CHAPTER 7
THE NORMAL PROBABILITY DISTRIBUTION
1.
A probability function must be non-negative. That is the curve should be above the x-axis.
Also, the total area under the curve must equal 1.
3.
The actual shape of a normal distribution depends on its mean and standard deviation.
Also, given the values of mean and standard deviation, a normal distribution is uniquely
defined. Thus, there is a normal distribution, and an accompanying normal curve, for a
mean of 7 and a standard deviation of 2. There is another normal curve for a mean of
$25,000 and a standard deviation of $1742, and so on.
5.
(a)
From the Z-table at the end of the book we find that :
area under the Z-curve between 0 and 0.36 is 0.1406;
area under the Z-curve between 0 and 1.82 is 0.4656.
Hence, area under the curve between –0.36 and 1.82 is 0.1406 + 0.4656 = 0.6062.
0.1406
0.4656
-0.36 0
(b)
1.82
From the Z-table we find that:
area under the Z-curve between 0 and 1.47 is 0.4292;
area under the Z-curve between 0 and 2.71 is 0.4966.
Hence, area under the curve between 1.47 and 2.71 is 0.4966 – 0.4292 = 0.0674.
0.4292
0.0674
0
(c)
1.47
2.71
From the Z-table, area under the Z-curve between 0 and 1.51 is 0.4345.
0.4345
0
7-1
1.51
(d)
From the Z-table, area under the Z-curve between 0 and 1.94 is 0.4738. Hence, area under
the curve to the right of 1.94 is 0.5 – 0.4738 = 0.0262.
0.4738
0.0262
0
(e)
1.94
From the Z-table, area under the Z-curve between 0 and 0.82 is 0.2939. Hence, area under
the curve to the right of –0.82 is 0.5 + 0.2939 = 0.7939.
0.2939
0.5
-0.82
7.
(a)
0
Area between 0 and the z-value = 0.5 – 0.025 = 0.475
From the Z-table, we have the corresponding z-value = 1.96
0.475
0.025
0
(b)
1.96
Area between 0 and the z-value = 0.95 – 0.5 = 0.45
From the Z-table, we find that the z-value is midway between 1.64 and 1.645 (we shall
take it to be 1.645)
0.45
0.5
1.645
(c)
From the Z-table, we get directly that the required z-value is slightly greater than 1.28. So,
it is approximately 1.281
0.4
7-2 1.281
9.
(a)
55  60
 1 .
5
65  60
 1.
z-value corresponding to 65 is z 
5
z-value corresponding to 55 is z 
Hence, the area between 55 and 65 is the area between ( - 1 ) and ( + 1 ) which
equals 0.6826.
(b)
50  60
 2 .
5
70  60
2.
z-value corresponding to 70 is z 
5
z-value corresponding to 50 is z 
Hence, the area between 50 and 70 is the area between ( - 2 ) and ( + 2 ) which equals
0.9544.
(c)
45  60
 3 .
5
75  60
 3.
z-value corresponding to 75 is z 
5
z-value corresponding to 45 is z 
Hence, the area between 45 and 75 is the area between ( - 3 ) and ( + 3 ) which
equals 0.9974.
11.
(a)
z-value corresponding to 10 is z  10  12.2  0.88 , and
2.5
z-value corresponding to 14.3 is z 
13.
14.3  12.2
 0.84 .
2.5
(b)
We have already calculated in part (a) that the z-value corresponding to 14.3 is 0.84.
We want area under the Z-curve between 0 and 0.84.
From the Z-table, we find this area to be 0.2995.
(c)
We have already calculated in part (a) that the z-value corresponding to 10 is –0.88
We want area under the Z-curve to the left of –0.88.
From the Z-table we find that the area under the Z-curve between 0 and 0.88 is 0.3106.
Hence, area under the curve to the left of –0.88 is 0.5 – 0.3106 = 0.1894.
(a)
z-value corresponding to 415 is z 
415  400
 1.5 .
10
We want area under the Z-curve between 0 and 1.5.
From the Z-table, we find this area to be 0.4332.
(b)
z-value corresponding to 395 is z 
395  400
 0.5 .
10
We want area under the Z-curve between –0.5 and 0.
From the Z-table, area under the Z-curve between 0 and 0.5 is 0.1915.
Hence, area under the curve between –0.5 and 0 is 0.1915.
7-3
15.
(c)
We want area under the normal curve to the left of 395. We have already calculated in part
(b) that the z-value corresponding to 395 is –0.5.
Thus, we want area under the Z-curve to the left of –0.5.
From the Z-table, area under the Z-curve between 0 and 0.5 is 0.1915. Hence, area under
the curve to the left of –0.5 is 0.5 – 0.1915 = 0.3085.
(a)
z-value corresponding to 75 is z 
75  80
 0.357 .
14
90  80
 0.714 .
z-value corresponding to 90 is z 
14
We want area under the normal curve between –0.357 and 0.714
From the Z-table,
area under the Z-curve between 0 and 0.357 is approximately 0.1394;
area under the Z-curve between 0 and 0.714 is approximately 0.2623.
Hence, area under the curve between –0.357 and 0.714 is approximately 0.1394 + 0.2623=
0.4017.
(b)
We have already calculated in part (a) that the z-value corresponding to 75 is –0.357 and
area under the Z-curve between 0 and 0.357 is approximately 0.1394.
Hence, area under the curve to the left of –0.357 is approximately 0.5 – 0.1394 = 0.3606.
(c)
z-value corresponding to 55 is z 
55  80
 1.786 .
14
70  80
 0.714 .
z-value corresponding to 70 is z 
14
We want area under the Z-curve between –1.786 and -0.714
From the Z-table,
area under the Z-curve between 0 and 1.786 is approximately 0.463;
area under the Z-curve between 0 and 0.714 is approximately 0.2623.
Hence, area under the curve between -1.786 and -0.714 is approximately 0.463 - 0.2623 =
0.2007.
17.
(a)
35000  36280
 0.388 .
3300
40000  36280
 1.127 .
z-value corresponding to 40000 is z 
3300
z-value corresponding to 35000 is z 
We want area under the Z-curve between –0.388 and 1.127.
From the Z-table,
area under the Z-curve between 0 and 0.388 is approximately 0.1510;
area under the Z-curve between 0 and 1.127 is approximately 0.37.
Hence, area under the curve between –0.388 and 1.127 is approximately 0.1510 + 0.370 =
0.5210.
(b)
z-value corresponding to 45000 is z 
45000  36280
 2.642 .
3300
We want area under the Z-curve to the right of 2.642.
From the Z-table, area under the Z-curve between 0 and 2.642 is approximately 0.4959.
7-4
Hence, area under the curve to the right of 2.642 is approximately 0.5 – 0.4959 = 0.0041.
19.
(c)
We have already calculated in part (a) that the z-value corresponding to 40000 is 1.127 and
area under the Z-curve between 0 and 1.127 is approximately 0.37. Also, we have already
calculated in part (b) that the z-value corresponding to 45000 is 2.642 and area under the Zcurve between 0 and 2.642 is approximately 0.4959.
Hence, area under the curve between 1.127 and 2.642 is approximately 0.4959 –
0.37 = 0.1259.
(a)
z-value corresponding to 253 is z 
253  250
 0.75 .
4
258  250
 2.0 .
z-value corresponding to 258 is z 
4
We want area under the Z-curve between 0.75 and 2.0.
From the Z-table,
area under the Z-curve between 0 and 0.75 is 0.2734;
area under the Z-curve between 0 and 2.0 is 0.4772.
Hence, area under the curve between 0.75 and 2.0 is 0.4772 - 0.2734 = 0.2038.
(b)
z-value corresponding to 260 is z 
260  250
 2.5 .
4
We want area under the Z-curve to the right of 2.5.
From the Z-table, area under the Z-curve between 0 and 2.5 is 0.4938.
Hence, area under the curve to the right of 2.5 is 0.5 – 0.4938 = 0.0062.
(c)
245  250
 1.25 .
4
255  250
 1.25 .
z-value corresponding to 255 is z 
4
z-value corresponding to 245 is z 
We want area under the Z-curve between -1.25 and 1.25.
From the Z-table, area under the Z-curve between 0 and 1.25 is 0.3944.
Hence, area under the curve between -1.25and 1.25 is 0.3944 + 0.3944 = 0.7888.
21.
We want a number x such that the area under the normal curve to the right of x is 0.8.
Hence, the area between x and  must be (0.8 – 0.5) = 0.3.
From the Z-table, we find that the z-value, such that the area under the Z-curve between z
and 0 is 0.3 equals approximately -0.842.
Hence, the required x value is approximately 80 - (0.842)(14) = 68.212.
23.
We want a number x such that the area under the normal curve to the left of x is 0.03.
Hence, the area between x and  must be (0.5 – 0.03) = 0.47.
From the Z-table, we find that the z-value, such that the area under the Z-curve between z
and 0 is 0.47 equals approximately –1.88.
Hence, the required x value is approximately ( - 1.88 ) = 3100 – (1.88)(250) = $2630.
25.
(a)
We want 1 – (area under the normal curve between 357 and 363).
7-5
357  360
 1 .
3
363  360
1.
z-value corresponding to 363 is z 
3
z-value corresponding to 357 is z 
From the Z-table, area under the Z-curve between 0 and 1 is 0.3413.
Hence, area under the curve between -1 and 1 is 0.3413 + 0.3413 = 0.6826.
Hence, fraction of bottles that will be discarded is (1 – 0.6826) = 0.3174.
On average, (1000)(0.3174) = 317.4 bottles will be discarded per day.
27.
(b)
We want a number x such that the area under the normal curve between x and 363 is 1 –
0.2 = 0.8.
z-value corresponding to 363 is 1, and area under the Z-curve between 0 and 1 is 0.3413.
Hence, area under the given normal curve between  (=360) and 363 is 0.3413.
Hence, area under the normal curve between x and  (=360) should be 0.8 – 0.3413 =
0.4587.
From the Z-table, we find that the value of z, such that area under the Z-curve between z
and 0 is 0.4587, equals approximately –1.736.
Hence, the required value of x is ( - 1.736 ) = 360 – (1.736)(3) = 354.792 ml.
(c)
We want area under the normal curve between 357 and 363 (that is, between 360-3 and
360+3) to be at least 0.8. So, the area between 360 and 363 should be at least 0.4.
From the Z-table, we find that the z-value, such that the area under the Z-curve between 0
and z is 0.4 equals approximately 1.282.
Hence 3  1.282 new or new  (3/1.282) = 2.34.
The standard deviation of the new machine should be no greater than 2.34.
(a)
  np  (40)(0.55)  22 ;   np(1  p)  40(0.55)(0.45)  3.15
(b)
Since n p = 22 > 5 and n (1 – p) = 40 (0.45) = 18 > 5, we can approximate the binomial
distribution by normal distribution with  = 22 and  = 3.15.
We want area under the normal curve to the right of 24.5.
z-value corresponding to 24.5 is z 
24.5  22
 0.794 .
3.15
Thus, we want area under the Z-curve to the right of 0.794.
From the Z-table, area under the Z-curve between 0 and 0.794 is approximately 0.2864.
Hence, area under the curve between 0 and 0.794 is approximately 0.5 – 0.2864 = 0.2136.
(c)
We want area under the normal curve to the left of 15.5.
z-value corresponding to 15.5 is z 
15.5  22
 2.063 .
3.15
Thus, we want area under the Z-curve to the left of –2.063.
From the Z-table, area under the Z-curve between 0 and 2.063 is approximately 0.4804.
Hence, area under the curve to the left of -2.063 is approximately 0.5 – 0.4804 = 0.0196.
(d)
We want area under the normal curve between 14.5 and 25.5.
z-value corresponding to 14.5 is z 
14.5  22
 2.381 .
3.15
7-6
z-value corresponding to 25.5 is z 
25.5  22
 1.111 .
3.15
Thus, we want area under the Z-curve between -2.381 and 1.111.
From the Z-table,
area under the Z-curve between 0 and 2.381 is approximately 0.4913;
area under the Z-curve between 0 and 1.111 is approximately 0.3667.
Hence, area under the curve between –2.381 and 1.111 is approximately 0.4913 + 0.3667 =
0.858.
29.
(a)
Let X be the number of installations, out of 50, that will take more than 30 minutes. Then X
follows binomial distribution.
E(X) =  = n p = (50)(0.2) = 10
(b)
n p = (50) (0.2) > 5 and n (1 – p) = 50 (0.8) = 40 > 5. Hence, we shall approximate the
binomial distribution by normal distribution with mean and standard deviation;
  np  (50)(0.2)  10 ;   np(1  p)  50(0.2)(0.8)  2.83
We want area under the normal curve to the left of 7.5.
z-value corresponding to 7.5 is z 
7.5  10
 0.883 .
2.83
Thus, we want area under the Z-curve to the left of –0.883
From the Z-table, area under the Z-curve between 0 and 0.883 is approximately 0.3114.
Hence, area under the curve to the left of –0.883 is approximately 0.5 – 0.3114 = 0.1886.
(c)
We want area under the normal curve to the left of 8.5.
z-value corresponding to 8.5 is z 
8.5  10
 0.53 .
2.83
Thus, we want area under the Z-curve to the left of –0.53
From the Z-table, area under the Z-curve between 0 and 0.53 is 0.2019.
Hence, area under the curve to the left of –0.53 is 0.5 – 0.2019 = 0.2918.
(d)
31.
We want area under the normal curve between 7.5 and 8.5.
We have already calculated in part (b) that the z-value corresponding to 7.5 is –0.883 and
area under the Z-curve between 0 and 0.883 is approximately 0.3114. Also, we have
already calculated in part (c) that the z-value corresponding to 8.5 is –0.53 and area under
the Z-curve between 0 and 0.53 is 0.2019.
Hence, area under the curve between –0.833 and –0.53 is approximately 0.3114 –
0.2019 = 0.1095.
Let X be the number of offenders, out of 100, who will commit another crime. Then X is a
binomial random variable with mean and standard distribution;
  np  (100)(0.38)  38 ;   np(1  p)  100(0.38)(0.62)  4.854
n p = (100) (0.38) = 38 > 5 and n (1 – p) = 100 (0.62) = 62 > 5. Hence, we can
approximate the binomial random variable by normal random variable with  = 38 and  =
4.854
(a)
We want area under the normal curve to the right of 29.5.
7-7
z-value corresponding to 29.5 is z 
29.5  38
 1.751 .
4.854
Thus, we want area under the Z-curve to the right of –1.751.
From the Z-table, area under the Z-curve between 0 and 1.751 is approximately 0.46.
Hence, area under the curve to the right of –1.751 is approximately 0.5 + 0.46 = 0. 96.
(b)
We want area under the normal curve to the left of 40.5
z-value corresponding to 40.5 is z 
40.5  38
 0.515 .
4.854
Thus, we want area under the Z-curve to the left of 0.515.
From the Z-table, area under the Z-curve between 0 and 0.515 is approximately 0.1968.
Hence, area under the curve to the left of 0.515 is approximately 0.5 + 0.1968 = 0.6968.
33.
(c)
We want area under the normal curve between 29.5 and 40.5.
We have already calculated in part (a) that the z-value corresponding to 29.5 is –1.751 and
area under the Z-curve between 0 and 1.751 is approximately 0.46. Also, we have already
calculated in part (b) that the z-value corresponding to 40.5 is 0.515 and area under the Zcurve between 0 and 0.515 is approximately 0.1968.
Hence, area under the curve between –1.751 and 0.515 is approximately 0.46 +
0.1968 = 0.6568.
(a)
z-value corresponding to Interline’s sales and number of employees are
z
(b)
170  180
1850  1500
 0.4 and z 
 2.9167 respectively.
25
120
Interline’s sales compared to the other fabricators;
From the Z-table, area under the Z-curve between 0 and 0.4 is 0.1554.
Hence, area under the curve to the right of –0.4 is 0.5 + 0.1554 = 0.6554. It means that
65.54% of fabricators have greater net sales compared with Interline.
Interline’s number of employees compared to the other fabricators;
From the Z-table, area under the Z-curve between 0 and 2.9167 is 0.4982.
Hence, area under the curve to the right of 2.9167 is 0.5 - 0.4982 = 0.0018. It means that
0.18% of fabricators have more employees than Interline.
35.
(a)
z-value corresponding to 5 is z 
5  4.2
 1.333 .
0.60
We want area under the Z-curve between 0 and 1.333.
From the Z-table, we find that this area equals approximately 0.4088.
(b)
We have already calculated in part (a) that the z-value corresponding to 5 is 1.333 and area
under the Z-curve between 0 and 1.333 is approximately 0.4088.
Hence, area under the curve to the right of 1.333 is approximately 0.5 - 0.4088 = 0.0912.
(c)
In part (a), we found that the z-value corresponding to 5 is 1.333 and area under Z-curve
between 0 and 1.333 is approximately 0.4088.
z-value corresponding to 6 is z 
6  4.2
 3.
0.60
7-8
From the Z-table, we get area under the Z-curve between 0 and 3 is 0.4987.
Hence, area under the curve between 1.333 and 3 is approximately 0.4987 - 0.4088 =
0.0899.
(d)
In part (c), we found that the z-value corresponding to 6 is 3 and area under the Z-curve
between 0 and 3 is 0.4987.
z-value corresponding to 4 is z 
4  4.2
 0.333 .
0.60
We want area under the Z-curve between –0.333 and 3.
From the Z-table, we get area under the Z-curve between 0 and 0.333 is approximately
0.1306.
Hence, area under the curve between –0.333 and 3 is approximately 0.1306 + 0.4987 =
0.6293.
37.
(e)
We want a number z such that area under the Z-curve to the right of z is 0.04. The area
under Z-curve between 0 and z is then 0.5 - 0.04 = 0.46.
From the Z-table, we find that z equals approximately 1.75.
Hence, the length of time, such that 4 percent of the calls last longer than 4.2 minutes is
approximately 4.2 + (1.75) 0.6 = 5.25 minutes.
(a)
z-value corresponding to 80 is z 
80  100
 1.0 .
20
110  100
 0.5 .
z-value corresponding to 110 is z 
20
We want area under the Z-curve between –1.0 and 0.5.
From the Z-table,
area under the Z-curve between 0 and 1.0 is 0.3413;
area under the Z-curve between 0 and 0.5 is 0.1915.
Hence, area under the curve between –1.0 and 0.5 is 0.3413 + 0.1915 = 0.5328.
(b)
z-value corresponding to 90 is z 
90  100
 0.5 .
20
We want area under the Z-curve to the left of –0.5.
From the Z-table, area under the Z-curve between 0 and 0.5 is 0.1915.
Hence, area under the curve to the left of -0.5 is approximately 0.5 - 0.1915 = 0. 3085.
(c)
In part (b), we found that the z-value corresponding to 90 is –0.5 and area under the Zcurve between 0 and 0.5 is 0.1915.
z-value corresponding to 130 is z 
130  100
 1.5 .
20
We want area under the Z-curve between –0.5 and 1.5.
From the Z-table, we get area under the Z-curve between 0 and 1.5 is 0.4332.
Hence, area under the curve between –0.5 and 1.5 is 0.1915 + 0.4332 = 0.6247
(d)
In part (a), we found that the z-value corresponding to 110 is 0.5 and area under the Zcurve between 0 and 0.5 is 0.1915.
7-9
z-value corresponding to 140 is z 
140  100
 2.
20
We want area under the Z-curve between 0.5 and 2.
From the Z-table, we get area under the Z-curve between 0 and 2 is 0.4772.
Hence, area under the curve between 0.5 and 2 is 0.4772 - 0.1915 = 0.2857.
(e)
39.
(a)
We want a number z such that area under the Z-curve to the left of z is 0.05. The area under
Z-curve between 0 and z is then 0.5 - 0.05 = 0.45.
From the Z-table we get z equals approximately 1.645.
Hence, the duration, such that the waiting time of only 5% of the patient is less than it is
approximately 100 - (1.645) 20 = 67.1 minutes.
We want area under the normal curve to the right of 55000.
z-value corresponding to 55000 is z 
55000  50000
 0.625 .
8000
Thus, we want area under the Z-curve to the right of 0.625.
From the Z-table, we get area under the Z-curve between 0 and 0.625 is approximately
0.2341.
Hence, area under the curve to the right of 0.625 is approximately 0.5 - 0.2341 = 0.2659.
(b)
We want area under the normal curve between 40000 and 65000.
40000  50000
 1.25 .
8000
65000  50000
 1.875 .
z-value corresponding to 65000 is z 
8000
z-value corresponding to 40000 is z 
Thus, we want area under the Z-curve between –1.25 and 1.875.
From the Z-table,
area under the Z-curve between 0 and 1.25 is 0.3944;
area under the Z-curve between 0 and 1.875 is approximately 0.4696.
Hence, area under the curve between –1.25 and 1.875 is approximately 0.3944 + 0.4696 =
0.8640.
(c)
We want area under the normal curve between 38000 and 45000.
38000  50000
 1.5 .
8000
45000  50000
 0.625 .
z-value corresponding to 45000 is z 
8000
z-value corresponding to 38000 is z 
Thus, we want area under the Z-curve between –1.5 and –0.625.
From the Z-table,
area under the Z-curve between 0 and 1.5 is 0.4332;
area under the Z-curve between 0 and 0.625 is approximately 0.2341.
Hence, area under the curve between –1.5 and -0.625 is approximately 0.4332 – 0.2341 =
0.1991.
(d)
We want a number x such that the area under the normal curve to the right of x is 0.2.
Let us find number z such that area under the Z-curve to the right of z is 0.2.
7-10
The area under the Z-curve between 0 and z is then 0.5 - 0.2 = 0.3.
From the Z-table, we find that z equals approximately 0.842.
Hence, the cutoff point x between those who earn a bonus and those who do not is 50000 +
(0.842)(8000) = $ 56736.
41.
43.
We want a number x such that the area under the normal curve to the right of x is 0.05.
Let us find number z such that area under the Z-curve to the right of z is 0.05.
The area under the Z-curve between 0 and z is then 0.5 - 0.05 = 0.45.
From the Z-table, we find that z equals approximately 1.645.
So, x =  + 1.645  = 4000 + (1.645) (60) = 4098.7.
Hence the bonus will be paid if the production rate exceeds 4098.7 units per week.
(a)
Let us find area under the normal curve, corresponding to income, to the right of 50400.
z-value corresponding to 50400 is z 
50400  48000
 2.0 .
1200
Thus, we want area under the Z-curve to the left of 2.0.
From the Z-table, we find that area under the Z-curve between 0 and 2.0 is 0.4772.
Hence, area under the curve to the left of 2.0 is 0.5 + 0.4772 = 0.9772.
Thus, the annual income of 97.72 percent of the supervisors is less than that of John.
(b)
Let us find area under the normal curve, corresponding to length of service, to the left of
10.
z-value corresponding to 10 is z 
10  20
 2.0 .
5
Thus, we want area under the Z-curve to the left of –2.0.
From the Z-table, we find that area under the Z-curve between 0 and 2.0 is 0.4772.
Hence, area under the curve to the left of –2.0 is 0.5 - 0.4772 = 0.0228.
Only 2.28% of the supervisors have length of service less than that of John.
45.
(c)
We want a number x such that area under the income curve to the left of x is 0.08.
Let us find number z such that area under the Z-curve to the left of z is 0.08. The area under
Z-curve between z and 0 is then 0.5 - 0.08 = 0.42.
From the Z-table, we find that z equals approximately -1.405.
Hence, the cutoff point between those who earn a bonus and those who do not is
approximately 48000 – (1.405)(1200) = $ 46314.
(a)
z-value corresponding to 8 is z 
8  10.3
 1.022 .
2.25
We want area under the Z-curve to the left of –1.022.
From the Z-table, we get area under the Z-curve between 0 and 1.022 is approximately
0.3466.
Hence, area under the curve to the left of –1.022 is approximately 0.5 – 0.3466 = 0.1534.
(b)
12  10.3
 0.756 .
2.25
14  10.3
 1.644 .
z-value corresponding to 14 is z 
2.25
z-value corresponding to 12 is z 
7-11
We want area under the Z-curve between 0.756 and 1.644.
From the Z-table,
area under the Z-curve between 0 and 0.756 is approximately 0.2752;
area under the Z-curve between 0 and 1.644 is approximately 0.4499.
Hence, area under the curve between 0.756 and 1.644 is approximately 0.4499 – 0.2752 =
0.1747.
(c)
We want area under the normal curve to the left of 0.
z-value corresponding to 0 is z 
0  10.3
 4.578 .
2.25
Thus, we want area under the Z-curve to the left of –4.578.
From the Z-table, we get area under the Z-curve between 0 and 4.578 is almost 0.5.
Hence, the area to the left of –4.578 is almost (0.5 – 0.5) = almost 0.
The chance of a day with no returns is very very small.
47.
(a)
Let X be number of defective bottles in a lot of 100. Then X follows binomial distribution
with n = 100 and p = 0.05.
Hence,  = n p = (100)(0.05) = 5. We would expect 5 bottles to be defective.
The standard deviation is   np(1  p)  100(0.05)(0.95)  2.179 .
(b)
The total number of bottles in the sample is 100, a fixed number. Every bottle is either
defective or non-defective; every bottle has the same probability (= 0.05) of being
defective; and one bottle being defective has no effect on the probability of another bottle
being defective. Thus, the trials are independent of each other.
(c)
n p = (100)(0.05) = 5; n (1 – p) = 100 (0.95) = 95 > 5.
Hence, we can approximate X by a normal random variable with  = 5 and  = 2.179. The
required probability is approximately equal to the area under the normal curve to the right
of 7.5.
z-value corresponding to 7.5 is z 
7.5  5
 1.15 .
2.179
Thus, we want area under the Z-curve to the right of 1.15.
From the Z-table, we get area under the Z-curve between 0 and 1.15 is 0.3749.
Hence, area under the curve to the right of 1.15 is 0.5 – 0.3749 = 0.1251.
(d)
This is approximately equal to the area under the normal curve between 7.5 and 10.5.
In part (c), we found that the z-value corresponding to 7.5 is 1.15 and area under the Zcurve between 0 and 1.15 is 0.3749.
z-value corresponding to 10.5 is z 
10.5  5
 2.52 .
2.179
From the Z-table, we find that area under the Z-curve between 0 and 2.52 is 0.4941.
Hence, area under the curve between 1.15 and 2.52 is 0.4941 – 0.3749 = 0.1192
(e)
This is approximately equal to the area under the normal curve between 7.5 and 8.5.
In part (c), we found that the z-value corresponding to 7.5 is 1.15 and area under the Zcurve between 0 and 1.15 is 0.3749.
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z-value corresponding to 8.5 is z 
8.5  5
 1.61 .
2.179
From the Z-table, we find that area under the Z-curve between 0 and 1.61 is 0.4463.
Hence, area under the curve between 1.15 and 1.61 is 0.4463 – 0.3749 = 0.0714.
(f)
This is approximately equal to the area under the normal curve to the left of 0.5.
z-value corresponding to 0.5 is z 
0.5  5
 2.06 .
2.179
So, we want area under the Z-curve to the left of -2.06
From the Z-table, we get area under the Z-curve between 0 and 2.06 is 0.4803.
Hence, area under the curve to the left of –2.06 is 0.5 – 0.4803 = 0.0197.
49.
(a)
Let X be the number of students, out of the trial of 80 students, who will fail. Then X
follows binomial random variable with n = 80 and p = 0.1.
Hence,  = n p = (100)(0.05) = 5. We would expect 8 students to fail.
The standard deviation is   np(1  p)  80(0.1)(0.9)  2.6833 .
(b)
n p = (80)(0.1) = 8 > 5 and n (1 – p) = 80(0.9) = 72 > 5.
Hence, we can approximate X by a normal random variable with  = 8 and  = 2.6833. The
required probability is approximately equal to the area under the normal curve between 9.5
and 10.5.
9.5  8
 0.559 .
2.6833
10.5  8
 0.932 .
z-value corresponding to 10.5 is z 
2.6833
z-value corresponding to 9.5 is z 
Thus, we want area under the Z-curve between 0.559 and 0.932.
From the Z-table,
area under the Z-curve between 0 and 0.559 is approximately 0.212;
area under the Z-curve between 0 and 0.932 is approximately 0.3243.
Hence, area under the curve between 0.559 and 0.932 is approximately 0.3243 – 0.212 =
0.1123.
(c)
This is approximately equal to the area under the normal curve to the right of 4.5.
z-value corresponding to 4.5 is z 
4.5  8
 -1.304.
2.6833
Thus, we want area under the Z-curve to the right of –1.304.
From the Z-table, we get area under the Z-curve between 0 and 1.304 is approximately
0.4039.
Hence, area under the curve to the right of –1.3 is approximately 0.5 + 0.4039 = 0.9039.
51.
Let X be the number of Canadians, out of a random sample of 50, who will show distrust in
water quality. Then X is approximately binomially distributed with the mean and standard
deviation,
  np  (50)(0.46) = 23 ;
  np(1  p)  50(0.46)(0.54)  3.5242
n p = (50)(0.46) = 23 > 5 and n (1 – p) = 50(0.54) = 27 > 5.
7-13
Hence, the distribution of X can be approximated by normal distribution with  = 23 and 
= 3.5242. The desired probability is approximately equal to the area under the normal
curve to the left of 19.5.
z-value corresponding to 19.5 is z 
19.5  23
 -0.99.
3.5242
Thus, we want area under the Z-curve to the left of -0.99.
From the Z-table, we find that area under the Z-curve between 0 and 0.99 is 0.3389.
Hence, area under the curve to the left of -0.99 is 0.5 – 0.3389 = 0.1611.
53.
(a)
We want area under the normal curve between 2.0 and 3.0.
2.0  3.1
 -3.67.
0.3
3.0  3.1
 -0.33.
z-value corresponding to 3.0 is z 
0.3
z-value corresponding to 2.0 is z 
Thus, we want area under the Z-curve between –3.67 and -0.33.
From the Z-table,
area under the Z-curve between 0 and 3.67 is approximately 0.5;
area under the Z-curve between 0 and 0.33 is 0.1293.
Hence, area under the curve between –3.67 and –0.33 is approximately 0.5 – 0.1293 =
0.3707.
(b)
We want area under the normal curve to the left of 2.0. From solution to part (a), we see
that the z-value corresponding to 2.0 is –3.67 and the area under the Z-curve between –3.67
and 0 is almost 0.5.
Hence, the area to the left of –3.67 is almost 0. The percentage of students on probation is
almost 0.
(c)
Let us find area under the normal curve to the right of 3.7.
z-value corresponding to 3.7 is z 
3.7  3.1
 2.0.
0.3
From the Z-table, area under the Z-curve between 0 and 2.0 is 0.4772.
Hence, area to the right of 2.0 is 0.5 – 0.4772 = 0.0228.
Hence, 2.28 percent of students are on Dean’s list, or (10000)(0.0228) = 228 students are
on Dean’s list.
55.
(d)
We want a number x such that area under the normal curve to the right of x is 0.1.
Let us find the z-value such that area under the Z-curve to the right of z is 0.1. The area
under the curve between 0 and z should then (0.5-0.1) = 0.4. From the Z-table, we find that
z = approximately 1.282.
Hence, x =  + 1.282  = 3.10 + (1.282)(0.3) = 3.4846.
A student must attain a GPA of 3.4846 to qualify for a Bell scholarship.
(a)
We want area under the normal curve to the left of 4.0.
z-value corresponding to 4.0 is z 
4.0  4.2
 0.5 .
0.4
From the Z-table, we find that the area under the Z-curve between 0 and 0.5 is 0.1915.
Hence, the area to the left of –0.5 is (0.5 – 0.1915) = 0.3085.
7-14
Therefore, 0.3085 fraction of hams actually weigh less than the amount claimed on the
label.
(b)
If  is increased to 4.3, then
z-value corresponding to 4.0 is z 
4.0  4.3
 0.75 .
0.4
From the Z-table, we find that the area under the Z-curve between 0 and 0.75 is 0.2734.
Hence, the area to the left of –0.75 is (0.5 – 0.2734) = 0.2266.
If  is decreased to 0.2 kg, then
z-value corresponding to 4.0 is z 
4.0  4.2
 1.0 .
0.2
From the Z-table, we find that the area under the Z-curve between 0 and 1.0 is 0.3413.
Hence, the area to the left of –1.0 is (0.5 – 0.3413) = 0.1587.
0.1587 is less than 0.2266. Thus decreasing  leads to a smaller probability of hams below
the label weight. Furthermore, decreasing  also reduces probability of some hams having
too large in weight. Hence, decreasing  is a better option.
57.
We are given that,
(i)
40 percent times the sales are more than 470000.
So, (50 – 40) = 10 percent times sales are between  and 470000.
(ii)
10 percent times the sales are more than 500000.
So, (50 – 10) = 40 percent times sales are between  and 500000.
From the Z-table, we find that,
- the z-value such that area under the Z-curve between 0 and z equals 0.1, is approximately
0.253, and
- the z-value such that area under the Z-curve between 0 and z equals 0.4, is approximately
1.282.
Hence,  + 0.253  = approximately 470000
 + 1.282  = approximately 500000
Hence, (1.282 – 0.253)  = (500000 – 470000) = 30000.
or  
30000
 29154.52
1.282  0.253
 = 500000- (1.282)() = 500000- (1.282)(29154.52) = 462623.9
59.
Let X be the number of business phones in the list of 150 selected phone numbers. Then, if
the manufacturer’s claim is correct, X is binomially distributed with n = 150 and p = 0.15.
n p = (150) (0.15) = 22.5 > 5 and n (1-p) = (150)(0.85) =
127 > 5.
Hence, we can approximate X by a normal random variable with
 = n p = 22.5 and   np(1  p)  150(0.15)(0.85)  4.37.
The desired probability is thus approximately equal to the area under the normal curve to
the right of 29.5.
7-15
z-value corresponding to 29.5 is z 
29.5  22.5
 1.602 .
4.37
Thus, we want area under the Z-curve to the right of 1.602.
From the Z-table, area under the Z-curve between 0 and 1.602 is approximately 0.4454.
Hence, area under the curve to the right of 1.602 is approximately 0.5 - 0.4454 = 0.0546.
Hence, if manufacturer’s claim is correct, likelihood that 30 or more phone numbers
selected will be business numbers is approximately 0.0546.
61.
Let X be the number of households, out of the total of 50 selected households, that would
own a personal computer. Then X is approximately binomially distributed with n = 50 and
p = 0.69.
 = n p = (50) (0.69) = 34.5;   np(1  p)  50(0.69)(0.31)  3.27.
Since, n p = 34.5 > 5 and n (1 - p) =(50)(0.31) = 15.5 > 5, we can approximate X by a
normal random variable with  = 34.5 and  = 3.27.
The desired probability is thus approximately equal to the area under the normal curve to
the right of 39.5.
z-value corresponding to 39.5 is z 
39.5  34.5
 1.529 .
3.27
Thus, we want area under the Z-curve to the right of 1.529.
From the Z-table, area under the Z-curve between 0 and 1.529 is approximately 0.4369.
Hence, area under the curve to the right of 1.529 is approximately 0.5 - 0.4369 = 0.0631.
Thus, probability that 40 or more would own personal computer is approximately 0.0631.
63.
Let X be the number of guests, out of 105, who will actually show up. Then X is
approximately binomially distributed with n = 105 and p = 0.1.
Since, n p = (105) (0.1) = 10.5 > 5 and n (1 - p) =(105)(0.9) = 94.5 > 5, we can
approximate X by a normal random variable with
 = n p = (105) (0.1) = 10.5 and   np(1  p)  105(0.1)(0.9)  3.074.
We want probability that X is greater than or equal to 5.
This is approximately equal to the area under the normal curve to the right of 4.5.
z-value corresponding to 4.5 is z 
4.5  10.5
 1.952 .
3.074
Thus, we want area under the Z-curve to the right of –1.952.
From the Z-table, area under the Z-curve between 0 and 1.952 is approximately 0.4745.
Hence, the area under the curve to the right of –1.952 is approximately 0.5 + 0.4745 =
0.9745.
The probability is approximately 0.9745 that all the arriving guests will receive a room.
65.
We want a number x such that area under the normal curve to the right of x is 0.04.
So, area under the curve between  and x is (0.5 – 0.04) = 0.46.
From the Z-table, we find that z-value, such that area under the Z-curve between 0 and z
equals 0.46, is approximately 1.751.
Hence, x =  + 1.751  = 800 + (1.751)(80) = 940.08.
The new order quantity should be approximately 940.08 kg.
7-16
67.
(a)
The Minitab and Excel outputs are given below. Note that Minitab does not compute
population standard deviation. It only computes the sample standard deviation. The
population mean and standard deviation are:   3.363, and   5.077
MINITAB OUTPUT
Descriptive Statistics: C1
Variable
C1
N
87
Mean
3.363
Median
2.715
TrMean
3.420
Variable
C1
Minimum
-11.852
Maximum
18.987
Q1
1.099
Q3
4.953
StDev
5.106
SE Mean
0.547
MEGASTAT OUTPUT
Descriptive statistics
#1
count
87
mean
3.363183
population sample variance 25.772775
population standard deviation
(b)
69.
5.076689
Assuming that the distribution is normal, and using Minitab or Excel, we find that
probability, that percentage change during a randomly selected year does not exceed two
percent, is approximately 0.394.
By sorting the data, we see that 39 out of 87 values are less than or equal to 2.0. Hence, the
fraction of values that do not exceed 2.0 are 39/87 = 0.448.
This is not a really good approximation. But given the small moderate size (= 87) of the
population data, this is expected.
Using Megastat, we get the values   9.5 and   59.81 . (See the printout of the output below.)
MEGASTAT OUTPUT
Descriptive statistics
count
100
mean
9.4983
population sample variance 3,576.7866
population standard deviation
59.8062
Again, using Excel, we get for a normal random variable, X, with   9.5 and   59.81 ,
probability that a random value, x, will be less than 2 equal to 0.45. Thus, the probability that the
value of x will be more than 2 equals (1.0-0.45)=0.55. Thus, using normal approximation, we get
the probability (that the one-year percentage return on common equity of a randomly chosen
company in the list will be greater than 2 percent) equal to 0.55.
7-17
By sorting the data, using Excel or Minitab, we find that 71 out of the companies have a one-year
percentage return on common equity more than 2 percent. Thus, (71/100 =) 0.71 fraction of the
companies have a one-year percentage return on common equity more than 2 percent.
The two values (0.55 and 0.71) are quite different. Use of normal distribution did not yield a good
approximation in this case.
7-18