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Tech Math 2
HW#42 Solutions
Section 22.4: #9, 10, 11, 12, 13, 14, 15, and 16
Page 1 of 2
9. 200 bags of cement have a normal distribution. How many are within one standard
deviation of the mean?
 34.13% are within one standard deviation above the mean, and 34.13% are within one
standard deviation below the mean. Therefore, 68.26% are within one standard
deviation.
 200(.6826) = 136.52  137 bags
10. 200 bags of cement have a normal distribution. How many are within two standard
deviations of the mean?
 68.26% are within one standard deviation. An additional 13.59% + 13.59% = 27.18%
are between one and two standard deviations from the mean. Therefore, 95.44% are
within two standard deviations.
 200(.9544) = 190.88  191 bags
11. 200 bags of cement have a normal distribution. How many are between the mean and
two standard deviations above the mean?
 95.44%  2 = 47.72% are between the mean and two standard deviations above the
mean.
 200(.4772) = 95.44  94 bags
12. 200 bags of cement have a normal distribution. How many are between one standard
deviation below the mean and three standard deviations above the mean?
 34.13% + 34.13% + 13.59% + 2.15% = 84.00% are between one standard deviation
below the mean and three standard deviations above the mean.
 200(.8400) = 168 bags
Tech Math 2
HW#42 Solutions
Section 22.4: #9, 10, 11, 12, 13, 14, 15, and 16
Page 2 of 2
13.The voltages of 500 batteries are normally distributed with  = 1.50 V and  = 0.05V.
How many batteries have voltages between 1.45V and 1.55V?
 This range of voltages is one standard deviation above and below the mean.
 34.13% + 34.13% = 68.26% are in this range
 500(.6826) = 341.3  341 batteries
14.The voltages of 500 batteries are normally distributed with  = 1.50 V and  = 0.05V.
How many batteries have voltages between 1.52V and 1.58V?
 The z-score for 1.52 V is z = (1.52V – 1.50V) / 0.05V = 0.4 standard deviations above
the mean
 The z-score for 1.58 V is z = (1.58V – 1.50V) / 0.05V = 1.6 standard deviations above
the mean.
 The area for z = 0.4 is 0.1554
 The area for z = 1.6 is 0.4452
 0.4452 – 0.1554 = 0.2898
 500(.2898) = 144.9  145 batteries
15.The voltages of 500 batteries are normally distributed with  = 1.50 V and  = 0.05V.
How many batteries have voltages below 1.54V ?
 The z-score for 1.54 V is z = (1.54V – 1.50V) / 0.05V = 0.8 standard deviations above
the mean
 The area for z = 0.8 is 0.2881
 The area for everything up to the mean is 0.5000
 0.5000 + 0.2881 = 0.7881
 500(.7881) = 394.05  394 batteries
16.The voltages of 500 batteries are normally distributed with  = 1.50 V and  = 0.05V.
How many batteries have voltages above 1.64V ?
 The z-score for 1.64 V is z = (1.64V – 1.50V) / 0.05V = 2.8 standard deviations above
the mean
 The area for z = 2.8 is 0.4974
 The area for everything up to the mean is 0.5000
 0.5000 + 0.4974 = 0.9974 (this is the percentage of batteries with voltages less than
1.64 V)
 The percentage of batteries with higher voltages is 1.000 – .9974 = 0.0026
 500(.0026) = 1.3  1 battery