Download Mann ‑ Introductory Statistics, Fifth Edition, Students Solutions

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.
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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
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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
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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
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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.
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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.
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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
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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