QUIZ 3 - Measures of Centre and Spread
Q1 Determine the mean and median of the following data.
-9



19
20
Mean:
Median:
Mode:
0
10
6.8
9
10
-2
9
10
4
(add all values and divide by # of values; 61/9)
(-9,-2,0,4,9,10,10,19,20; order values and pick middle #)
(# that is most common; can be multiple #s)
Q2 Determine the distribution of the data pictured below
Ans: Skewed-left
Q3 The mean (x-bar) of the 5 observations in the following table is 6.
∑x = 30 (add all the x values; -9+-4+10+1+32=30)
∑(x−x) = 0 (add all values from 2nd column)
∑(x−x)2 = 1042 (add all values from 3rd column)
Sample variance, s2 = 260.5 (sum of 3rd column / (n-1))
 1042/5-1 = 1042/4 = 260.5
Sample standard deviation, s = 16.14 (sqrt of s2)
 Sqrt(260.5) = 16.14
Q4 The table below gives the number of hours spent watching TV last week by a sample of 22
children.
23
13
13
25
10
16
36
20
35
38
31
11
33
19
19
18
39
38
17
11
32
9
Range:
30.00 (highest – lowest value; 39-9 = 30)
Sample variance:
108.19 (Excel: =VAR.S(highlight all values) = 108.19)
Sample standard deviation:
10.40 (Excel: =STDEV.S(highlight all values) = 10.40)
Q5 Given the following table of unsorted values, calculate the indicated locator and percentile.
Do not round your results.
Determine the locator for the 80th percentile, L80
 L80 = 64.8
34
55
35
23
38
75
12
71
𝑃
80
Lp = (n+1)100  L80 = (80+1)100 = 64.8
44
59
51
8
30
36
81
54
37
29
22
16
96
93
6
64
76
65
41
60
15
91
85
92
47
79
10
86
11
57
48
53
68
94
62
45
61
52
13
77
20
9
49
74
95
73
18
100
25
46
80
98
88
69
19
87
31
40
90
99
67
27
21
14
43
82
5
33
72
32
83
17
Find the 80th percentile, P80
 P80 = 81.8
P80 is between 64th and 65th values (after arranging data in
order); those are 81 and 82.
81+0.8(82-81) = 81.8
(The 0.8 is the decimal from the 64.8 from Lp)
Approximately, what % of the scores in a dataset are
below the 80th percentile?
 Ans: 80%
Q6 The Acme Company manufactures widgets. The distribution of widget weights is bellshaped. The widget weights have a mean of 47 ounces and a standard deviation of 7 ounces.
Use the Empirical Rule and a sketch of the normal distribution in order to answer these
questions.
68% of the widget weights lie between 40 and 54
 68% of the data is within one standard deviation of the mean; x-bar ± s
 47 ± 7  47-7 = 40 & 47+7 = 54
95% of the widget weights lie between 33 and 61
 95% of the data is within two standard deviations of the mean; x-bar ± 2s
 47 ± 2(7)  47-14 = 33 & 47+14 = 61
99.7% of the widget weights lie between 26 and 68
 99.7% of the data is within three standard deviations of the mean; x-bar ± 3s
 47 ± 3(7)  47-21 = 26 & 47+21 = 68