Question Set E

CBSE 11th Mathematics Statistics Model Questions Papers Study Material - QB365

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Questions:

Generic 5

12 x 5 = 60

Find the mean deviation about the median for the data :
36, 72, 46, 42, 60, 45, 53, 46, 51, 49

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

Find the mean deviation about median for the following data:

Marks	0-10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Number of Girls	6	8	14	16	4	2

Calculate the mean deviation about median age for the age distribution of 100 persons gives below:

Age	16-20	21-25	26-30	31-35	36-40	41-45	46-50	51-55
Number	5	6	12	14	26	12	16	9

[Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

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

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 variance for each of the data :

x_i	92	93	97	98	102	104	109
f_i	3	2	3	2	6	3	3

Find the C.V of the following data:

Size (in m)	10-15	15-20	20-25	25-30	30-35	35-40
No.of items	2	8	20	35	20	15

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

10.

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.

11.

The mean and standard deviation of 20 observation are found to be 10 and 2 respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted
(ii) If it is replaced by 12.

12.

The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, physics and chemistry are given below:

Subject	Mathematics	Physics	Chemistry
Mean	42	32	40.9
Standard deviation	12	15	20

Which of the three subjects shows the highest variability in marks and which shows the lowest?

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Answers Key:

Generic 5

12 x 5 = 60

Arrange the data in ascending order, we have
36,42,45,46,46,49,51,53,60,72
Here n = 10 (which is even)
So median is average of 5th and 6th observation
∴ Median=\(\frac{46+49}{2}=\frac{95}{2}=47.5\)

x_i	\|x_i-M\|
36	11.5
42	5.5
45	2.5
46	1.5
46	1.5
49	1.5
51	3.5
53	5.5
60	12.5
72	24.5
Total	70

M.D. about median=\(\frac { 1 }{ n } \sum _{ i=1 }^{ n }{ \left| { x }_{ i }-M \right| } \)
\(=\frac{1}{10}\times70=7\)

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

Marks	Mid values x_i	f_i	c.f	\|x_i- 27.86\|	f_i\|x_i- 27.86\|
0-10	5	6	6	22.86	137.16
10 - 20	15	8	14	12.86	102.88
20 - 30	25	14	28	2.86	40.04
30 - 40	35	16	44	7.14	114.24
40 - 50	45	4	48	17.14	68.56
50 - 60	55	2	50	27.14	54.28
		50			517.16

\(\frac{N}{2}=\frac{50}{2}=25\)∴ Median class is 20 - 30
∴ Median=\(20+\frac { 25-14 }{ 14 } \times 10=20+7.86=27.86\)
M.D. about median =\(\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 50 } \times 517.16=10.34\)

Age	Mid values x_i	f_i	c.f	\|x_i-38\|	f_i\|x_i-38\|
16-20	18	5	5	20	100
21-25	23	6	11	15	90
26-30	28	12	23	10	120
31-35	33	14	37	5	70
36-40	38	26	63	0	0
41-45	43	12	75	5	60
46-50	48	16	91	10	160
51-55	53	9	100	15	135
		100			735

\(\frac{N}{2}=\frac{100}{2}=50\)
∴ Median class is 35.5-40.5
∴ Median =\(35.5+\frac { 50-37 }{ 26 } \times 5=35.5+2.5=38\)
M.D. about median
\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-M \right| } =\frac { 1 }{ 100 } \times 735=7.35\)

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

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

x_i	f_i	f_ix_i	(x_i-100)	(x_i-100)²	f_i(x_i-100)²
92	3	276	-8	64	192
93	2	186	-7	49	98
97	3	291	-3	9	27
98	2	196	-2	4	8
102	6	612	2	4	24
104	3	312	4	16	48
109	3	327	9	81	243
	22	2200			640

Mean \((\overline { x } )=\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 22 } \times 2200=100\)
Variance = \({ \sigma }^{ 2 }=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }{ \left( { x }_{ i }-\overline { x } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 22 } \times 640=29.09\)

Size	Mid Value x_i	f_i	\(u=\frac { x-27.5 }{ 5 } \)	fu	fu²
10-15	12.5	2	-3	-6	18
15-20	17.5	8	-2	-16	32
20-25	22.520	-1	-20	20	25-30
25-30	27.5	35	0	0	0
30-35	32.5	20	1	20	20
35-40	37.5	15	2	30	60
		100		8	150

Mean \(\left( \bar { x } \right) =A+\frac { \sum { fu } }{ N } \times h=27.5+\frac { 8 }{ 100 } \times 5=27.5+0.4=27.9\)
Standard deviation \(\left( \sigma \right) \) \(=\frac { h }{ N } \sqrt { N\sum { { fu }^{ 2 }-{ (\sum { fu) } }^{ 2 } } } \)
\(\sigma =\frac { 5 }{ 100 } \sqrt { 100\times 150-{ (8) }^{ 2 } } =\frac { 1 }{ 20 } \sqrt { 15000-64 } =\frac { 1 }{ 24 } \times 122.21=6.11\)
\(\therefore \ C.V.=\frac { \sigma }{ x } \times 100=\frac { 6.11 }{ 27.9 } \times 100=21.89\)

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.

10.

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.

11.

Here n = 20, \(\bar { x } \) = 10 and \(\sigma \) = 2
\(\therefore \bar { x } =\frac { 1 }{ n } \sum { { x }_{ i } } \Rightarrow \sum { { x }_{ i } } =n\times \bar { x } =20\times 10=200\)
\(\therefore\) Incorrect \(\sum { { x }_{ i } } =200\)
Now \(\frac { 1 }{ n } \sum { { x }_{ i }^{ 2 }-{ (\bar { x } ) }^{ 2 }=4 } \)
\(\Rightarrow \frac { 1 }{ 20 } \sum { { x }_{ i }^{ 2 } } -{ (10) }^{ 2 }=4\quad \Rightarrow \sum { { x }_{ i }^{ 2 } } =2080\)
(i) If wrong item is omitted.When wrong item 8 is omitted from the data then we have 19 observations.
\(\therefore \ Correct\ \sum { { x }_{ i }=Incorrect\ \sum { { x }_{ i } } -8 } \)
\(Correct\quad \sum { { x }_{ i }=200-8=192 } \)
\(\therefore Correct\quad mean=\frac { 192 }{ 19 } =10.1\)
Also correct \(\sum { { x }_{ i }^{ 2 }=Incorrect\quad \sum { { x }_{ i }^{ 2 }-{ (8) }^{ 2 } } } \)
\(\Rightarrow\) Correct \(\sum { { x }_{ i }^{ 2 } } =2080-64=2016\)
\(\therefore\) correct variance
\(=\frac { 1 }{ 19 } (correct\ \sum { { x }_{ i }^{ 2 } } )-{ (correct\ mean) }^{ 2 }\)
\(=\frac { 1 }{ 19 } \times 2016-{ \left( \frac { 192 }{ 19 } \right) }^{ 2 }=\frac { 2016 }{ 19 } -\frac { 36864 }{ 361 } \)
\(=\frac { 38304-36864 }{ 361 } =\frac { 1440 }{ 361 } \)
Correct S.D. = \(\sqrt { \frac { 1440 }{ 361 } } =\sqrt { 3.99 } =1.997\)
(ii) If it is replaced by 12
When wrong item 8 is replaced by 12
\(\therefore\) Correct \(\sum { { x }_{ i } } \) = Incorrect \(\sum { { x }_{ i } } \) - 8 + 12
= 200 - 8 + 12 = 204
\(\therefore \ Correct\ mean=\frac { 204 }{ 20 } =10.2\)
Also correct \(\sum { { x }_{ i }^{ 2 } } \) = Incorrect \(\sum { { x }_{ i }^{ 2 } } \) - (8)² + (12)²
= 2080 - 64 + 144 = 2160
\(\therefore\) Correct variance
\(=\frac { 1 }{ 20 } \left( correct\quad \sum { { x }_{ i }^{ 2 } } \right) -({ correct\quad mean) }^{ 2 }\)
\(=\frac { 2160 }{ 20 } -{ \left( \frac { 204 }{ 20 } \right) }^{ 2 }=\frac { 2160 }{ 20 } -\frac { 41616 }{ 400 } \)
\(=\frac { 43200-41616 }{ 400 } =\frac { 1584 }{ 400 } \)
Correct S.D. = \(\sqrt { \frac { 1584 }{ 400 } } =\sqrt { 3.96 } =1.989\)

12.

For Mathematics
\(\bar { x } =42\ and\ \sigma =12\)
\(\therefore \ C.V\ of\ Mathematics=\frac { 12 }{ 42 } \times 100=28.57%\)%
For Physics
\(\bar { x } =32\ and\ \sigma =15\)
\(\therefore \ C.V.\ of\ physics=\frac { 15 }{ 32 } \times 100=46.88\)%
For chemistry
\(\bar { x } =40.9\ and\ \sigma =20\)
\(\therefore C.V.\ of\ Chemistry=\frac { 20 }{ 40.9 } \times 100=48.9\)%
Thus chemistry with hightest C.V. shows highest variability and mathematics with lowest C.V. shows lowest variability.