Solution | unit 3 statistics

Math

Unit-3StatisticsQuestion-1: Find the arithmetic mean for the following distribution:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	7	8	20	10	5

Solution. Let assumed mean (a) = 25

Class

Midvalue

Frequency

40— 50

Total

Question-2: Find the median of the following dataset-

Sol.

Class interval	Frequency	Cumulative frequency
0 - 10	3	3
10 – 20	5	8
20 – 30	7	15
30 – 40	9	24
40 – 50	4	28

So that median class is 20-30.Now putting the values in the formula-

So that the median is 28.57Question-3: Find the mode from the following data:

Age	0-6	6-12	12-18	18-24	24-30	30-36	36-42
Frequency	6	11	25	35	18	12	6

Solution.

Age

Frequency

Cumulative frequency

0-6

6-12

12-18

24-30

30-36

36-42

107

113

Mode

Question-4: Find the quartile deviation of the following data-

Class interval	0-10	10-20	20-30	30-40	40-50
Frequency	3	5	7	9	4

Sol.Here N/4 = 28/4 = 7 so that the 7’th observation falls in class 10 – 20.And 3N/4 = 21, and 21’st observation falls in the interval 30 – 40 which is the third quartile.The quartiles can be calculated as below-

And

Hence the quartile deviation is-

Question-5: Find the mean deviation from the mean of the following data-

Class interval	0-6	6-12	12-18	18-24	24-30
Frequency	8	10	12	9	5

Sol.

Class interval	Mid-value	Frequency	d = x - a	f.d	\|x - 14\|	f \|x - 14\|
0-6	3	8	-12	-96	11	88
6-12	9	10	-6	-60	5	50
12-18	15	12	0	0	1	12
18-24	21	9	6	54	7	63
24-30	27	5	12	60	13	65
Total		44		-42		278

Then mean deviation from mean-

Question-6: Calculate the standard deviation of the following frequency distribution-

Weight	60 – 62	63 – 65	66 – 68	69 – 71	72 – 74
item	5	18	42	27	8

Sol.

Weight	Item (f)	X	d = x – 67	f.d
60 – 62	5	61	-6	-30	180
63 – 65	18	64	-3	-54	162
66 – 68	42	67	0	0	0
69 – 71	27	70	3	81	243
72 – 74	8	73	6	48	288
Total	100			45	873

Question-7: Calculate coefficient variation for the following frequency distribution.

Wages in Rupees earned per day	0-10	10-20	20-30	30-40	40-50	50-60
No. of Labourers	5	9	15	12	10	3

Solution:

We already calculated

Now,

A.M

Coefficient of Variation

Question-8: Calculate Karl Pearson’s coefficient of skewness of marks obtained by 150 students.

Sol. The mode is not well defined so that first we calculate mean and median-

Class	f	x	CF		fd
0-10	10	5	10	-3	-30	90
10-20	40	15	50	-2	-80	160
20-30	20	25	70	-1	-20	20
30-40	0	35	70	0	0	0
40-50	10	45	80	1	10	10
50-60	40	55	120	2	80	160
60-70	16	65	136	3	48	144
70-80	14	75	150	4	56	244

Now,

And

Standard deviation-

Then-

Question-9: By the method of least squats, find the straight line that best fits the following data:

	1	2	3	4	5
	14	27	40	55	68

Solution: Let the equation of the straight line best fit be

…. (1)


1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25

Normal equations are

Putting the values of

in (2) and (3), we have

On solving (4) and (5), we get

On Substituting the values of a and b in (1), we get

Question-10: Fit a second-degree parabola to the following data by the least-squares method.

	1929	1930	1931	1932	1933	1934	1935	1936	1937
	352	356	357	358	360	361	361	360	359

Solution: Taking

Taking

The equation

is transformed into


1929	-4	352	-5	20	16	-80	-64	256
1930	-3	360	-1	3	9	-9	-27	81
1931	-2	357	0	0	4	0	-8	16
1932	-1	358	1	-1	1	1	-1	1
1933	0	360	3	0	0	0	0	0
1934	1	361	4	4	1	4	1	1
1935	2	361	4	8	4	16	8	16
1936	3	360	3	9	9	27	27	81
1937	4	359	2	8	16	32	64	256
Total

Normal equations are

On solving these equations, we get

Question-11: Find the correlation coefficient between age and weight of the following data-

Age	30	44	45	43	34	44
Weight	56	55	60	64	62	63

Sol.

X	Y					())
30	56	-10	100	-4	16	40
44	55	4	16	-5	25	-20
45	60	5	25	0	0	0
43	64	3	9	4	16	12
34	62	-6	36	2	4	-12
44	63	4	16	3	9	12
Sum= 240	360	0	202	0	70	32

Karl Pearson’s coefficient of correlation-

Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that as age increases, the weight also increasesQuestion-12: Compute the Spearman’s rank correlation coefficient of the dataset given below-

Person	A	B	C	D	E	F	G	H	I	J
Rank in test-1	9	10	6	5	7	2	4	8	1	3
Rank in test-2	1	2	3	4	5	6	7	8	9	10

Sol.

Person	Rank in test-1	Rank in test-2	d =
A	9	1	8	64
B	10	2	8	64
C	6	3	3	9
D	5	4	1	1
E	7	5	2	4
F	2	6	-4	16
G	4	7	-3	9
H	8	8	0	0
I	1	9	-8	64
J	3	10	-7	49
Sum				280

Question-13: Find the regression line of y on x for the given dataset.

X	4.3	4.5	5.9	5.6	6.1	5.2	3.8	2.1
Y	12.6	12.1	11.6	11.8	11.4	11.8	13.2	14.1

Sol.Let y = a + bx is the line of regression of y on x, where ‘a’ and ‘b’ are given as-

We will make the following table-

x	y	Xy
4.3	12.6	54.18	18.49
4.5	12.1	54.45	20.25
5.9	11.6	68.44	34.81
5.6	11.8	66.08	31.36
6.1	11.4	69.54	37.21
5.2	11.8	61.36	27.04
3.8	13.2	50.16	14.44
2.1	14.1	29.61	4.41
Sum = 37.5	98.6	453.82	188.01

Using the above equations we get-

On solving these both equations, we get-a = 15.49 and b = -0.675So that the regression line is – y = 15.49 – 0.675x

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