-
7/28/2019 Answer Scheme Exercise Topic 5
1/5
Mathematics 2
MTH1022
Page 1 of5
Topic 5Measures of central tendency and measures of
dispersion
Ungrouped Data:
The quantities ordered (in million units) for the first 10 weeks
for a certain product in Maziah Company aregiven as follows:
14.25 19.00 11.00 28.00 24.00 23.00 43.20 14.00 27.00 25.00
Find:
a. Mean
b. Median
11.00 14.00 14.25 19.00 23.00 24.00 25.00 27.00 28.00 43.20
c. Mode
d. Range e. Variance
x x^2
11.00 121
14.00 196
14.25 203.0625
19.00 36123.00 529
24.00 576
25.00 625
27.00 729
28.00 784
43.20 1866.24
228.45 5990.30
()
N
xx
N
2
22 1
-
7/28/2019 Answer Scheme Exercise Topic 5
2/5
Mathematics 2
MTH1022
Page 2 of5
f. Standard deviation
g. What is the shape of the distribution?Mean < Median22.85
< 23.5 negatively skewed or skew to the left.
Grouped Data (with class interval):
The table below shows the age of cars (in years) sold by a
second-hand car dealer for the last two months:
Age of carNumber of
cars, fcumulative
frequency, Fmidpoint, x fx x^2 fx^2
14 16 16 2.5 40 6.25 100
58 20 36 6.5 130 42.25 845
912 28 64 10.5 294 110.25 3087
1316 24 88 14.5 348 210.25 5046
1720 16 104 18.5 296 342.25 5476
2124 11 115 22.5 247.5 506.25 5568.75
2528 5 120 26.5 132.5 702.25 3511.25
120 1488 1919.75 23634
Find:
a. Mean
b. MedianMedian class =
(observation)
median class = 912 (8.512.5)
c. ModeMode class = 912 (8.512.5)
ffxX
-
7/28/2019 Answer Scheme Exercise Topic 5
3/5
Mathematics 2
MTH1022
Page 3 of5
()() d. Range e. Variance
f. Standard deviation
g. What is the shape of the distribution?Mean = 12.4Median =
11.93Mode = 11.17
Mode < Median < Mean positively skewed or skew to the
right.
f
fxfx
f
2
22 1
-
7/28/2019 Answer Scheme Exercise Topic 5
4/5
Mathematics 2
MTH1022
Page 4 of5
Grouped Data (without class interval):
The number of cigarettes consumed per day for a random sample of
80 respondents is given in the tablebelow:
Number of
cigarettes, x
Number of
smokers, f
cumulative
frequency fx x^2 fx^2
12 5 5 60 144 720
13 10 15 130 169 1690
14 20 35 280 196 3920
15 10 45 150 225 2250
16 15 60 240 256 3840
17 10 70 170 289 2890
18 5 75 90 324 1620
19 5 80 95 361 1805
80 1215 18735
Find:
a. Mean
b. MedianMedian class =
(observation)
Median = 15
c. ModeMode class = 20 (the highest frequency)Mode = 14
d. RangeRange = max valuemin value = 1912 = 7
e. Variance
f. Standard deviation
f
fxX
f
fxfx
f
2
22 1
-
7/28/2019 Answer Scheme Exercise Topic 5
5/5
Mathematics 2
MTH1022
Page 5 of5
g. What is the shape of the distribution?Mean = 15.19Median =
15Mode = 14
Mode < Median < Mean positively skewed or skew to the
right.
