Section 9.3
This section continues from 9.2, using the mean (and getting the median) to
calculate the variance and finally the standard deviation. The variance
(represented as S2) equals the sum of all the values squared minus the square of
all the values added together divided by the number of values and then dividing
all that by one less than the number of values: 𝑆 2 =
(∑ 𝑥)
𝑛
∑ 𝑥2−
2
𝑛−1
And of course, the standard deviation (represented by S) is the square root of the
variance.
One other thing is introduced, the range approximation for S, based on one of 4
equations (depending on how large n is): for n<16, Range/√n; for n between 16
and 49, Range/4; for range between 50 and 199, Range/5; and for n>199,Range/6
Problem 32
68+70+72+73+74+75+76+84+85+87
Sample mean 𝑋̅ =
= 𝟕𝟔. 𝟓
10
For the Median, first put in order:
1
position
2
3
4
5
6
7
8
9 10
68, 70, 72, 73, 74, 75, 76, 84, 85, 87
Position for the Median is (n+1)/2 = (10+1)/2 = 5.5 (so half way between 5 and 6)
Median = (74+75)/2 = 74.5
Problem 33
∑ 𝑥 2 = 682 + 702 + 722 + 732 + 742 + 752 + 762 + 842 + 852 + 872 = 58901
2
Sample variance, 𝑆 =
(765)2
10
58901−
10−1
= 𝟒𝟐. 𝟎𝟓𝟓𝟔
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √42.0556 = 𝟔. 𝟒𝟖𝟓𝟎
Problem 34
Range = 87 – 68 = 18
Range Approximation of S =
18
√10
= 5.6921
Problem 35
Sample mean 𝑋̅ =
200+215+232+285
4
= 𝟐𝟑𝟑
For the Median, first put in order:
position
1
2
3 4
200, 215, 232, 285
Position for the Median is (n+1)/2 = (4+1)/2 = 2.5 (so half way between 2 and 3)
Median = (215+232)/2 = 223.5
Problem 36
∑ 𝑥 2 = 2002 + 2152 + 2322 + 2852 = 221274
Sample variance, 𝑆 2 =
221274−
(932)2
4
4−1
= 𝟒𝟏𝟏𝟖
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √4118 = 𝟔𝟒. 𝟏𝟕𝟏𝟔
Problem 37
Range = 285 – 200 = 85
Range Approximation of S =
85
√4
= 42.5
Problem 38
22+25+26+27
Sample mean 𝑋̅ =
= 𝟐𝟓
4
For the Median, first put in order:
position
1 2 3 4
22, 25, 26, 27
Position for the Median is (n+1)/2 = (4+1)/2 = 2.5 (so half way between 2 and 3)
Median = (25+26)/2 = 25.5
Problem 39
∑ 𝑥 2 = 222 + 252 + 262 + 272 = 2514
Sample variance, 𝑆 2 =
(100)2
4
2514−
4−1
= 𝟒. 𝟔𝟔𝟔𝟕
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √4.6667 = 𝟐. 𝟏𝟔𝟎𝟐
Problem 40
Range = 27 – 22 = 5
Range Approximation of S =
5
√4
= 2.5
Problem 41
∑ 𝑥 = 6.45 + 6.78 + 5.75 + 6.35 + 6.75 + 6.00 + 7.15 + 5.95 + 7.60 + 6.85 + 7.55
+ 6.75 + 7.45 + 6.45 + 6.50 + 7.20 + 7.85 + 6.50 + 6.85 + 7.25 + 6.50
+ 6.45 + 7.45 + 7.24 = 163.62
∑ 𝑥 2 = 6.452 + 6.782 + 5.752 + 6.352 + 6.752 + 6.002 + 7.152 + 5.952 + 7.602 + 6.852
+ 7.552 + 6.752 + 7.452 + 6.452 + 6.502 + 7.202 + 7.852 + 6.502 + 6.852
+ 7.252 + 6.502 + 6.452 + 7.452 + 7.242 = 1122.616
Sample variance, 𝑆 2 =
1122.616−
(163.62)2
24
24−1
= 𝟎. 𝟑𝟏𝟎𝟐
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √0.3102 = 𝟎. 𝟓𝟓𝟕𝟎
Problem 42
Range = 7.85 – 5.75 = 2.1
Range Approximation of S =
2.1
4
= 0.525
Problem 43
∑ 𝑥 = 4.45 + 5.35 + 4.86 + 5.50 + 4.00 + 4.25 + 3.95 + 4.12 + 4.85 + 4.35 + 3.85
+ 4.15 + 3.90 + 4.20 + 4.56 + 4.40 + 5.35 + 4.95 + 4.65 + 4.35 + 4.35
+ 4.85 + 5.15 + 4.95 = 109.34
∑ 𝑥 2 = 4.452 + 5.352 + 4.862 + 5.502 + 4.002 + 4.252 + 3.952 + 4.122 + 4.852 + 4.352
+ 3.852 + 4.152 + 3.902 + 4.202 + 4.562 + 4.402 + 5.352 𝑐𝑐 + 4.652 + 4.352
+ 4.352 + 4.852 + 5.152 + 4.952 = 503.5076
Sample variance, 𝑆 2 =
503.5076−
(109.34)2
24
24−1
= 𝟎. 𝟐𝟑𝟔𝟐
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √0.2362 = 𝟎. 𝟒𝟖𝟔𝟎
Problem 44
Range = 5.5 – 3.85 = 1.65
Range Approximation of S =
1.65
4
= 0.4125
Problem 45
∑ 𝑥 = 21 + 25 + 34 + 18 + 17 + 3 + 21 + 24 + 32 + 20 + 17 + 25 + 27 + 26 + 25
+ 12 + 10 + 23 + 30 + 9 + 24 + 19 + 16 + 17 + 26 + 23 + 17 + 29 = 590
∑ 𝑥 2 = 212 + 252 + 342 + 182 + 172 + 32 + 212 + 242 + 322 + 202 + 172 + 252 + 272
+ 262 + 252 + 122 + 102 + 232 + 302 + 92 + 242 + 192 + 162 + 172 + 262
+ 232 + 172 + 292 = 13800
Sample variance, 𝑆 2 =
(590)2
28
13800−
28−1
= 𝟓𝟎. 𝟔𝟔𝟏𝟕
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √50.6617 = 𝟕. 𝟏𝟏𝟕𝟕
Range = 34 – 3= 31
Range Approximation of S =
31
4
= 7.75
Problem 46
∑ 𝑥 = 175 + 189 + 167 + 178 + 184 + 166 + 176 + 177 + 159 + 169 + 163 + 154
+ 158 + 170 + 176 + 159 + 174 + 170 + 184 + 166 + 160 + 149 + 190
+ 177 + 168 + 165 + 160 + 162 + 178 + 168 + 182 + 177 + 177 + 170
= 5797
∑ 𝑥 2 = 1752 + 1892 + 1672 + 1782 + 1842 + 1662 + 1762 + 1772 + 1592 + 1692 + 1632
+ 1542 + 1582 + 1702 + 1762 + 1592 + 1742 + 1702 + 1842 + 1662 + 1602
+ 1492 + 1902 + 1772 + 1682 + 1652 + 1602 + 1622 + 1782 + 1682 + 1822
+ 1772 + 1772 + 1702 = 991585
2
Sample variance, 𝑆 =
991585−
(5797)2
34
34−1
= 𝟗𝟔. 𝟖𝟔𝟑𝟎
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √96.8630 = 𝟗. 𝟖𝟒𝟏𝟗
Problem 47
Range = 190 – 149 = 41
Range Approximation of S =
41
4
= 10.25
Problem 48
5797
Sample mean 𝑋̅ =
= 170.5
34
1 standard deviation is 9.8419
2 standard deviations is 19.6838
3 standard deviations is 29.3476
So between 170.5 - 9.8419 and 170.5 + 9.8419 would be the probability of being
within one standard deviation of the mean: between 160.6581 and 180.3419; and
there are 24 of the 34 values between those numbers or 0.7059 (which is a bit
greater than the basic 68%)
So between 170.5 - 19.6838 and 170.5 + 19.6838 would be the probability of
being within two standard deviation of the mean: between 150.8162 and
190.1838; and there are 33 of the 34 values between those numbers or 0.9706
(which is a bit greater than the basic 95%)
So between 170.5 - 29.3476 and 170.5 + 29.3476 would be the probability of
being within two standard deviation of the mean: between 140.9743 and
200.0251; and there are 34 of the 34 values between those numbers or 1 (which
is a bit greater than the basic 99.7%)
Problem 49
∑ 𝑥 = 87 + 76 + 113 + 98 + 145 + 123 + 95 + 134 + 45 + 78 = 994
∑ 𝑥 2 = 872 + 762 + 1132 + 982 + 1452 + 1232 + 952 + 1342 + 452 + 782 = 106962
Sample variance, 𝑆 2 =
(994)2
10
106962−
10−1
= 𝟗𝟎𝟔. 𝟒𝟗𝟏𝟕
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐷𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛, 𝑆 = √906.4917 = 𝟑𝟎. 𝟏𝟎𝟖𝟎
Problem 47
Range = 145 – 45 = 100
Range Approximation of S =
100
√10
= 31.6228