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Chapter Six 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. 6.3 Since P(a) = 0 and P(b) = 0 for a continuous random variable, P ( a x b) P(a x b) . 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. In other words, 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. The z value for a point on the horizontal axis gives the distance between the mean and that point in terms of the standard deviation. 6.7 As its standard deviation decreases, the width of a normal distribution curve decreases and its height increases. 6.9 For a standard normal distribution, z gives the distance between the mean and the point represented by z in terms of the standard deviation. 6.11 Area between 1.5 and 1.5 is the area from z 1.50 to z 1.50 , which is: P(1.5 z 1.5) P( z 1.5) P( z 1.5) .9332 .0668 .8664 TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two z values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate your numbers with commas! For this example, after normalcdf( we would enter -1.5, 1.5, 0, 1) and then press ENTER. The result is shown below. 101 102 Chapter Six Normalcdf( -1.5, 1.5, 0, 1) .8663855426 Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to represent the missing endpoint. To type in E99 (positive infinity) to represent the missing endpoint press the 2nd key followed by the , key and then typing in 99. To type in - E99 (negative infinity) to represent the missing endpoint press (−) then press the 2 nd key followed by the , key and then typing in 99. MINITAB: Enter the two z values in a column in your worksheet. For this example we will use column C1. Now select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0, and the Standard deviation says 1.0. Beside the words Input column type C1 or the number of whatever column your z vales are in and then click on OK. The result appears in the Session box. Once we have the probabilities actually have P(z < a) and P (z < b), but we desire P(a < z < b) so we P (z < b) − P(z < a) to get the answer. For this example a is -1.5 and b is 1.5. P (z < 1.5) − P(z < -1.5) = .933193 − .066807 which equals 0.866386. Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press ENTER. Where x is the z value such that we calculate P(z < x ), µ is 0, σ is 1, and True mean this is a cumulative distribution. Since we want P(-1.5 < z < 1.5) we calculate P (z < 1.5) − P(z <−1.5), so in the first cell x is 1 and in the cell directly below it we repeat the process but this time we type in x as −1. Then all we have to do is subtract the second cell from the first. The results are shown below where labels were added for easier reading. Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 6.13 103 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 .9876 6.15 a. P(0 z 1.95) P( z 1.95) P( z 0) .9744 .5000 .4744 b. P(1.85 z 0) P( z 0) P( z 1.85) .5000 .0322 .4678 c. P(1.15 z 2.37 ) P( z 2.37 ) P( z 1.15) .9911 .8749 .1162 d. P(2.88 z 1.53) P( z 1.53) P( z 2.88) .0630 .0020 .0610 e. P(1.67 z 2.44) P( z 2.44) P( z 1.67 ) .9927 .0475 .9452 6.17 a. P( z 1.56) 1 P( z 1.56) 1 .9406 .0594 b. P( z 1.97 ) .0244 c. P( z 2.05) 1 P( z 2.05) P( z 2.05) .9798 d. P( z 1.86) P( z 1.86) .9686 TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two z values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate your numbers with commas! For part a of this example, after normalcdf( we would enter 1.56, E99, 0, 1) and then press ENTER. For part b of this example, after normalcdf( we would enter − E99, -1.97, 0, 1) and then press ENTER. The results shown below display the answer for part a and b. Normalcdf( 1.56, E99, 0, 1) .0593799504 Normalcdf( -E99,−1.97, 0, 1) .0244191152 Note: To type in E99 (positive infinity) to represent the missing endpoint press the 2nd key followed by the , key and then typing in 99. To type in − E99 (negative infinity) to represent the missing endpoint press (−) then press the 2nd key followed by the , key and then typing in 99. MINITAB: With only one end point we can enter it into a worksheet or directly into the pop-up menu. In the later case, select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0, and the Standard deviation says 1.0. This time click beside the words Input constant and enter your z value in the box beside those words. Lastly click on OK. The result appears in the Session box. Once we have the probabilities actually have P(z < a) but we desire P(a > z) so we have two choices we can 104 Chapter Six insert a and the do a little math 1 − P (z < a) = P(z > a) to get the answer. Or since we know the distribution is symmetric and the P(z > a) = P(z < −a), we can enter −a into this menu instead. For part a, a is 1.56 so −1.56. 1 − P(z < 1.56) = 1 − .940620 = .0593799 and P(z < − 1.56) = .0593799. Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press ENTER. Where x is the z value such that we calculate P(z < x ), µ is 0, σ is 1, and True mean this is a cumulative distribution. Since we want P( z > 1.56) and Excel will calculate P(z <1.56)we can precede in one of two ways. The first method to get the appropriate probability is to insert right after the = a 1− before the formula and inserting 1.56 for our x. We also know that P( z > 1.56) = P(z < x ), so we could insert -1.56 for our x and come up with the same answer. The results for part a are shown below where labels were added for easier reading. 6.19 a. P(0 z 4.28) P( z 4.28) P( z 0) 1 .5 .5 approximately b. P(3.75 z 0) P( z 0) P( z 3.75) .5 .0000 .5 approximately c. P( z 7.43) P( z 7.43) .0000 approximately d. P( z 4.49) .0000 approximately 6.21 a. P(1.83 z 2.57 ) P( z 2.57 ) P( z 1.83) .9949 .0336 .9613 b. P(0 z 2.02) P( z 2.57 ) P( z 0) 9783 .5 .4783 c. P(1.99 z 0) P( z 1.99) P( z 0) .9767 .5 .4767 d. P( z 1.48) P( z 1.48) .0694 6.23 a. P( z 2.14) .0162 b. P(.67 z 2.49) P( z 2.49) P( z .67 ) .9936 .7486 .2450 Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 105 c. P(2.07 z .93) P( z .93) P( z 2.07 ) .1762 .0192 .1570 d. P( z 1.78) .9625 6.25 a. P( z .98) P( z .98) .8365 b. P(2.47 z 1.19) P( z 1.19) P( z 2.47 ) .8830 .0068 .8762 c. P(0 z 4.25) P( z 4.25) P( z 0) 1 .5000 .5 approximately d. P(5.36 z 0) P( z 0) P( z 5.36) .5000 .0000 .5 approximately e. P( z 6.07 ) P( z 6.07 ) .0000 approximately f. 6.27 P( z 5.27 ) .0000 approximately Here, 30 and 5 a. z ( x ) / (39 30) / 5 9 / 5 1.80 b. z ( x ) / (17 30) / 5 13 / 5 2.60 c. z ( x ) / (22 30) / 5 8 / 5 1.60 d. z ( x ) / (42 30) / 5 12 / 5 2.40 6.29 Here, 20 and 3 a. For x 20 : z ( x ) / (20 20) / 4 0 For x 27 : z ( x ) / (27 20) / 4 7 / 4 1.75 P(20 x 27 ) P(0 z 1.75) P( z 1.75) P( z 0) .9599 .5000 .4599 b. For x 23 : z ( x ) / (23 20) / 4 3 / 4 .75 For x 25 : z ( x ) / (25 20) / 4 5 / 4 1.25 P(23 x 25) P(.75 z 1.25) P( z 1.25) P( z .75) .8944 .7734 .1210 c. For x 9.5 : z ( x ) / (9.5 20) / 4 10.5 / 4 2.63 For x 17 : z ( x ) / (17 20) / 4 3 / 4 .75 P(9.5 x 17 ) P(2.63 z .75) P( z .75) P( z 2.63) .2266 .0043 .2223 TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two x values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate 106 Chapter Six your numbers with commas! For the first part of this problem, after normalcdf( we would enter 20, 27, 20, 4) and then press ENTER. The result is shown below. Normalcdf( 20, 27, 20, 4) .4599408859 Normalcdf( 23, 25, 20, 4) .1209774405 Normalcdf( 9.5, 17, 20, 4) .2222947931 Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to represent the missing endpoint. To type in E99 (positive infinity) to represent the missing endpoint press the 2nd key followed by the , key and then typing in 99. To type in - E99 (negative infinity) to represent the missing endpoint press (−) then press the 2 nd key followed by the , key and then typing in 99. MINITAB: Enter the two z values in a column in your worksheet. For this example we will use column C1. Now select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0, and the Standard deviation says 1.0. Replace the standard normal mean and standard deviation to match the µ and σ which is now stated in terms of xs. Beside the words Input column type C1 or the number of whatever column your z vales are in and then click on OK. The result appears in the Session box. Once we have the probabilities actually have P(z < a) and P (z < b), but we desire P(a < z < b) so we P (z < b) − P(z < a) to get the answer. For this example µ is 20, σ is 4, a is 20 and b is 27. P (x <27) − P(x < 27) = .959941 − .5000000 which equals 0.459941. Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press ENTER. Where x is the z value such that we calculate P(z < x ), µ is mean in terms of xs, σ is the standard deviation in terms of xs, and True mean this is a cumulative distribution. Since we want P(20 < x < 27) we calculate P (x < 27) − P(x <20), so in the first cell x is 27 and in the cell directly below it we repeat the process but this time we type in x as 20. For both cells in this example µ is 20 Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 107 and σ is 4. Then all we have to do is subtract the second cell from the first. The results are shown below where labels were added for easier reading. 6.31 Here, 55 and 7 a. For x 58 : z (58 55) / 7 3 / 7 .43 P( x 58) P( z .43) P( z .43) .3336 b. For x 43 : z (43 55) / 7 12 / 7 1.71 P( x 43) P( z 1.71) P( z 1.71) .9564 c. For x 67 : z (67 55) / 7 12 / 7 1.71 P( x 67 ) P( z 1.71) .9564 d. For x 24 : z (24 55) / 7 31 / 7 4.43 P( x 24) P( z 4.43) .0000 approximately TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to 2: normalcdf( and press ENTER. Next enter the smaller of the two z values, followed by the larger one, the µ, the σ, the ) symbol, and press ENTER. Remember to separate your numbers with commas! For part a of this example, after normalcdf( we would enter 58, E99, 55, 7) and then press ENTER. For part c of this example, after normalcdf( we would enter − E99, 67, 55, 7) and then press ENTER. The results shown below display the answer for part a and c in the first screen with b and d in the second screen. Normalcdf( 58, E99, 55, 7) .3341176023 Normalcdf( -E99,67, 55, 7) .956761902 Normalcdf(43, E99, 55, 7) .956761902 Normalcdf( -E99, 24, 55, 7) 4.74696936E−6 Note: To type in E99 (positive infinity) to represent the missing endpoint press the 2 nd key followed by the , key and then typing in 99. To type in − E99 (negative infinity) to represent the missing endpoint press (−) then press the 2nd key followed by the , key and then typing in 99. MINITAB: With only one end point we can enter it into a worksheet or directly into the pop-up menu. In the later case, select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Cumulative Probability is filled in, the Mean says 0.0, and the Standard deviation says 1.0. This time click beside the words Input constant and enter your z 108 Chapter Six value in the box beside those words. Lastly click on OK. The result appears in the Session box. Once we have the probabilities actually have P(x < a) but we desire P(a > x) so we can insert a and the do a little math 1 − P (x < a) = P(x > a) to get the answer. For part a, a is 58 so P(58 > x)= 1 − P(x < 58) = 1 − .665882 = .334118. We do the same for part b, while in c and d with areas to the left of µ MINITAB will do the entire calculation for us. Excel: In the first empty cell type in the formula =NORMDIST(x, µ, σ, cumulative) and then press ENTER. Where x is the z value such that we calculate P(z < x ), µ is mean in terms of xs, σ is the standard deviation in terms of xs, and True mean this is a cumulative distribution. Since we want P( z > 58) and Excel will calculate P(z <58)we can precede in one of two ways. The first method to get the appropriate probability is to insert 1− right after the = and before the formula. In the formula we then insert 58 for our x, 55 for µ and 7 for σ. The second way is to find = P(z < x ) and in an empty box simply subtract that probability from 1. The results for part a are shown below where labels were added for easier reading. 6.33 Here, 25 and 6 a. For x 29 : z (29 25) / 6 4 / 6 .67 For x 36 : z (36 25) / 6 11 / 6 1.83 P(29 x 36) P(.67 z 1.83) P( z 1.83) P( z .67 ) .9664 .7486 .2178 b. For x 22 : z (22 25) / 6 3 / 6 .50 For x 33 : z (33 25) / 6 8 / 6 1.33 P(22 x 33) P(.50 z 1.33) P( z 1.33) P( z .50) .9082 .3085 .5997 Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 6.35 6.37 109 Here, 80 and 12 a. For x 69 : z (69 80 ) / 12 11 / 12 .92 P( x 69) P( z .92) P( z .92) .8212 b. For x 74 : z (74 80) / 12 6 / 12 .50 P( x 74) P( z .50) .3085 c. For x 101 : z (101 80) / 12 21 / 12 1.75 P( x 101) P( z 1.75) P( z 1.75) .0401 d. For x 88 : z (88 80) / 12 8 / 12 .67 P( x 88) P( z .67 ) .7486 Here, 190 minutes and 21 minutes a. For x 150 : z (150 190 ) / 21 40 / 21 1.90 P( x 150 ) P( z 1.90) .0287 b. For x 205 : z (205 190) / 21 15 / 21 .71 For x 245 : z (245 190) / 21 2.62 P(205 x 245 ) P(.71 z 2.62) P( z 2.62) P( z .71) = .9956 – .7611 = .2345 Using technology, for a detailed description of how to use technology to solve these problems see the answers to 6.29 and 6.31. 6.39 Here, = $10 and = $1.10 a. For x 12.00 : z (12 .00 10.00) / 1.10 1.82 P( x 12.00) P( z 1.82) P( z 1.82) .0344 b. For x 8.50 : z (8.50 10.00) / 1.10 1.36 For x 10 .80 : z (10 .80 10.00) / 1.10 .73 P(8.50 x 10.80) P(1.36 z .73) P( z .73) P( z 1.36) = .7673 − .0869 = .6804 6.41 Here, =40 miles per hour and = 4 miles per hour a. For x 46 : z (40 46) / 4 6 / 4 1.50 P( x 0) P( z 1.50) P( z 1.50) = .9332 Thus, about 93.32% of the cars traveling in the construction zone are exceeding the 40 miles per hour posted speed limit. 110 Chapter Six b. For x 50 : z (50 46) / 4 4 / 4 1.00 For x 55 : z (55 46) / 4 9 / 4 2.25 P(50 x 55) P(1.00 z 2.25) P( z 2.25) P( z 1.00) =.9878 – .8413 = .1465 Thus, about 14.65% of the cars traveling in the construction zone have a speed of 50 to 55 miles per hour. 6.43 Here, = $2084 and = $300 a. For x 2500 : z (2500 2084 ) / 300 1.39 P( x 2500 ) P( z 1.39) P( z 1.39) .0823 b. For x 1800 : z (1800 2084 ) / 300 .95 For x 2400 : z (2400 2084 ) / 300 1.05 P(1800 x 2400 ) P(.95 z 1.05) P( z 1.05) P( z .95) = .8531 − .1711 = .6820 or 68.20% 6.45 Here, = 1650 kwh and = 320 kwh a. For x 1850 : z (1850 1650 ) / 320 .63 P( x 1850 ) P( z .63) .7357 b. For x 900 : z (900 1650 ) / 320 2.34 For x 1340 : z (1340 1650 ) / 320 .97 P(900 x 1340 ) P(2.34 z .97 ) P( z .97 ) P( z 2.34) =.1660 - .0096 = .1564 or 15.64% 6.47 Here, = $43,070 and = $3000 a. For x 40,000 : z (40,000 43,070 ) / 3000 1.02 P( x 40,000 ) P( z 1.02) .1539 or 15.39% b. For x 45,000 : z (45,000 43,070 ) / 3000 .64 P( x 45,000 ) P( z .64) P( z .64) = .2611 or 26.11% 6.49 Here, = 516 and = 90 a. For x 600 : z (600 516 ) / 90 .93 P( x 600 ) P( z .93) P( z .93) .1762 or 17.62% Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual b. For x 450 : z (450 516 ) / 90 .73 111 P( x 450 ) P( z .73) = .2327 or 23.27% c. For x 700 : z (700 516 ) / 90 2.04 P( x 700 ) P( z 2.04 ) P( z 2.04) = .0207 or 2.07% 6.51 Here, = 3.0 inches and = .009 inch For x 2.98 : z (2.98 3.00) / .009 2.22 For x 3.02 : z (3.02 3.00) / .009 2.22 P( x 2.98) P( x 3.02) P( z 2.22) P( z 2.22) {P( z 2.22)} {1 P( z 2.22)} {.0132 } {1 .9868 } .0132 .0132 .0264 Thus, about 2.64% of the nails produced by this machine are unusable. 6.53 a. z = 2.00 b. z = -2.02 approximately c. z = -.37 approximately d. z = 1.02 approximately TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to3: invNorm( and press ENTER. Next enter the percentile, which is the area to the left of z, followed by the µ, the σ, the ) symbol, and press ENTER. Remember to separate your numbers with commas! Note: if you are looking for a z score enter 0 for µ and 1 for σ. For this example, after invNorm( we would enter .9772, 0, 1) and then press ENTER. Caution: In part a the area is between 0 and z but since we want all of the area to the left of z we simply add .5 to the percentile described in part a. Note you can do the necessary calculations like adding .5 either before entering the formula or during the calculation in the formula by setting the addition problem between parenthesis. The results of both methods as well as the solution to parts b, c, and d are shown below. invNorm(.9772, 0, 1) 1.999077216 invNorm((.5+.4772), 0, 1) 1.999077216 invNorm((.5-.4785), 0, 1) -2.023709992 invNorm(.3564, 0, 1) -.3678300036 invNorm((1-.1530), 0, 1) 1.023651314 Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to represent the missing endpoint. MINITAB: Select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Inverse Cumulative Probability is filled in, the Mean says 112 Chapter Six 0.0, and the Standard deviation says 1.0. Beside the words Input constant, type in your percentile and then click on OK. The result appears in the Session box. Note: if you have multiple percentiles to calculate you can input them into your worksheet, click beside the words Input column, and find your zs all at once. Caution: unlike Excel or MINITAB you can not do arithmetic within formulas so for ease of computation convert all percentiles to P(z < a) first! The results for parts a-d of this problem appear below. Excel: In the first empty cell type in the formula =NORMINV(percentile µ, σ) and then press ENTER. Where probability is the percentile for P(z < x ), µ is 0, and σ is 1 when you are looking for z values. Since we want P( z < x) we convert any percentiles into this format either before inserting the probability into the formula or within parenthesis doing the calculation within the formula. In the solutions presented below the first method is used to generate the z values. 6.55 6.57 a. z = 1.65 approximately b. z = -1.96 c. z = -2.33 approximately d. z = 2.58 approximately Here, = 200 and = 25 a. Area to left of x = .6330 z for area of .6630 = .34 approximately x = + z = 200 + (.34)(25) = 208.50 approximately b. Area to the right of x = .05 Area to the left of x = 1 - .05 = .9500 z for area of .9500 = 1.65 approximately x = + z = 200 + (1.65)(25) = 241.25 approximately Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 113 c. x z 200 (.86)( 25) 178 .50 d. x z 200 (2.17 )( 25) 145 .75 e. x z 200 (1.67 )( 25) 158 .25 f. x z 200 (2.05)( 25) 251 .25 approximately TI-83: Start by pressing the 2nd key followed by the VARS key which bring up the distribution menu. Highlight DISTR, scroll down to3: invNorm( and press ENTER. Next enter the percentile, which is the area to the left of z, followed by the µ, the σ, the ) symbol, and press ENTER. Remember to separate your numbers with commas! Note: remember to enter the µ and σ stated in terms of xs to find the x value, however if you are looking for a z score enter 0 for µ and 1 for σ,. For part a of this example, after invNorm( we would enter .6330, 200, 25) and then press ENTER. Caution: In part e the area is between 0 and z but since we want all of the area to the left of z we simply add .5 to the percentile described in part e. Note you can do the required calculations like adding .5 to our probability either before entering the formula or during the calculation in the formula by setting the addition problem between parentheses. The results are shown below. invNorm(.6330, 200, 25) 208.4952374 invNorm((1−.05), 200, 25) 241.1213406 invNorm((1−.8051), 200, 25) 178.500497 invNorm(.015, 200, 25) −.3678300036 invNorm((.5−.4525), 200, 25) 158.2601856 invNorm((.5 + .4800), 200, 25) 251.3437228 Note: if this interval has only 1 end point, use –E99 (negative infinity) or E99 (positive infinity) to represent the missing endpoint. MINITAB: Select Calc, the Probability Distribution, and Normal which will cause a new menu to pop up. In the new menu make sure box beside Inverse Cumulative Probability is filled in. Replace the standard normal mean and standard deviation to match the µ and σ which is now stated in terms of xs. Beside the words Input constant, type in your percentile and then click on OK. The result appears in the Session box. Note: if you have multiple percentiles to calculate you can input them into your worksheet, click beside the words Input column, and find your zs all at once. Caution: it is unlike Excel or MINITAB you can not do arithmetic within formulas so for ease of computation convert all percentiles to P(z < a) first! The results for parts a thru d appear below in order. 114 Chapter Six Excel: In the first empty cell type in the formula =NORMINV(percentile µ, σ) and then press ENTER. Where probability is the percentile for P(x < a), where a is some number and where µ and σ are stated in terms of xs when you are looking for the x values. Since we want P(x < a) we convert any percentiles into this format either before inserting the probability into the formula or within parenthesis doing the calculation within the formula. In the solutions presented below the first method is used to generate the z values. 6.59 Here, = 15 minutes and = 2.4 minutes Let x denote the time to service a randomly chosen car. Then, we are to find x so that the area in the right tail of the normal distribution curve is 5%. Thus, x z 15 (1.65)( 2.4) 18.96 19 minutes. 6.61 Here, = 1650 kwh and = 320 kwh Bill Johnson’s monthly electric consumption is: x z 1650 (1.28)(320 ) 2059 .6 2060 kwh 6.63 We are given that: = $9.50 and P( x 90 ) .20 Hence, the area to the left of x = 90 is .8000. The z value for .8000 is .84 approximately. Since x = + z , we obtain: 6.65 x z 90 (.84)(9.50) $82.02 approximately = 8 minutes and = 2 minutes a. For x = 3: z = (3 – 8) / 2 = -2.50 P( x 3) P( z 2.50) .0062 Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 115 b. For x = 10: z = (10 – 8) / 2 = 1.00 For x = 13: z = (13 – 8) / 2 = 2.50 P(10 x 13) P(1.00 z 2.50) P( z 2.50) P( z 1.00) = .9938 – .8413 = .1525 or 15.25% c. For x = 6: z = (6 – 8) / 2 = -1.00 For x = 12: z = (12 – 8) / 2 = 2.00 P(6 x 12) P(1.00 z 2.00) P( z 2.00) P( z 1.00) = .9772 − .1587 = .8185 or 81.85% d. For x = 16: z = (16 – 8) / 2 = 4.00 P( x 16) P( z 4.00) P( z 4.00) .0000 approximately Although it is possible for a customer to wait for more than 16 minutes, the probability of this is very close to zero. 6.67 = 16 ounces and = .18 ounce a. For x = 16.20: z = (16.20 – 16) / .18 = 1.11 For x = 16.50: z = (16.50 – 16) / .18 = 2.78 P(16.20 x 16.50) P(1.11 z 2.78) P( z 2.78) P( z 1.11) = .9973 - .8665 = .1308 b. For x = 15.70: z = (15.70 – 16) / .18 = -1.67 P( x 15.70) P( z 1.67 ) .0475 or 4.75% c. For x = 15.20: z = (15.20 – 16) / .18 = -4.44 P( x 15.20) P( z 4.44) .0000 approximately Although it is possible for a carton to contain less than 15.20 ounces, its probability is very close to zero. 6.69 = 28 minutes and = 5 minutes Area in the right tail of the normal curve is given to be 1%, which gives z =2.33 approximately. x = + z = 28 + (2.33)(5) = 39.65 minutes approximately. Thus Jenn must leave approximately 40 minutes before 8:30AM, so she must leave by 7:50 AM. 6.71 We are given: = $350 and P( x 2500 ) = .15 Hence, the area to the left of x = 2500 is 1 - .15 = .8500, which gives z = 1.04 approximately. 116 Chapter Six Consequently, 6.73 x z 2500 – (1.04)(350) = $2136 a. If $3500 = $1000 + $12 x , where x is depth in feet, then x =208.33, i.e., Company B charges ft more for depths of more than 208.33 ft. = 250, = 40 For x = 208.33: z = (208.33 – 250) / 40 = -1.04 P( x 208 .33) P( z 1.04) P( z 1.04) .8508 Company B charges more in 85.08% of all wells. b. μ = 250, so the mean amount charged by Company B is $1000 + 6.75 $12 250 ft = $4000. ft = 45,000; = 2000 For x = 46,000: z = (46,000 – 45,000) / 2000 = 1 = .50 2 P( x 46,000 P( z .50) P( z .50) .3085 So the probability of one tire lasting at least 46,000 miles is .3085. Hence, the probability of all four lasting more than 46,000 miles is (.3085) 4 = .0091, i.e., less than 1%. 6.77 = 0, = 2 mph a. Let x be the error of these estimates in mph. For x = 5: z = (5 – 0) / 2 = 2.50 P( x 5) P( z 2.50) P( z 2.50) .0062 So the probability that an estimate is at least 5 mph higher than the actual speed is .0062 or .62%. Hence, the probability that a car traveling at 60 mph or less will be cited for speeding is at most .0062. b. We are given that the area to the left of x is .99. So, the area between and x is .99 -.5=.49, which gives z 2.33. x z 0 2.33(2) 4.66 5 mph So the minimum estimate of speed at which a car should be cited for speeding is 60 + 5 = 65 mph. 6.79 We are given: = .07 ounces and P( x 8) = .99. Hence the area to the left of x = 8 is 1 − .99 = .01, which gives z 2.33 . Consequently, x z 8 (2.33)(. 07 ) 8.16 Mann - Introductory Statistics, Fifth Edition, Students Solutions Manual 117 The mean should be set to 8.16 ounces. 6.81 a. = 2 minutes and = .5 minute For x = 1: z = (1 – 2) / .5 = -2.00 P( x 1) P( z 2.00) .0228 P(two customers take less than one minute each) = P(first takes less than one minute) P (second takes less than one minute)= (.0228)(.0228) = .0005 b. For x = 2.25: z = (2.25 – 2) / .5 = .50 P( x 2.25) P( z .50) .P( z .50) .3085 Using complementary events: P( x 2.25) .6915 P(at least one of the four needs 2.25 minutes)= 1 – P (each of the four needs 2.25 minutes or less) = 1 – (.6915)4 = .7714 Self-Review Test for Chapter Six 1. a 8. 2. a 3. d 4. b 5. a 6. c 7. b a. P(.85 z 2.33) P( z 2.33) P( z .85) .9901 .8203 .1878 b. P(2.97 z 1.49) P( z 1.49) P( z 2.97 ) .9319 .0015 .9304 c. P( z 1.29) .0985 d. P( z .74) P( z .74) .7704 9. 10. a. -1.28 approximately b. .61 c. 1.65 approximately d. -1.07 approximately = 16 hours and = 3.5 hours a. For x = 12: z = (12 – 16) / 3.5 = -1.14 For x = 22: z = (22 – 16) / 3.5 = 1.71 P(12 x 22) P(1.14 z 1.71) P( z 1.71) P( z 1.14) = .9564 − .1271 = .8293 b. For x = 10: z = (10 – 16) / 3.5 = -1.71 P( x 10) P( z 1.71) .0436 118 Chapter Six c. For x = 20: z = (20 – 16) / 3.5 = 1.14 P( x 20) P( z 1.14) P( z 1.14) .1271 d. For x = 20: z = (20 – 16) / 3.5 = 1.14 For x = 25: z = (25 – 16) / 3.5 = 2.57 P(20 x 25) P(1.14 z 2.57 ) P( z 2.57 ) P( z 1.14) = .9949 – .8729 = .1220 11. = 16 hours and = 3.5 hours a. For .10 area in the left tail of the normal distribution curve, z 1.28 . Hence, x z 16 (1.28)(3.5) 11.52 hours b. For .05 area in the right tail of the normal distribution curve, z 1.65 . Hence, x z 16 (1.65)(3.5) 21.78 hours