = (26 − 20)/4 = 1.
- P (23 ≤ x ≤ 26) = P (.75 ≤ z ≤ 1.50) = P ( z ≤ 1.50) − P ( z ≤ .75) = .9332 − .7734 =.
- c. For x = 9.5: z = ( x − μ )/ σ = (9.5 − 20)/4 = −2.
- For x = 17: z = ( x − μ )/ σ = (17 − 20)/4 = −.
- P (9.5 < x < 17) = P (−2.63 < z < −.75) = P ( z < −.75) − P ( z < −2.63) = .2266 − .0043 =.
- 6.30 μ = 12 and σ =
- a. For x = 7.76: z = ( x − μ )/ σ = (7.76 − 12)/2 = −2.
- For x = 12: z = ( x − μ )/ σ = (12 − 12)/2 =
- P (7.76 < x < 12) = P (−2.12 < z < 0) = P ( z < 0) − P ( z < −2.12) = .5000 − .0170 =.
- b. For x = 14.48: z = ( x − μ )/ σ = (14.48 − 12)/2 = 1.
- For x = 16.54: z = ( x − μ )/ σ = (16.54 − 12)/2 = 2.
- P (14.48 < x < 16.54) = P (1.24 < z < 2.27) = P ( z < 2.27) − P ( z < 1.24) = .9884 − .8925 =.
- c. For x = 8.22: z = ( x − μ )/ σ = (8.22 − 12)/2 = −1.
- For x = 10.06: z = ( x − μ )/ σ = (10.06 − 12)/2 = −.
- P (8.22 ≤ x ≤ 10.06) = P (−1.89 ≤ z ≤ −.97) = P ( z ≤ −.97) − P ( z ≤ −1.89) = .1660 − .0294 =.
- 6.31 μ = 55 and σ =
- a. For x = 58: z = ( x − μ )/ σ = (58 − 55)/7 =.
- P ( x > 58) = P ( z > .43) = 1 − P ( z ≤ .43) = 1 − .6664 =.
- b. For x = 43: z = ( x − μ )/ σ = (43 − 55)/7 = −1.
- P ( x > 43) = P ( z > −1.71) = 1 − P ( z ≤ −1.71) = 1 − .0436 =.
- c. For x = 68: z = ( x − μ )/ σ = (68 − 55)/7 = 1.
- P ( x < 68) = P ( z < 1.86) =.
- d. For x = 22: z = ( x − μ )/ σ = (22 − 55)/7 = −4.
- 6.32 μ = 18.3 and σ = 3. P ( x < 22) = P ( z < −4.71) = 0 approximately
- a. For x = 10.9: z = ( x − μ )/ σ = (10.9 − 18.3)/3.4 = −2.
- P ( x < 10.9) = P ( z < −2.18) =.
- b. For x = 14: z = ( x − μ )/ σ = (14 − 18.3)/3.4 = −1.
- P ( x > 14) = P ( z > −1.26) = 1 − P ( z ≤ −1.26) = 1 − .1038 =.
- c. For x = 22.7: z = ( x − μ )/ σ = (22.7 − 18.3)/3.4 = 1.
- P ( x < 22.7) = P ( z < 1.29) =.
- d. For x = 29.2: z = ( x − μ )/ σ = (29.2 − 18.3)/3.4 = 3.
- P ( x > 29.2) = P ( z > 3.21) = 1 − P ( z ≤ 3.21) = 1 − .9993 =.
- 6.33 μ = 25 and σ =
- a. For x = 29: z = ( x − μ )/ σ = (29 − 25)/6 =.
- For x = 36: z = ( x − μ )/ σ = (36 − 25)/6 = 1.
- P (29 < x < 36) = P (.67 < z < 1.83) = P ( z < 1.83) − P ( z < .67) = .9664 − .7486 =.
- b. For x = 22: z = ( x − μ )/ σ = (22 − 25)/6 = −.
- For x = 35: z = ( x − μ )/ σ = (35 − 25)/6 = 1.
- P (22 < x < 35) = P (−.50 < z < 1.67) = P ( z < 1.67) − P ( z < −.50) = .9525 − .3085 =.
- 6.34 μ = 117.6 and σ = 14.
- a. For x = 77.9: z = ( x − μ )/ σ = (77.9 − 117.6)/14.6 = −2.
- For x = 98.3: z = ( x − μ )/ σ = (98.3 − 117.6)/14.6 = −1.
- P (77.9 < x < 98.3) = P (−2.72 < z < −1.32) = P ( z < −1.32) − P ( z < −2.72) = .0934 − .0033 =.
- Section 6.3 Applications of the Normal Distribution
- b. For x = 85.3: z = ( x − μ )/ σ = (85.3 − 117.6)/14.6 = −2.
- For x = 142.6: z = ( x − μ )/ σ = (142.6 − 117.6)/14.6 = 1.
- P (85.3 < x < 142.6) = P (−2.21 < z < 1.71) = P ( z < 1.71) − P ( z < −2.21) = .9564 − .0136 =.
- 6.35 μ = 80 and σ =
- a. For x = 69: z = ( x − μ )/ σ = (69 − 80)/12 = −.
- P ( x > 69) = P ( z > −.92) = 1 − P ( z ≤ −.92) = 1 − .1788 =.
- b. For x = 73: z = ( x − μ )/ σ = (73 − 80)/12 = −.
- P ( x < 73) = P ( z < −.58) =.
- c. For x = 101: z = ( x − μ )/ σ = (101 − 80)/12 = 1.
- P ( x > 101) = P ( z > 1.75) = 1 − P ( z ≤ 1.75) = 1 − .9599 =.
- d. For x = 87: z = ( x − μ )/ σ = (87 − 80)/12 =.
- P ( x < 87) = P ( z < .58) =.
- 6.36 μ = 65 and σ =
- a. For x = 45: z = ( x − μ )/ σ = (45 − 65)/15 = −1.
- P ( x < 45) = P ( z < −1.33) =.
- b. For x = 79: z = ( x − μ )/ σ = (79 − 65)/15 =.
- P ( x > 79) = P ( z > .93) = 1 − P ( z ≤ .93) = 1 − .8238 =.
- c. For x = 54: z = ( x − μ )/ σ = (54 − 65)/15 = −.
- P ( x > 54) = P ( z > −.73) = 1 − P ( z ≤ −.73) = 1 − .2327 =.
- d. For x = 70: z = ( x − μ )/ σ = (70 − 65)/15 =.
- P ( x < 70) = P ( z < .33) =.
- Section 6.
- a. For x = 160: z = ( x − μ )/ σ = (160 − 190)/21 = −1. 6.37 μ = 190 minutes and σ = 21 minutes
- b. For x = 215: z = ( x − μ )/ σ = (215 − 190)/21 = 1. P ( x < 160) = P ( z < −1.43) = .0764 or 7.64%
- For x = 245: z = ( x − μ )/ σ = (245 − 190)/21 = 2.
- 6.38 μ = $297 and σ = $74. P (215 ≤ x ≤ 245) = P (1.19 ≤ z ≤ 2.62) = P ( z ≤ 2.62) − P ( z ≤ 1.19) = .9956 − .8830 = .1126 or 11.26%
- a. For x = 400: z = ( x − μ )/ σ = (400 − 297)/74.42 = 1.
- b. For x = 200: z = ( x − μ )/ σ = (200 − 297)/74.42 = −1. P ( x > 400) = P ( z > 1.38) = 1 − P ( z ≤ 1.38) = 1 − .9162 = .0838 or 8.38%
- For x = 340: z = ( x − μ )/ σ = (340 − 297)/74.42 =.
- 6.39 μ = 616.13 and σ = $ P (200 < x < 340) = P (−1.30 < z < .58) = P ( z < .58) − P ( z < −1.30) = .7190 − .0968 = .6222 or 62.22%
- a. For x = 450: z = ( x − μ )/ σ = (450 − 616.13)/120 = −1.
- b. For x = 500: z = ( x − μ )/ σ = (500 − 616.13)/120 = −. P ( x < 450) = P ( z < −1.38) = .0838 or 8.38%
- For x = 750: z = ( x − μ )/ σ = (750 − 616.13)/120 = 1. - Section 6.5 The Normal Approximation to the Binomial Distribution
- Section 6.
- 6.66 a. From Table I of Appendix C, for n = 20 and p = .60, P ( x = 14) =. nq > 5.
- b. μ = np = 20(.60) = 12 and σ = npq = 20(.60)(.40)= 2.
- For x = 13.5: z = (13.5 – 12)/2.19089023 =.
- For x = 14.5: z = (14.5 – 12)/2.19089023 = 1.
- P (13.5 ≤ x ≤ 14.5) = P (.68 ≤ z ≤ 1.14) = P ( z ≤ 1.14) − P ( z ≤ .68) = .8729 − .7517 =.
- The difference between this approximation and the exact probability is .1244 – .1212 =. - = .1200 + .1511 + .1612 + .1465 + .1140 + .0760 =. P (8 ≤ x ≤ 13) = P ( x = 8) + P ( x = 9) + P ( x = 10) + P ( x = 11) + P ( x = 12) + P ( x = 13)
- b. μ = np = 25(.40) = 10 and σ = npq = 25(.40)(.60)= 2. - For x = 7.5: z = (7.5 – 10)/2.44948974 = −1. - For x = 13.5: z = (13.5 – 10)/2.44948974 = 1. - P (7.5 ≤ x ≤ 13.5) = P (−1.02 ≤ z ≤ 1.43) = P ( z ≤ 1.43) − P ( z ≤ −1.02) = .9236 − .1539 =. - The difference between this approximation and the exact probability is .7697 – .7688 =.
- 6.68 a. μ = np = 80(.50) = 40 and σ = npq = 80(.50)(.50)= 4.
- b. For x = 41.5: z = (41.5 – 40)/4.47213596 =. - P ( x ≥ 41.5) = P ( z ≥ .34) = 1 − P ( z < .34) = 1 − .6331 =.
- c. For x = 40.5: z = (40.5 – 40)/4.47213596 =. - For x = 48.5: z = (48.5 – 40)/4.47213596 = 1. - P (40.5 ≤ x ≤ 48.5) = P (.11 ≤ z ≤ 1.90) = P ( z ≤ 1.90) − P ( z ≤ .11) = .9713 − .5438 =.
- 6.69 a. μ = np = 120(.60) = 72 and σ = npq = 120(.60)(.40)= 5.
- b. For x = 69.5: z = (69.5 – 72)/5.36656315 = −. - P ( x ≤ 69.5) = P ( z ≤ −.47) =.
- c. For x = 66.5: z = (66.5 – 72)/5.36656315 = −1. - For x = 73.5: z = (73.5 – 72)/5.36656315 =. - P (66.5 ≤ x ≤ 73.5) = P (−1.02 ≤ z ≤ .28) = P ( z ≤ .28) − P ( z ≤ −1.02) = .6103 − .1539 =.
- 6.70 a. μ = np = 140(.45) = 63 and σ = npq = 140(.45)(.55)= 5. - For x = 66.5: z = (66.5 – 63)/5.88642506 =. - For x = 67.5: z = (67.5 – 63)/5.88642506 =. - P (66.5 ≤ x ≤ 67.5) = P (.59 ≤ z ≤ .76) = P ( z ≤ .76) − P ( z ≤ .59) = .7764 − .7224 =.
- b. μ = np = 100(.55) = 55 and σ = npq = 100(.55)(.45)= 4. - For x = 51.5: z = (51.5 – 55)/4.97493719 = −. - For x = 60.5: z = (60.5 – 55)/4.97493719 = 1. - P (51.5 ≤ x ≤ 60.5) = P (−.70 ≤ z ≤ 1.11) = P ( z ≤ 1.11) − P ( z ≤ −.70) = .8665 − .2420 =.
- c. μ = np = 90(.42) = 37.8and σ = npq = 90(.42)(.58)= 4. - For x = 39.5: z = (39.5 – 37.80)/4.68230712 =. - P ( x ≥ 39.5) = P ( z ≥ .36) = 1 − P ( z ≤ .36) = 1 − .6406 =.
- d. μ = np = 104(.75) = 78 and σ = npq = 104(.75)(.25)= 4. 136 Chapter 6 Continuous Random Variables and the Normal Distribution
- For x = 72.5: z = (72.5 – 78)/4.41588043 = −1.
- P ( x ≤ 72.5) = P ( z ≤ −1.25) =.
- 6.71 a. μ = np = 70(.30) = 21 and σ = npq = 70(.30)(.70)= 3. - For x = 17.5: z = (17.5 – 21)/3.83405790 = −. - For x = 18.5: z = (18.5 – 21)/3.83405790 = −. - P (17.5 ≤ x ≤ 18.5) = P (−.91 ≤ z ≤ −.65) = P ( z ≤ −.65) − P ( z ≤ −.91) = .2578 − .1814 =.
- b. μ = np = 200(.70) = 140 and σ = npq = 200(.70)(.30)= 6.
- For x = 132.5: z = (132.5 – 140)/6.48074070 = −1.
- For x = 145.5: z = (145.5 – 140)/6.48074070 =.
- P (132.5 ≤ x ≤ 145.5) = P (−1.16 ≤ z ≤ .85) = P ( z ≤ .85) − P ( z ≤ −1.16) = .8023 − .1230 =.
- c. μ = np = 85(.40) = 34 and σ = npq = 85(.40)(.60)= 4.
- For x = 29.5: z = (29.5 – 34)/4.51663592 = −1.
- P ( x ≥ 29.5) = P ( z ≥ −1.00) = 1 − P ( z < −1.00) = 1 − .1587 =.
- d. μ = np = 150(.38) = 57 and σ = npq = 150(.38)(.62)= 5.
- For x = 62.5: z = (62.5 – 57)/5.94474558 =.
- P ( x ≤ 62.5) = P ( z ≤ .93) =.
- 6.72 μ = np = 400(.0544) = 21.76and σ = npq = 400 .0544( ) (.9456 ) =4.
- For x = 22.5: z = (22.5 – 21.76)/4.536105819 =.
- For x = 27.5: z = (27.5 – 21.76)/4.536105819 = 1.
- P (22.5 ≤ x ≤ 27.5) = P (.16 ≤ z ≤ 1.27) = P ( z ≤ 1.27) − P ( z ≤ .16) = .8980 − .5636 =.
- 6.73 μ = np = 800(.39) = 312 and σ = npq = 800(.39)(.61) =13.
- For x = 330.5: z = (330.5 − 312)/13.76565149 = 1.
- P ( x > 330) = P ( z > 1.34) = 1 − P ( z ≤ 1.34) = 1 − .9099 =.
- 6.74 μ = np = 400(.49) = 196 and σ = npq = 400(.49)(.51) =9.
- a. For x = 204.5: z = (204.5 − 196)/9.9979998 =.
- For x = 205.5: z = (205.5 − 196)/9.9979998 =.
- P (204.5 ≤ x ≤ 205.5) = P (.85 ≤ z ≤ .95) = P ( z ≤ .95) − P ( z ≤ .85) = .8289 − .8023 =.
- b. For x = 189.5: z = (189.5 − 196)/9.9979998 = –.
- P ( x ≤ 189.5) = P ( z ≤ –.65) =.
- c. For x = 209.5: z = (209.5 − 196)/9.9979998 = 1.
- For x = 220.5: z = (220.5 − 196)/9.9979998 = 2.
- P (209.5 ≤ x ≤ 220.5) = P (1.35 ≤ z ≤ 2.45) = P ( z ≤ 2.45) − P ( z ≤ 1.35) = .9929 − .9115 =.
- 6.75 μ = np = 600(.07) = 42 and σ = npq = 600(.07)(.93) =6.
- a. For x = 44.5: z = (44.5 – 42)/6.249799997 =.
- For x = 45.5: z = (45.5 – 42)/6.249799997 =.
- P (44.5 ≤ x ≤ 45.5) = P (.40 ≤ z ≤ .56) = P ( z ≤ .56) − P ( z ≤ .40) = .7123 − .6554 =.
- b. For x = 53.5: z = (53.5 – 42)/6.249799997 = 1.
- P ( x ≤ 53.5) = P ( z ≤ 1.84) =.
- c. For x = 29.5: z = (29.5 – 42)/6.249799997 = –2.
- For x = 50.5: z = (50.5 – 42)/6.249799997 = 1.
- P (29.5 ≤ x ≤ 50.5) = P (−2.00 ≤ z ≤ 1.36) = P ( z ≤ 1.36) − P ( z ≤ −2.00) = .9131 − .0228 =.
- Chapter 6 Supplementary Exercises
- 6.76 μ = np = 500(.92) = 460 and σ = npq = 200(.92)(.08) =6.
- a. For x = 444.5: z = (444.5 − 460)/6.066300355 = −2.
- For x = 445.5: z = (445.5 − 460)/6.066300355 = −2.
- P (444.5 ≤ x ≤ 445.5) = P (−2.56 ≤ z ≤ −2.39) = P ( z ≤ −2.39) − P ( z ≤ −2.56) = .0084 − .0052 =.
- b. For x = 449.5: z = (449.5 − 460)/6.066300355 = −1.
- P ( x ≥ 449.5) = P ( z ≥ −1.73) = 1 − P ( z < −1.73) = 1 − .0418 =.
- c. For x = 439.5: z = (439.5 − 460)/6.066300355 = −3.
- For x = 470.5: z = (470.5 − 460)/6.066300355 = 1.
- P (439.5 ≤ x ≤ 470.5) = P (−3.38 ≤ z ≤ 1.73) = P ( z ≤ 1.73) − P ( z ≤ −3.38) = .9582 − .0004 =.
- 6.77 μ = np = 100(.80) = 80 and σ = npq = 100(.80)(.20)=
- a. For x = 74.5: z = (74.5 – 80)/4 = −1.
- For x = 75.5: z = (75.5 – 80)/4 = −1.
- P (74.5 ≤ x ≤ 75.5) = P (−1.38 ≤ z ≤ −1.13) = P ( z ≤ −1.13) − P ( z ≤ −1.38) = .1292 − .0838 =.
- b. For x = 73.5: z = (73.5 – 80)/4 = −1.
- P ( x ≤ 73.5) = P ( z ≤ −1.63) =.
- c. For x = 73.5: z = (73.5 – 80)/4 = −1.
- For x = 85.5: z = (85.5 – 80)/4 = 1.
- P (73.5 ≤ x ≤ 85.5) = P (−1.63 ≤ z ≤ 1.38) = P ( z ≤ 1.38) − P ( z ≤ −1.63) = .9162 − .0516 =.
- 6.78 μ = np = 500(.05) = 25 and σ = npq = 500(.05)(.95)= 4.
- a. For x = 28.5: z = (28.5 – 25)/4.87339717 =.
- For x = 29.5: z = (29.5 – 25)/4.87339717 =.
- P (28.5 ≤ x ≤ 29.5) = P (.72 ≤ z ≤ .92) = P ( z ≤ .92) − P ( z ≤ .72) = .8212 − .7642 =.
- b. For x = 26.5: z = (26.5 – 25)/4.87339717 =.
- P ( x ≥ 26.5) = P ( z ≥ .31) = 1 − P ( z < .31) = 1 − .6217 =.
- c. For x = 14.5: z = (14.5 – 25)/4.87339717 = −2.
- For x = 22.5: z = (22.5 – 25)/4.87339717 = −.
- P (14.5 ≤ x ≤ 22.5) = P (−2.15 ≤ z ≤ −.51) = P ( z ≤ −.51) − P ( z ≤ −2.15) = .3050 − .0158 =.
- 6.79 μ = np = 100(.05) = 5 and σ = npq = 100(.05)(.95)= 2.
- a. For x = 6.5: z = (6.5 – 5)/2.17944947 =.
- P (shipment is accepted) = P ( x ≤ 6.5) = P ( z ≤ .69) =.
- b. P (shipment is not accepted) = 1 – P (shipment is accepted) = 1 – .7549 =.
- a. For x = 3: z = (3 – 8)/2 = −2. 6.80 μ = 8 minutes and σ = 2 minutes
- P ( x < 3) = P ( z < −2.50) =.
- b. For x = 10: z = (10 – 8)/2 = 1.
- For x = 13: z = (13 – 8)/2 = 2.
- c. For x = 6: z = (6 – 8)/2 = −1. P (10 ≤ x ≤ 13) = P (1.00 ≤ z ≤ 2.50) = P ( z ≤ 2.50) − P ( z ≤ 1.00) = .9938 − .8413 = .1525 or 15.25%
- For x = 12: z = (12 – 8)/2 = 2.
Chapter 6 Supplementary Exercises 139
6.86 μ = 9.125 inches and σ = .06 inch For x = 9: z = (9 – 9.125)/.06 = −2. For x = 9.25: z = (9.25 – 9.125)/.06 = 2. P ( x < 9) + P ( x > 9.25) = 1 − [ P (9 ≤ x ≤ 9.25)] = 1 − [ P (−2.08 ≤ z ≤ 2.08)] = 1 − [ P ( z ≤ 2.08) − P ( z ≤ −2.08)] = 1 − [.9812 − .0188] = 1 −. = .0376 or 3.76%
6.87 μ = $98,620 and σ = $18, Being in the top 10% is the same as being greater than 90%. Thus, the area to the left of x is .90, which gives z ≈ 1.28. Then, x = μ + zσ = 98,620 + (1.28)(18,000) = $121,660. An actuary would have to be paid $121,660 in order to be in the highest paid 10% of all actuaries.
6.88 μ = np = 100(.80) = 80 and σ = npq = 100(.80)(.20)= 4
a. For x = 84.5: z = (84.5 – 80)/4 = 1. For x = 85.5: z = (85.5 – 80)/4 = 1.
P (23 ≤ x ≤ 26) = P (.75 ≤ z ≤ 1.50) = P ( z ≤ 1.50) − P ( z ≤ .75) = .9332 − .7734 =.
b. For x = 74.5: z = (74.5 – 80)/4 = −1. P ( x ≤ 74.5) = P ( z ≤ −1.38) =. c. For x = 74.5: z = (74.5 – 80)/4 = −1. For x = 87.5: z = (87.5 – 80)/4 = 1. P (74.5 ≤ x ≤ 87.5) = P (−1.38 ≤ z ≤ 1.88) = P ( z ≤ 1.88) − P ( z ≤ −1.38) = .9699 − .0838 =. d. For x = 71.5: z = (71.5 – 80)/4 = −2. For x = 77.5: z = (77.5 – 80) / 4 = −. P (71.5 ≤ x ≤ 77.5) = P (−2.13 ≤ z ≤ −.63) = P ( z ≤ −.63) − P ( z ≤ −2.13) = .2643 − .0166 =.
6.89 μ = np = 200(.80) = 160 and σ = npq = 200(.80)(.20)= 5.
a. For x = 149.5: z = (149.5 – 160)/5.65685425 = −1. For x = 150.5: z = (150.5 – 160)/5.65685425 = −1. P (149.5 ≤ x ≤ 150.5) = P (−1.86 ≤ z ≤ −1.68) = P ( z ≤ −1.68) − P ( z ≤ −1.86) = .0465 −. =. b. For x = 169.5: z = (169.5 – 160)/5.65685425 = 1. P ( x ≥ 169.5) = P ( z ≥ 1.68) = 1 − P ( z ≤ 1.68) = 1 − .9535 =. c. For x = 165.5: z = (165.5 – 160)/5.65685425 =. P ( x ≤ 165.5) = P ( z ≤ .97) =. d. For x = 163.5: z = (163.5 – 160)/5.65685425 =. For x = 172.5: z = (172.5 – 160)/5.65685425 = 2. P (163.5 ≤ x ≤ 172.5) = P (.62 ≤ z ≤ 2.21) = P ( z ≤ 2.21) − P ( z ≤ .62) = .9864 − .7324 =.
6.90 σ = $350 and P ( x > 2500) =. The area to the left of x = 2500 is 1 – .15 = .85 and z = 1.04 approximately. Then, μ = x − zσ = 2500 − (1.04)(350) = $2136. Thus, the mean monthly mortgage is approximately $2136.
6.91 σ = .18 ounce and P ( x > 16)=. The area to the left of x = 16 is 1 – .90 = .10 and z = –1.28 approximately. Then, μ = x − zσ = 16 − (–1.28)(.18) = 16.23 ounces. Thus, the mean amount of ice cream put into these cartons by this machine should be approximately 16.23 ounces.
140 Chapter 6 Continuous Random Variables and the Normal Distribution
6.92 a. If $3500 = $1000 + $12 , x where x is depth in feet, then x =208.33. Hence, Company B charges more for depths of more than 208.33 ft. μ = 250 and σ = 40 For x = 208.33: z = (208.33 – 250)/40 = –1. P ( x > 208.33) = P ( z > –1.04) = 1 − P ( z ≤ –1.04) = 1 − .1492 =. The probability that Company B charges more than Company A to drill a well is .8508. b. μ = 250, so the mean amount charged by Company B is $1000 + $12 250⋅ ft = $4000.
6.93 μ = 253 feet and σ = 8. The simplest solution to this exercise is obtained by using complementary events; i.e., P (at least one throw is 270 feet or longer) = 1 – P (all three throws are less than 270 feet). First, we find P (any one throw is less than 270 feet). For x = 270: z = (270 – 253)/8.4 = 2. P ( x < 270) = P ( z < 2.02) =. Since the three throws are independent, P (all three throws are less than 270 feet ) = (.9783) 3 = .9363. Then, P (at least one throw is 270 feet or longer) = 1 – .9363 = .0637.
6.94 μ = 45,000 and σ = 2000 First, we find the probability that one tire lasts at least 46,000 miles. For x = 46,000: z = (46,000 – 45,000)/2000 =. P ( x ≥ 46,000) = P ( z ≥ .50) = 1 − P ( z ≤ .50) = 1 − .6915 =. So, the probability of one tire lasting at least 46,000 miles is .3085. Then, P (all four tires last more than 46,000 miles) = (.3085)^4 = .0091.
6.95 Plant A: μ = 20 and σ = 2 For x = 18: z = (18 – 20)/2 = −1. For x = 22: z = (22 – 20)/2 = 1. P (18 ≤ x ≤ 22) = P (−1.00 ≤ z ≤ 1.00) = P ( z ≤ 1.00) − P ( z ≤ −1.00) = .8413 − .1587 =. Plant B: μ = 19 and σ = 1 For x = 18: z = (18 – 19)/1 = −1. For x = 22: z = (22 – 19)/1 = 3. P (18 ≤ x ≤ 22) = P (−1.00 ≤ z ≤ 3.00) = P ( z ≤ 3.00) − P ( z ≤ −1.00) = .9987 − .1587 =. Plant B produces the greater proportion of pints that contain between 18% and 22% air.
6.96 μ = 0 and σ = 2 mph
a. Let x be the error of these estimates in mph. For x = 5: z = (5 – 0)/2 = 2.
P (9.5 < x < 17) = P (−2.63 < z < −.75) = P ( z < −.75) − P ( z < −2.63) = .2266 − .0043 =.
b. We are given that the area to the left of x is .99, which gives z = 2.33 approximately. Then, x = μ + zσ = 0 + (2.33)(2) = 4.66 mph ≈ 5 mph. So, the minimum estimate of speed at which a car should be cited for speeding is 60 + 5 = 65 mph.
6.97 μ = 45 minutes and σ = 3 minutes Let x be the amount of time Ashley spends commuting to work. We are given that the area to the left of x is .95, which gives z = 1.65 approximately. Then, x = μ + zσ = 45 + (1.65)(3) = 49.95 ≈ 50 minutes. Ashley should leave home at about 8:10 am in order to arrive at work by 9 am 95% of the time.
6.98 σ = .07 ounce and P ( x ≥ 8) =. The area to the left of x = 8 is 1 – .99 = .01 and z = −2.33 approximately. Then, μ = x − zσ = 8 − (−2.33)(.07) = 8.16 ounces. Thus, the mean should be set at approximately 8.16 ounces.
142 Chapter 6 Continuous Random Variables and the Normal Distribution
b. Let n = number of tickets sold Since μ and σ both depend on n , it is not easy to solve for n directly. Instead, use trial and error. In part a, n = 65 was too large, since P ( x ≤ 60.5) = P ( z ≤ .83) = .7967 < .95.
Try n = 62: μ = np = 62(.90) = 55.8and σ = npq = 62(.90)(.10)= 2.
For x = 60.5: z = (60.5 – 55.8)/2.36220236 = 1. P ( x ≤ 60.5) = P ( z ≤ 1.99) = .9767 > .95. Thus, n = 62 satisfies the requirement. To see if n may be increased, try n = 63.
μ = np = 63(.90) = 56.7and σ = npq = 63(.90)(.10)= 2.
For x = 60.5: z = (60.5 – 56.7)/2.38117618 = 1. P ( x ≤ 60.5) = P ( z ≤ 1.60) = .9452 <. Thus, n = 63 is too large. Therefore, the largest number of tickets the company can sell and be at least 95% sure that the bus can hold all ticket holders who show up is 62.
6.103 a. μ = 2 minutes and σ = .5 minute First, we find the probability that it will take less than 1 minute for one customer to be served. For x = 1: z = (1 – 2)/.5 = −2. P ( x < 1) = P ( z < −2.00) =. Since customers are independent, P (two customers take less than one minute each) = (.0228)(.0228) =. b. Using complementary events, P (at least one of the four needs more than 2.25 minutes) = 1 – P (each of the four needs 2.25 minutes or less), so we first find the probability that one customer needs 2.25 minutes or less to be served. For x = 2.25: z = (2.25 – 2)/.5 =. P ( x ≤ 2.25) = P ( z ≤ .5) =. Since customers are independent, P (at least one of the four needs more than 2.25 minutes) = 1 – P (each of the four needs 2.25 minutes or less) = 1 – (.6915) 4 = .7714.
6.104 μ = np = 15(.02) = .30and σ = npq = 15(.02)(.98)=.
Since np < 5, the normal approximation to the binomial is not appropriate. The Empirical Rule requires a bell-shaped distribution, and this distribution is skewed right. By the Empirical Rule, approximately 68%
of the observations fall in the interval μ ± σ, approximately 95% fall in the interval μ ± 2 σ, and about
99.7% fall in the interval μ ± 3 σ. These intervals are −.24 to .84, −.78 to 1.38, and −1.33 to 1.93,
respectively. Using the normal approximation with continuity correction, For x = −.74: z = (−.74 – .3)/.54221767 = –1. For x = 1.34: z = (1.34 – .3)/.54221767 = 1. P (−.74 ≤ x ≤ 1.34) = P (–1.92 ≤ z ≤ 1.92) = P ( z ≤ 1.92) − P ( z ≤ –1.92) = .9726 − .0274 = .9452 >. For x = −1.28: z = (−1.28 – .3)/.54221767 = –2. For x = 1.88 z = (1.88 – .3)/.54221767 = 2. P (−1.28 ≤ x ≤ 1.88) = P (–2.91 ≤ z ≤ 2.91) = P ( z ≤ 2.91) − P ( z ≤ –2.91) = .9982 − .0018 = .9964 >. For x = −1.83: z = (−1.83 – .3)/.54221767 = –3. For x = 2.43: z = (2.43 – .3)/.54221767 = 3. P (−1.83 ≤ x ≤ 2.43) = P (–3.93 ≤ z ≤ 3.93) = P ( z ≤ 3.93) − P ( z ≤ –3.93) = 1 − 0 ≈ 1 >.
6.105 a. P (.3 ≤ z ≤ .4) = P ( z ≤ .4) − P ( z ≤ .3) = .6554 − .6179 =.
b. P (–.1 ≤ z ≤ .4) = P ( z ≤ .4) − P ( z ≤ –.1) = .6554 − .4602 =.
Chapter 6 Self-Review Test 143
c. P (at least one out of five games, the ball lands in the required slot) = 1 − P (none) For one game, the probability the ball lands outside the required slot is 1 − P (–.1 ≤ z ≤ .4) = 1 − .1952 = .8048. Since the games are independent, 1 − P (none) = 1 − (.8048) 5 = 1 − .3376 =. d. Note that 100 out of 500 games is equivalent to 1 out of 5 games. Then, the probability that the ball lands outside the required slot for one game is 1 − P (.4 ≤ z ≤ .5) = 1 − [ P ( z ≤ .5) − P ( z ≤ .4)] = 1 − [.6915 − .6554] = 1 − .0361 =. Since the games are independent, 1 − P (none) = 1 − (.9639)^5 = 1 − .8321 =.
6.106 μ = 8 and P ( x > 8.2) = .03. Then, z = 1.88 approximately.
Now x = μ + z σ,so σ = ( x − μ )/ z = (8.2 − 8)/(1.88) = .1064.
Self-Review Test
1. a 2. a 3. d 4. b 5. a 6. c 7. b 8. b 9. a. P (.85 ≤ z ≤ 2.33) = P ( z ≤ 2.33) − P ( z ≤ .85) = .9901 − .8023 =.
b. P (−2.97 ≤ z ≤ 1.49) = P ( z ≤ 1.49) − P ( z ≤ −2.97) = .9319 − .0015 =. c. P ( z ≤ −1.29) =. d. P ( z > −.74) = 1 − P ( z ≤ −.74) = 1 − .2296 =.
10. a. z = −1.28 approximately b. z =.
c. z = 1.65 approximately d. z = −1.07 approximately
11. μ = 45,000 miles and σ = 2360 miles
a. For x = 42,000: z = (42,000 – 45,000)/2360 = −1. For x = 46,000: z = (46,000 – 45,000)/2360 =. P (42,000 < x < 45,000) = P (−1.27 < z < .42) = P ( z < .42) − P ( z < −1.27) = .6628 − .1020 =. b. For x = 38,000 z = (38,000 – 45,000)/2360 = −2. P ( x < 38,000) = P ( z < −2.97) =. c. For x = 50,000: z = (50,000 – 45,000)/2360 = 2. P ( x > 50,000) = P ( z > 2.12) = 1 − P ( z ≤ 2.12) = 1 − .9830 =. d. For x = 46,500: z = (46,500 – 45,000)/2360 =. For x = 47,500: z = (47,500 – 45,000)/2360 = 1. P (46,500 < x < 47,500) = P (.64 < z < 1.06) = P ( z < 1.06) − P ( z < .64) = .8554 − .7389 =.
12. μ = 45,000 miles and σ = 2360 miles
a. For .06 area in the right tail of the normal distribution curve, z ≈ 1.55 approximately. Then, x = μ + zσ = 45,000 + (1.55)(2360) = 48,658 miles b. For .02 area in the left tail of the normal distribution curve, z ≈ −2.05. Then, x = μ + zσ = 45,000 + (−2.05)(2360) = 40,162 miles
13. μ = np = 800(.15) = 120 and σ = npq = 800(.15)(.85)= 10.
a. i. For x = 114.5: z = (114.5 – 120)/10.09950494 = −. For x = 115.5: z = (115.5 – 120)/10.09950494 = −. P (114.5 ≤ x ≤ 115.5) = P (−.54 ≤ z ≤ −.45) = P ( z ≤ −.45) − P ( z ≤ −.54) = .3264 − .2946 =. ii. For x = 102.5: z = (102.5 – 120)/10.09950494 = −1. For x = 142.5: z = (142.5 – 120)/10.09950494 = 2. P (102.5 ≤ x ≤ 142.5) = P (−1.73 ≤ z ≤ 2.23) = P ( z ≤ 2.23) − P ( z ≤ −1.73) = .9871 −. =.
