Question

X	f
0-9	3
10-19	14
20-29	11
30-39	5

1. The mean [3 marks]
2. The median [4 marks]
3. The mode [4 marks]
4. The first quartile [4 marks]
5. The standard deviation.

Answer 1

Solution x-A x - 24.5 10 | d= 1 Class Frequency ( Mid value (r) = 24.5, h = 10 (5) = (2) x (4) (6)-(5) x (4) (7) 0-9 10 19 20 - 29 30 - 39 4.5 14.5 24.5 A 34.5 -6 12 14 -14 17 28 233 2- 31 Part (a) 15 = 24.5+-- .10 - 24.5+ - 0.4545 10 - 24.5+ - 4.5455 - 19.9545

Part (b) To find Median Class er value of(33)-observation value of - value of 16 observation From the column of cumulative frequency cf, we find that the 16 observation lies in . The median class is 9.5 195 Now, 10-19 lower boundary point of median class95 n Total frequency33 fCumulative frequency of the class preceding the median class3 Frequency of the median class14 class length of median class10 Median M Ic 16-3 -95-10 13 14 95 10 -9.5 9.2857 - 19.1429

Part (c) To find Mode Class Here, maximum frequency is 14 . The mode class is 9.5-195 Ilower boundary point of mode class9.3 frequency of the mode class14 frequency of the preceding class3 = frequency of the succedding class = 11 class length of mode class10 14-3 2 14-3-11 9.5- 10 14 9.5 + 73571 - 17.3571

Part (e) Population Standard deviation ơ-11- k -15)2 31- 10 31 6.8182 24.1318 0.7328 10 -0.856-10 -8.5603