Question

1- Suppose there are only 38 students in advance statistics class and final scores of these 38 students are: 10 34 45 55 63 75 class lit 23 34 49 56 64 82 24 36 50 58 65 82 26 41 51 59 71 95- 32 41 52 60 73 33 45 54 62 74 s lin 46 54 63 74 a- Construct grouped frequency distribution table for this dataset with relative and cumulative frequencies (with closed intervals). b- Find mean for this grouped dataset. c- Find median and mode for this grouped dataset. d- Find standard deviation. e- Draw a histogram for this grouped dataset. f- Draw ogive for the grouped dataset. What does cumulative frequency in 3rd d.umalative frequeney in Jrd class indicate? g-According to Chebyshev's Theorem, in which range (or interval) does % of the data values change in?

Answer 1

(a)
CI	f	Relative f	Cumulative f
10 - 30	4	0.105	4
30 - 50	11	0.289	15
50 - 70	15	0.395	30
70 - 90	7	0.184	37
90 - 110	1	0.026	38
Sums =	38	1
(b)
x	f	fx
20	4	80
40	11	440
60	15	900
80	7	560
100	1	100
Sums =	38	2080
Mean = ∑fx/∑f = 2080/38 = 54.74
(c) Median class is 50 - 70
L = 50, N/2 = 19, CF = 15, F = 15, H = 20
Median = L + ((N/2) - CF) * H/F = 50 + (19 - 15) * 20/15 = 55.33
Modal class is 50 - 70
L = 50, F1 = 15, F0 = 11, F2 = 7, H = 20
Mode = L + ((F1 - F0)/(2F1 - F0 - F2)) * H = 50 + ((15 - 11)/(2 * 15 - 11 - 7)) * 20 = 60
(d)
x	f	d = x - Mean	d^2	fd^2
20	4	-34.74	1206.868	4827.470
40	11	-14.74	217.268	2389.944
60	15	5.26	27.668	415.014
80	7	25.26	638.068	4466.473
100	1	45.26	2048.468	2048.468
Sums =	38			14147.37
Standard deviation = √[∑fd^2 / ∑f] = √ [14147.37/38] =					19.295

(e) Histogram

(f)
Less than	CF
30	4
50	15
70	30
90	37
110	38

Cumulative frequency in the third class indicates the number of students who scored less than 70
(g) 1 - (1/k)^2 = 0.75
Solving, k = 2
Mean - 2 * Standard deviation = 54.74 - 2 * 19.295 = 16.15
Mean + 2 * Standard deviation = 54.74 + 2 * 19.295 = 93.33
The interval is [16.15, 93.33]