Section 9.5
This section is basically the Central Limit Theorem, which basically means that for
any shaped probability distribution, that if you take a sampling of 30 or more and
average each of the samples of 30 or more, until all the data is gone from the
population that the resulting distribution will be a normal distribution with the
same mean, but the standard deviation will be changed by the square root of the
number in the sample: 𝜎𝑥̅ =
𝜎𝑥 .
√𝑛
Problem 71
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
4
=𝟒
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √4 = 2
Problem 72
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
8
=𝟐
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √2 = 1.4142
Problem 73
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
16
=𝟏
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √1 = 1
Problem 74
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
32
= 𝟎. 𝟓
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √0.5 = 0.7071
Problem 75
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
64
= 𝟎. 𝟐𝟓
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √0.25 = 0.5
Problem 76
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
128
= 𝟎. 𝟏𝟐𝟓
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √0.125 = 0.3236
Problem 77
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟒𝟎
Variance 𝜎𝑥̅ 2 =
𝜎𝑥 2
𝑛
=
16
4
=𝟒
Standard Deviation 𝜎𝑥̅ = √𝜎𝑥̅ 2 = √4 = 2
Problem 78
Standard Deviation 𝜎𝑥̅ = 1 =
𝜎𝑥
√𝑛
=
25
√𝑛
𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 √𝑛 = 25, 𝑠𝑜, 𝑛 = 625
Problem 79
Standard Deviation 𝜎𝑥̅ = 3 =
𝜎𝑥𝑛𝑒𝑤
̅̅̅̅̅̅̅̅ =
𝜎𝑥
√𝑛
𝑤ℎ𝑖𝑐ℎ 𝑒𝑞𝑢𝑎𝑙𝑠 𝜎𝑥 = 3√𝑛 𝑠𝑜, 𝑖𝑓
𝜎𝑥
√4𝑛
𝑡ℎ𝑒𝑛 𝑒𝑞𝑢𝑎𝑙𝑠 𝜎𝑥𝑛𝑒𝑤
̅̅̅̅̅̅̅̅ =
3 √𝑛
√4𝑛
=
Problem 80
From problem #77, 𝜇𝑥̅ = 20 and Standard Deviation 𝜎𝑥̅ = 0.5
Now to do the rest of the problem just like we did in section 8.7:
3√ 𝑛
2√ 𝑛
=
𝟑
𝟐
19
20
21
Z19 = (19-20)/0.5 = -2.00 and Z21 = (21-20)/0.5 = 2.00
From the table they are both a probability of 0.4772
𝑠𝑜, (𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒𝑠 𝑎𝑑𝑑)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.4772 + 0.4772= 0.9544
Problem 81
From problem #77, 𝜇𝑥̅ = 20 and Standard Deviation 𝜎𝑥̅ = 0.5
Now to do the rest of the problem just like we did in section 8.7:
Since 20 is the MEAN (and that divides the probability distribution in half,
𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 = 0.5
Problem 82
From problem #77, 𝜇𝑥̅ = 20 and Standard Deviation 𝜎𝑥̅ = 0.5
Now to do the rest of the problem just like we did in section 8.7:
20
21
Z21 = (21-20)/0.5 = 2.00
From the table it has a probability of 0.4772
𝑠𝑜, (𝑡𝑎𝑖𝑙 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 0.5)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 − 0.4772= 0.0228
Problem 83
From problem #77, 𝜇𝑥̅ = 20 and Standard Deviation 𝜎𝑥̅ = 0.5
Now to do the rest of the problem just like we did in section 8.7:
18.5
20
Z = (18.5-20)/0.5 = -3.00
From the table it has a probability of 0.4987
𝑠𝑜, (𝑡𝑎𝑖𝑙 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 0.5)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 − 0.4987= 0.0013
Problem 84
𝜇 = 𝜇𝑥̅ = 160 and √𝜎 2 = √400 𝑠𝑜, 𝜎𝑥̅ =
20
√30
= 3.6515
Next since it crashes at 4600 pounds, then if there were 30 people in the elevator
at that time the average weight of a person in the elevator would be:
4600/30 = 153.33
Now to do the rest of the problem just like we did in section 8.7:
153.3333
160
Z = (153.3333-160)/3.6515 = -1.83
From the table it has a probability of 0.4664
𝑠𝑜, (𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑎𝑑𝑑)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 + 0.4664= 0.9664
Problems 85-87
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟖𝟎𝟎
Standard Deviation 𝜎𝑥̅ =
𝜎𝑥
√𝑛
=
450
√225
= 30
Problem 85
Now to do the rest of the problem just like we did in section 8.7:
770
800
890
Z770 = (770-800)/30 = -1.00 and Z890 = (890-800)/30 = 3.00
From the table they are (respectively) a probability of 0.3413 and 0.4990
𝑠𝑜, (𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒𝑠 𝑎𝑑𝑑)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.3413 + 0.4987= 0.8400
Problem 86
Now to do the rest of the problem just like we did in section 8.7:
740
800
860
Z740 = (740-800)/30 = -2.00 and Z860 = (860-800)/30 = 2.00
From the table they are (respectively) a probability of 0.4772 and 0.4772
𝑠𝑜, (𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒𝑠 𝑎𝑑𝑑)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.4772 + 0.4772= 0.9544
Problem 87
Now to do the rest of the problem just like we did in section 8.7:
710
800
Z = (710-800)/30 = -3.00
From the table it has a probability of 0.4987
𝑠𝑜, (𝑡𝑎𝑖𝑙 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 0.5)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 − 0.4987= 0.0013
Problems 88-90
𝑀𝑒𝑎𝑛 𝜇𝑥̅ = 𝜇𝑥 = 𝟏𝟔
Standard Deviation 𝜎𝑥̅ =
𝜎𝑥
√𝑛
=
1
√64
= 0.125
Problem 88
Now to do the rest of the problem just like we did in section 8.7:
16
16.2
Z = (16.2-16)/0.125 = 1.60
From the table the probability is 0.4452
𝑠𝑜, (𝑡𝑎𝑖𝑙 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 0.5)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 − 0.4452= 0.0548
Problem 89
Now to do the rest of the problem just like we did in section 8.7:
15.7
16
16.5
Z15.7 = (15.7-16)/0.125 = -2.40 and Z16.5 = (16.5-16)/0.125 = 4.00
From the table they are (respectively) a probability of 0.4918 and 0.5 (or 0.49997)
𝑠𝑜, (𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒𝑠 𝑎𝑑𝑑)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.4918 + 0.5= 0.9918
Problem 90
Now to do the rest of the problem just like we did in section 8.7:
15.5
16
Z = (15.5-16)/0.125 = -4.00
From the table it has a probability of 0.5 (or 0.49997)
𝑠𝑜, (𝑡𝑎𝑖𝑙 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑠 𝑓𝑟𝑜𝑚 0.5)𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑖𝑠 0.5 − 0.0.5= 0 (or 0.00003)