Unit-3

Statistics

Question-1: Find the arithmetic mean for the following distribution:

Solution. Let assumed mean (a) = 25

Class	Midvalue	Frequency
40— 50
Total

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

Sol.

So that median class is 20-30.

Now putting the values in the formula-

So that the median is 28.57

Question-3: Find the mode from the following data:

Solution.

Age	Frequency	Cumulative frequency
0-6 6-12 12-18 24-30 30-36 36-42	6 11 25 35 12 6	6 17 42 77 95 107 113

Mode

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

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-

Sol.

Then mean deviation from mean-

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

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.

Solution:

We already calculated

Now, 

A.M    

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-

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 increases

Question-12: Compute the Spearman’s rank correlation coefficient of the dataset given below-

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.

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 = -0.675

So that the regression line is –

y = 15.49 – 0.675x 

Question-14: Discuss the Reliability of Regression Estimates:

Solution:

For A,

	45	38	59	64	72
	2025	1444	3481	4096	5184

For B,

	60	48	82	93	45
	2025	1444	3481	4096	5184

Now,

	45	38	59	64	72
	60	48	82	93	45
	2700	1824	4838	5952	3240

The standard error of Regression of estimates of y on x is

…..(Standard error of Regression of estimates of y on x is )