Statistics

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Questions:

Generic 3

5 x 3 = 15

Find the variance and standard deviation for the following data:
65, 68, 58, 44, 48, 45, 60, 62, 60, 50

Calculate the mean deviation from the median of the following data

Wages per week in Rs	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of workers	4	6	10	20	10	6	4

Find the mean and standard deviation for the following data.

Age(in years)	Number of teachers
25-30 30-35 35-40 40-45 45-50 50-55	30 23 20 14 10 3

Find the mean and standard deviation of the following frequency distribution

x_i	6	10	14	18	24	28	30
f_i	2	4	7	12	8	4	3

Calculate the mean deviation from the mean of the following distribution.

Marks	0-10	10-20	20-30	30-40	40-50
Numbers of students	5	8	15	16	6

Generic 5

6 x 5 = 30

The mean and variance of eight observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

From the prices of shares X and Y below, find out which is more stable in value:

X	35	54	52	53	56	58	52	50	51	49
Y	108	107	105	105	106	107	104	103	104	101

Find the mean and variance for each of the data :

x_i	6	10	14	18	24	28	30
f_i	2	4	7	12	8	4	3

Find the mean and standard deviation of the following distribution:

Marks	20-30	30-40	40-50	50-60	60-70	70-80	80-90
Number of students	3	6	13	15	14	5	4

10.

Find the mean deviation about the mean for the data

Height in cm	95-105	105-115	115-125	125-135	135-145	145-155
Number of boys	9	13	26	30	12	10

11.

Find the mean deviation about the mean for the data

Income per day	0-100	100-200	200-300	300-400	400-500	500-600	600-700	700-800
Number of persons	4	8	9	10	7	5	4	3

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Answers Key:

Generic 3

5 x 3 = 15

\(x={65+68+58+44+48+45+60+62+60+50\over 10}={560\over 10}=56\)

x	x-56	(x-56)²
65	9	81
58	2	4
68	12	144
44	-12	144
48	-8	64
45	-11	121
60	4	16
62	6	36
60	4	16
50	-6	36
		662

Now variance = \({1\over n}(\sum x-\bar x)^2={662\over10}=66.2\)
S.D. = \(\sqrt{variance}=\sqrt{66.2}=8.13\)

Wages per weeks in Rs	Mid value x	Frequency f	.f.	\|x-45\|	f\|x-45\|
10-20	15	4	4	30	120
20-30	25	6	10	20	120
30-40	35	10	20	10	100
40-50	45	20	40	0	0
50-60	55	10	50	10	100
60-70	65	6	56	20	120
70-80	75	4	60	30	120
		60			680

Here N = 60 \(∴\ {N\over 2}={60\over 2}=30\)
∴ Median class is 40 - 50
∴ Median = \(40+{(30-20)\over 20}\times10=40+5=45\)
∴ Mean deviation from me\({f|x-45|\over N}={680\over60}=11.33\)dian =

Let assume mean, a=425
Let us make the table from the given data.

Class	Mid value (x_i)	f_i	\({ u }_{ i }=\frac { { x }_{ i }-A }{ 5 } \)	\({ u }_{ i }^{ 2 }\)	f_iu_i	f_i\({ u }_{ i }^{ 2 }\)
25-30 30-35 35-40 40-45 45-50 50-55	27.5 32.5 37.5 42.5 47.5 52.5	30 23 20 14 10 3	-3 -2 -1 0 1 2	9 4 1 0 1 4	-90 -46 -20 0 10 6	270 92 20 0 10 12
Total		\(\sum { { f }_{ i }=100 } \)			\(\sum { { u }_{ i }^{ 2 }=-140 } \)	\(\sum { { { f }_{ i }u }_{ i }^{ 2 }=404 } \)

\(\therefore \ Mean,\bar { x } =a+\frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i }\times h=425+\frac { 1 }{ 100 } (-140)\times 5 } \)
=35.5 yr
\(Standard\ deviation=[\sqrt { \frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i }^{ 2 } } -(\frac { 1 }{ N } \sum { { f }_{ i }{ u }_{ i } } )^{ 2 } } ]\times h\)
\( =[\sqrt { \frac { 404 }{ 100 } -(\frac { -140 }{ 100 } )^{ 2 }] } \times 5=7.21\ yr\)

Let us make the following table from the given data.

x_i	f_i	f_ix_i	\(\left( { x }_{ i }-\bar { x } \right) =\left( { x }_{ i }-19 \right) \)	\(\left( { x }_{ i }-\bar { x } \right) ^{ 2 }\)	\({ f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 }\)
6 10 14 18 24 28 30	2 4 7 12 8 4 3	12 40 98 216 192 112 90	-13 -9 -5 -1 5 9 11	169 81 25 1 25 81 121	338 324 175 12 200 324 363
	\(\sum { { f }_{ i } } =40\)	\(\sum { { f }_{ i }{ x }_{ i }=760 } \)			\(\sum { { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 }=1736 } \)

\(\therefore \ (\bar { x } )=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ N } =\frac { 760 }{ 40 } =19\)
\(and\ standard\ deviation=\sqrt { \frac { 1 }{ N } [\sum { { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 }] } } \)
\(=\sqrt { \frac { 1736 }{ 40 } } =\sqrt { 434 } =6.59\)
Ans.19,6.59

We make the table from the given data.

Class	Mid value(x_i)	f_i	\({ u }_{ i }=\frac { { x }_{ i }-45 }{ 10 } \)	f_iu_i	\(\|{ x }_{ i }-\bar { x } \|=\|\bar { x } -27\|\)	\({ f }_{ i }\|{ x }_{ i }-\bar { x } \|\)
0-10	5	5	-2	-10	22	110
10-20	15	8	-1	-8	12	96
20-30	25	15	0	0	2	30
30-40	35	16	1	16	8	128
40-50	45	6	2	12	18	108
Total		50		10		472

Here, \(\sum { f_{ i }=50,\quad a=25 }\)
\(\sum { { f }_{ i }{ u }_{ i }=10,\quad h=10 } \)
\(\therefore Mean,\bar { x } =a+\frac { \sum { f_{ i }{ u }_{ i } } }{ \sum { f_{ i } } } \times h=25+\frac { 10 }{ 50 } \times 10=27\)
\(\therefore\) Mean deviation from median = \(\frac { \sum { f_{ i }|{ x }_{ i }-\bar { x } | } }{ \sum { f_{ i } } } \)
\([\because \sum { f_{ i }|{ x }_{ i }-\bar { x } } ]=472\)
\(=\frac { 472 }{ 50 } =9.44\)

Generic 5

6 x 5 = 30

Let two remaining observations be x and y. Then
\(\frac { 6+7+10+12+12+13+x+y }{ 8 } =9\)
\(\therefore\) 60 + x + y = 72 \(\Rightarrow\) x + y = 12 ......(i)
Also \(\frac { 1 }{ 8 } ({ 6 }^{ 2 }+{ 7 }^{ 2 }+{ 10 }^{ 2 }+{ 12 }^{ 2 }+{ 12 }^{ 2 }+{ 13 }^{ 2 }+{ x }^{ 2 }+{ y }^{ 2 })-{ (9) }^{ 2 }=9.25\)
\(\Rightarrow \frac { 1 }{ 8 } (36+49+100+144+144+169+{ x }^{ 2 }+{ y }^{ 2 })-81=9.25\)
\(\Rightarrow\) 642 + x² + y² = 722
\(\Rightarrow\) x² + y² = 80 ......(ii)
Now (x+y)² + (x-y)² = 2(x²+y²)
\(\Rightarrow\) (12)²+ (x - y)² = 2 x 80
\(\Rightarrow\) (x - y)² = 160 - 144
\(\Rightarrow\) (x - y)² = 16 \(\Rightarrow\) x - y = \(\pm \) 4
When x - y = 4
Solving x + y = 12 and x - y = 4 we get x = 8 and y = 4
When x - y = -4
Solving x + y = 12 and x - y = -4 we get x = 4 and y = 8.

X	Y	(X-\(\bar { X } \))	(Y-\(\bar { Y } \))	(X-\(\bar { X } \))²	(Y-\(\bar { Y } \))²
35	108	-16	3	256	9
54	107	3	2	9	4
52	105	1	0	1	0
53	105	2	0	4	0
56	106	5	1	25	1
58	107	7	2	49	4
52	104	1	-1	1	1
50	103	-1	-2	1	4
51	104	0	-1	0	1
49	101	-2	-4	4	16
510	1050			350	40

\(\bar { x } =\frac { 510 }{ 10 } =51,\ \bar { y } =\frac { 1050 }{ 10 } =105\)
\({ \sigma }_{ x }=\sqrt { \frac { \sum { ({ x-\bar { x } ) }^{ 2 } } }{ n } } =\sqrt { \frac { 350 }{ 10 } } =5.92\)
\({ \sigma }_{ y }=\sqrt { \frac { \sum { { (y-\bar { y } ) }^{ 2 } } }{ n } } =\sqrt { \frac { 40 }{ 10 } } =2\)
\(C.V.\ of\ x=\frac { 5.92 }{ 51 } \times 100=11.61\)
C.V. of y=\(\frac { 2 }{ 105 } \times 100=1.9\)
C.V. or Y < c.V. of X
Thus prices of share Y are more stable.

x_i	f_i	f_ix_i	(x_i-19)	(x_i-19)²	f_i(x_i-19)²
6	2	12	-13	169	338
10	4	40	-9	81	324
14	7	98	-5	25	175
18	12	216	-1	1	12
24	8	192	5	25	200
28	4	112	9	81	324
30	3	90	11	121	363
	40	760			1736

Mean \((\overline { x) } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 40 } \times 760=19\)
Variance = \(\\ { \sigma }^{ 2 }=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }{ \left( { x }_{ i }-\overline { x } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 40 } \times 1736=43.4\)

Marks	Mid value x	f	\(u=\frac { x-55 }{ 10 }\)	fu	fu²
20-30	25	3	-3	-9	27
30-40	35	6	-2	-12	24
40-50	45	13	-1	-13	13
50-60	55	15	0	0	0
60-70	65	14	1	14	14
70-80	75	5	2	10	20
80-90	85	4	3	12	36
		60		2	134

Mean\(\left( \overline { x } \right) =A+\frac { \Sigma fu }{ N } \times h=55+\frac { 2 }{ 60 } \times 10=55+0.33=5.33\)
\(S.D.\left( \sigma \right) =\frac { \sqrt { N\Sigma { fu }^{ 2 }-{ (\Sigma fu) }^{ 2 } } }{ N } \times h=\frac { \sqrt { 60\times 130-{ (2) }^{ 2 } } }{ 60 } \times 10\)
\(=\frac { \sqrt { 8036 } }{ 60 } \times 10=\frac { 89.64 }{ 60 } \times 10=14.94\)

10.

Height in cms	Mid values x_i	f_i	f_ix_i	\|x_i-125.3\|	f_i\|x_i-125.3\|
95 - 105	100	9	900	25.3	227.7
105 - 115	110	13	1430	15.3	198.9
115 - 125	120	26	3120	5.3	137.8
125 - 135	130	30	3900	4.7	141
135 - 145	140	12	1680	14.7	176.4
145 - 155	150	10	1500	24.7	247
		100	12530		1128.8

Mean \(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 100 } \times 12530=125.3\)
Mean deviation about mean=\(\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } =\frac { 1 }{ 100 } \times 1128.8=11.28\)

11.

Income per day	Mid values x_i	f_i	f_ix_i	\|x_i-358\|	f_i\|x_i-358\|
0-100	50	4	200	308	1232
100-200	150	8	1200	208	1664
200-300	250	9	2250	108	972
300-400	350	10	3500	8	80
400-500	450	7	3150	92	644
500-600	550	5	2750	192	960
600-700	650	4	2600	292	1168
700-800	750	3	250	392	1176
		50	17900		7896

Mean\(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 50 } \times 17900=358\)
Mean deviation about mean\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{50}\times7896 = 157.92\)