Statistics Class 11
Master Statistics Class 11 with comprehensive NCERT Solutions, Practice Questions, MCQs, Sample Papers, Case Based Questions, and Video lessons.
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Statistics Class 11 – NCERT Solutions
Each question below opens its complete step-by-step Teachoo solution.
Ex 13.1
12 questionsEx 13.1, 1
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Ex 13.1, 1
Find the mean deviation about the mean for the data
4,7, 8,9, 10, 12, 13, 17
Mean of the given data = —Sumof ali terms _
Total number of terms
eae ttZ +8494 10412 413 417
ee
_ 20 4 4-10=-6 |-6|=6
8 7 7-10=-3 |-3|=3
=10 8 8-10=-2 |-2|=2
9 9-10=-1 |-1]=1
Mean deviation about mean 10 10-10=0 10| =0
42 12-10=2 |2] =2
_ De, 7 2 13 13-10=3 |3| =3
8 17. 17-10=7 |7| =7
24 8
8 Y-#l= 24
=3 1
Ex 13.1, 2
Ex 13.1, 2 teachoo.com
Find the mean deviation about the mean for the data
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data = Fotet neeber oy teens
= 3B 470 448 + 40 4 42 + 55 + 63 + 46-4 54 4 44
° Era
_ 500 38 = 38-50=-12 |-12|=12
70 70 70-50=20 |20] = 20
= 50 48 48-50=-2 = |-2|=2
40 40-S0=-10 |—-10|/=10
Mean deviation about mean 42 42—50=-8 |-8]=8
55 §5-50=5 [5|=5
=k 63 63-50=13 = [13| =13
os 46 46-50=-4 |-4| =4
“To 54 54-S0=4 |4|=4
=8.4 44 44-50=-6 |-6|=6
10
Yix-zl- 34
1
Ex 13.1, 3
Ex 13. 1, 3 teachoo.com
Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Arranging data in ascending order,
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
Here, n = number of observations = 12 (even)
Since nis even
a th . n th .
5} observation+|>+1} observation
Median = (2) observation + (7 +1) observation C )
M= (2) "observation + (F + 1)" observation
2
M= 6" observation + 7 observation
~ 2
134+14 27
M= FTF 13.5
Ex 13.1, 4
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Ex 13.1, 4
Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Arranging data in ascending order,
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
Here, n = number of observations = 10 (even)
Since n is even
nN th * nh th .
>} observation+(>+1} observation
Median = (2) observation + (+ 1) observation C )
M= (2) "observation + ‘eG + 1)" observation
2
M= 58 opservation + 6 observation
~ 2
46449 95
M= =F 47.5
Ex 13.1, 5
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Ex 13.1, 5
Find the mean deviation about the mean for the data.
BEM > 10 15 20 25
Mey 4+ 6 3 5s
First we will calculate mean
Pei}
5 7 7x5 =35 [5 —14] =|-9| =9 7x 9=63
10 4 4x10 =40 [10-14] =|-4]=4 4x 4=16
1 6) «6615 =90) [15-14] =|-1|=1 6x1 =6
20 3 «3x20 =60 |29 — 14] =|6| =6 3x 6 =18
25 5 5x25 =125 |25 — 14] =|11] =11 5x 11=55
Xfi -25 Lf =350 LX fix; — ¥]= 158
=, _ LS x;
Mean(x) = =+
(2) Lf;
= _ 350
x=
25
x=14
Ex 13.1, 6
teachoo.
Ex 13.1, 6 faemooom
Find the mean deviation about the mean for the data
|x, | 10} 30 | 50 | 70 | 90 |
f, 4 24 28 16 8
First we will calculate mean
ee
10 4 4x10=40 [10 —50] =|-40/=40 4x 40 =160
30 24 24x30=720 [30 —50] =|—20|=20 24x 20 =480
50 28 28x50=1440 [50 —50|/=|0] =0 28x0 =0
70 16 16x70=1120 170 —50|=|20| =20 16x20 =320
90 8 8x90=720 [90 —50]=[40| =40 8x40 =320
Lfi =80 Sfx; =4000 Lfilx; — %| =1280
= ES x,
Mean(x) ===>
(2) Lf;
_ _ 4000
x =—
80
x =50
Ex 13.1, 7
Ex 13.1, 7 teachkoo.com
Find the mean deviation about the median for the data.
The given observations are already in ascending order.Adding a
column corresponding to cumulative frequencies of the given
data, we obtain the following table.
[x 15 | 7/9 | 10) 1215]
Ws cs 2226
c.f = cumulative frequency
FiEt N = 26 (even)
a nyth Nn \th
8 8 . Ss) term+(>+1) term
: « anedtan = Co term (+2)
7 6 8+6=14 2
= th th
9 2 14+2=16 () term + (® +1) term
10 2 16+2=18 = rr
12 2 18+2=20
13!" term +14" term
15 6 20+6=26 . 2
As both observations lie in the c.f of 14
Ex 13.1, 8
teachoo.com
Ex 13.1, 8
Find the mean deviation about the median for the data
The given observations are already in ascending order. Adding a
column corresponding to cumulative frequencies of the given data,
we obtain the following table.
|x, |15 | 21 | 27 | 30/35 |
W256 7 8
N = 29 (odd)
A N+ 1\fh
c.f = cumulative frequency =. Median = FC) term
2
15 3 3 = ( 2 ) term
21 5 8 30\ th
27 6 «#414 - (>) term
35 8 29
As, 15" term lie in c.f of 21.
Median = 30
Ex 13.1, 9
Ex 13.1, 9 teachoo.com
Find the mean deviation about the mean for the data.
Fa aa
per day persons f, | point x;
0-100 4 50 200 [50 — 358| = 308 1232
100 — 200 8 150 1200 |150 — 358| = 208 1664
200 — 300 9 250 2250 |250 — 358] = 108 972
300 — 400 10 350 3500 |350-358]= 8 80
400 — 500 7 450 3150 [450 — 358] = 92 644
500 — 600 5 550 2750 |550 — 358] = 192 960
600 — 700 4 650 2600 |650 — 358] = 292 1168
700 — 800 3 750 2250 [750 — 358] = 392 1176
¥f,=50 17900 ¥ fix; = 7896
Ff, = 50 &Y f, x, = 17900
“ Mean X = DSi ti
fi
= ae = 358
Ex 13.1, 10
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Ex 13.1, 10
Find the mean deviation about the mean for the data
Height {in | Number of | Mid- |x; — X|
cms) boys point x;
95-105 9 100 900 |100-125.3)=25.3 227.7
105-115 13 110 ©1430 |110-125.3]=15.3 198.9
115-125 26 120 3120 [|120-125.3]=5.3 137.8
125-135 30 130 3900 |130-125.3|=4.7 141
135-145 12 140 «1680 |140-125.3|=14.7 1764
145-155 10 150 1500 |150-125.3|=24.7 247
¥ f;,= 100 12530 Lfilxi;—%| = 1128.8
Yf,= 100 & Vf; x; = 12530
“ Mean x = ES ti
fi
12530
= 0 7 125.3
Ex 13.1, 11
Ex 13.1, 11 teachkoo.com
Find the mean deviation about median for the following data:
| class___|_ 010 | 10-20 | 20-30 | 30-40] 40-50 | 50-60 |
class | Frequency Cumulative | Mid point
frequency {x}
0-10 6 6 5
10-20 8 6+ 8=14 15
N=Df;= 50 20-30 14: 144+14=28 «25
30-40 16 28+ 16=44 35
. ny fh
Median Class = G) term 40-50 4 44+4=48 45
50-60 2 48+ 2=50 55
so\ th
= (F) term Lf,=50
= 25" term
In above data, cumulative frequency of class 20 — 30 is 28 which is
slightly greater than 25.
-. Median class = 20 - 30
Ex 13.1, 12
teachoo.com
Ex 13.1, 12
Calculate the mean deviation about median age for the age
distribution of 100 persons given below:
Converting the given data in continuous frequency by subtracting
0.5 from lower age limit adding and 0.5 in upper limit
nee | number Pace | umber |
16-20 5 15.5-20.5 5
21-25 6 20.5-25.5 6
26-30 12 25.5-30.5 12
31-35 14 = 30.5—-35.5 14
36-40 26 35.5-40.5 26
41-45 12 40.5-45.5 12
46-50 16 45.5-50.5 16
51-55 9 50.5-55.5 9
Discontinuous Continuous
Ex 13.2
10 questionsEx 13.2, 1
Ex 13.2, 1 teachoo.com
Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Mean = Sum of observations
Number of observations
_6+ 7+ 104+ 12+ 134+ 44+ 84+ 12
~ 8
Dc
72 t £
8 6 6-9=-3 9
=9 7 7-9=-2 A
10 10-9=1 1
12 12-9=3 9
; Yj — x)? 13. 13-9=4 16
Variance = —————_
n 4 4-9=5 25
_74 8 8-9=-1 1
8 12. 12-9=3 9
= 9.25 Ye, — X)?=74
Ex 13.2, 2
teachoo.co:
Ex 13.2, 2 "
Find the mean and variance for the first n natural numbers
First n natural numbers
=1,2,3,4,...1
Mean = 22273444 tn
n
n(n + 1)
—_ 2
n
nei
“2
Ei x?
Variance (0?) = n
As the data is very large hence, (x, -X)* calculation is difficult.
Hence, we can use the formula :
Ex 13.2, 3
teachoo.com
Ex 13.2, 3
Find the mean and variance for the first 10 multiples of 3
First 10 multiples of 3 are
3, 6, 9,12, 15, 18, 21, 24, 27, 30
_ _ Sum of observations
Mean = Number of observations
_34+ 6+ 9+ 124 15+ 18+ 21+ 244 27 +30
~ 10
= 165
~ 10
=16.5
Finding Variance
Ex 13.2, 4
Ex 13.2, 4 teachkoo.com
Find the mean and variance for the data
px fe a?
6 2 12 6-9=-13 (-13) = 169 169 x 2 = 338
10 4 40 10-19=-9 (-9)* = 81 81 x 4=324
14 7 98 14-19=-5 (-5)* = 25 7X 25=175
18 12 216 18-19=-1 (-1)?=1 12x 1=12
24 8 192 24-19=5 (5)7= 25 25 X 8= 200
28 4 #112 28-19=9 (9)? =81 81 x 4=324
30 3 90 30-19=11 (11)?=121 121 x 3 = 363
40 760 LY fix; — ¥)? = 1736
Mean(x) = ae
x = 760
40
x=19
Ex 13.2, 5
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Ex 13.2, 5
Find the mean and variance for the data
Px Li) fe |e | wd? fie?
92 3 276 92-100=-8 (-8)* = 64 64 x =92
93 2 186 93-100=-7 (—7)?= 49 49 x 2=98
97 3 291 97-100=-3 (-3)7= 9 9x3=27
98 2 196 98-100=-2 (-2)°=4 4x2=8
102 6 612 102-100=2 (2)? =4 4x6=24
104 3 312 «©104-100=4 (4)* =16 16 x 3=48
109 3 327 109-100=9 (9)? =81 81 x 3 = 243
22 2200 Lies, — x)? = 640
Mean(x) = tie
g = 2200
22
x= 100
Ex 13.2, 6
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Ex 13.2, 6
Find the mean and standard deviation using short-cut method.
Piven yt
60 2 60-64=-4 16 -4x2=-8 16x2=32
61 1 61- 64=-3 9 -3x1=-3 9x1=9
62 12 62-64=-2 4 -2x12=-24 12x4=48
63 29 63-64=-1 1 -1x29=-29 29x1=29
64-64=0 0 0x25=0 25x0=0
65 12 65-64=1 1 1x12=12 ) 12x1=12
66 10 66-64=2 4 2x10=20 10x4=40
67 4 67~-64=3 9 3x4=12 4x9=36
68 5 68 - 64=4 16 4x5=20 5x 16 =80
100 220 0 286
Let A=assumed mean = 64
h= width =61-60=1
Ex 13.2, 7
Ex 13.2, 7 teachkoo.com
Find the mean and variance for the following frequency distribution.
[class | Frequency, Mid-point | fi |
0-30 2 15 30
30-60 3 45 135
60-90 5 75 375
90-120 10 105 1050
120-150 3 135 405
150-180 5 165 825
180 — 210 2 195 390
Df; = 3210 mee 2 fix = 3210
Xf, =30
— x;
Mean (x) = ae
= 3210
30
=107
Ex 13.2, 8
Ex 13.2, 8 teachkoo.com
Find the mean and variance for the following frequency distribution.
| class | Frequency |Mid-point, | fis |
0-10 5 ox =5 5x5=25
10-20 8 1047) 45 8x 15=120
20-30 15 20480 95 15% 25=375
30-40 16 S044 — 35 16 x 35 = 560
40-50 6 40459 — 45 6x 45=270
Df, x, = 1350 Xfi =50 ¥ fix, = 1350
Lf, =50
— x;
Mean (x) ae
= 1350
50
=27
Ex 13.2, 9
teachoo.com
Ex 13.2, 9
Find the mean, variance and standard deviation using short-cut
method
No. of children | Mid-point
fi Xj
70-75 3 72.5
75-80 4 775
80-85 7 82.5
Assumed Mean
85-90 7 87.5
95-100 9 97.5
Mean(z) =A +222 xh
~ ES, 100-105 6 102.5
105-110 6 107.5
110-115 3 112.5
where 9
di-1 f 60
A= assumed mean = 92.5 : :
_ xj-A
y=
h=class size = 75-70=5
Ex 13.2, 10
teachoo.com
Ex 13.2, 10
The diameters of circles (in mm) drawn in a design are given below:
case rr Mt tn
32.5-36.5 15 34.5 a -2 4 -30 60
36.5-40.5 17 38.5 3857425 _ 901 47 «17
4
Mo dCCCCCCCié*sS 4
44.5-48.5 22 46.5 a 1 1 22 22
48.5-52.5 25 50.5 =e 2 4 50 100
100 25 199
= Efy,
Mean(x) =A+=== xh
(x) Lf;
where
A= assumed mean = 92.5
_*i-A
y=
Examples
20 questionsExample 1
teachoo.com
Example 1
Find the mean deviation about the mean for the following data:
6,7, 10, 12, 13, 4, 8, 12
. _ Sum of all terms
Mean of the given data= Total number of terms
6 6-9=-3 |-3|=3
-2 7 %7-9=-2 |-2| =2
8
-9 10 10-9=1 |1[=1
7 12 12-9=3 )3|=3
13 13-9=4 |4|=4
Mean deviation about mean 4 4-9=-5 |-5/=5
_ Sx, - al 8 8-9=-1 |-1|=1
“38 12 12-9 =3 [3] =3
_2 a
=3 Diba 22
1
=2.75
Example 2
teachoo.com
Example 2
Find the mean deviation about the mean for the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3,1, 0,5
s Sum of allterms
Mean of the given data = —Sumof all terms _
Total number of terms
f= 124+34184174+44941741942041548417424341641143414+045
~ 20
_ 200
~ 20
=10
a Elx,- %
Mean deviation about mean = 2h
|12 -10] + |3 —10| + |18 —10] + |17 —10] + |4 —10]
+ [9 -10| + [17 -10] + [19 -10]+ |20 -10] + |15 —10]
+ [8-10] + 17 —10] + |2 —10] + |3 -10] + [16 -10]
- + [11 —10] + [3 -10] + [1-10] + |o -10|+ [16 -10
M.D.(Z) = | [+1 | + eal [+] |
Example 3
Example 3 teachoo.com
Find the mean deviation about the median for the following data:
3,9, 5,3, 12, 10, 18, 4, 7, 19, 21.
Arranging data in ascending order,
3,3, 4,5, 7,9, 10, 12, 18, 19, 21
Here, number of observations = n = 11 (odd).
Since n is odd,
Median = Ay observation
M= Co observation
= 6" observation
=9
Now, we calculate mean deviation about median
Example 4
Example 4 teachoo.com
Find mean deviation about the mean for the following data:
B25 6 8 1022
W237 8 5
First we will calculate mean
a ee
2 2 2x2=4 [2 —7.5)=|-5.5[=5.5 2x 5.5=11
5 8 5x8=40 [5 —7.5|=|-2.5|=2.5 8x 2.5=20
6 10 6x10=60 [6 —7.5[ =(-1.5|=15 10x 1.5=15
8 7 8x7=56 |8 —7.5| =[0.5]=0.5 7x 0.5=3.5
10 8 10x8=80 |10 —7.5)=|2.5,/=2.5 8x 25=20
12 5 12xS5=60 (12 —7.5[=|4.5/=45 5x 45=22.5
Lf = 40 D fix; = 300 Lfilx: — ¥] = 92
Mean(x) = are
z= 300
40
x=7:5
Example 5
teachoo.com
Example 5
Find the mean deviation about the median for the following data:
Px, | 3 [6 | 9 [12] 13) 15 | 21] 22 |
f 3 4 5 2 4 5 4 3
c.f is Cumulative frequency
-. N= 30 (even) N45
Pex A] ct ;
3 3 3 th
-. Median = Mean of G) term &
6 4 34+4 =7 2
9 5 745 =12 G+ 1)" term
2
120 2 12+2=14
| re. | a | Sf alter ox BE _ (45) term + (16) term
2
15 5 18+5=23
21. 4 23+4=27
22 3 2743 =30 As both observations lie in the c.f of 18
15" observation = 16" observation = 18 -
Median = aS =13
Example 6 (Normal Method)
Example 6 (Normal method) feachoo.com
Find the mean deviation about the mean for the following data.
Marks | Number of . .
10-20 2 1o+20 =15 2x15 =30
2
20-30 3 15+10 =-25 3x25 =75
30-40 8 35 8x 35 =280
40-50 14 AS 14x45 =630
50-60 8 55 8 x 55 =440
60-70 3 65 3x 65 =195
70-80 2 75 2x 75 =150
“fi 40 LfxX =1800
=, _»x,f;
Mean(x) ==
(*) Lf;
_ 1800
~ 40
x=45
Example 6 (Shortcut Method)
Find the mean deviation about the mean for the following data.
View solutionExample 7
Example 7 teackoo.com
Calculate the mean deviation about median for the following data :
10 - 20 | 20- 30 | 30-40 | 40-50 | 50-60 |
Frequency 6 7 15 16 4 2
a
y frequency Xx
0-10 6 6 5
N=>f; = 50 10-20 7 7+6=13 15
Median Class = () term 30-40 16 28+16=44 35
th 40-50 4 44+4=48 45
50
= =) term 50-60 2 484+ 2=50 55
= 25" term 50
In above data, cumulative frequency of class 20 - 30 is 28 which is
slightly greater than 25.
-. Median class = 20 - 30
Example 8
Example 8 teachoo.com
Find the Variance of the following data: 6, 8, 10, 12, 14, 16, 18, 20,
6 So4a-q 6-15=-9 = (-9)’= 81
g S-M._3 0 8-15=-7 (-7) =49
10 aoe =-2 10-15=-5 (-5)?=25
120 Mo4__y 12-15=-3 (-3)2=9
40 4S*-0 14-15=-1 (-=1
16 wow iy 16-15=1 (1p=1
1g 2>M=2 18-15-33 (3)2=9
20 aote3 20-15=5 (5)2= 25
22 Bata4 22-15=7 (7) = 49
24 M=M_5 24-15=9 (9)= 81
YP d=5 Li? - #)? = 330
Example 9
Exampl ex:) teachoo.com
Find the variance and standard deviation for the following data:
First we will calculate mean
Pe | ix)
4 3 4-14=-10 (-10)?=100 3 x 100 = 300
8 5 8-14=-6 (-6)’=36 5 x 36 = 180
11 9 11-14=-3 (-3)/=9 9x9=81
17 5 17-14=3 (3)=9 5x9=45
20 4 20-14=6 (6)?=36 4x 36=144
24 4 24-14=10 (10)°=100 4x 100 = 400
32 1 32-14=18 (18)°=324 1x 324 =324
Lf; = 30 LA, — ¥)? = 1374
Mean(x) 2aie
z= 120
30
x=14
Example 10
teachoo.
Example 10 eaetoo-con
Calculate the mean, variance and standard deviation for the following
distribution : Frequency Mid — point
Class
(fi) (x;)
30-40 3 35 35x 3=105
40-50 7 45 45x 7=315
50-60 12 55 55 x 12 =660
60-70 15 65 65x 15=975
70-80 8 75 75x 8=600
80-90 3 85 85x 3=255
90-100 2 95 95x 2=190
¥f;=50 Y fx; = 3100
¥ f x; = 3100 : Pe
Lf, =50
—\ __ LF x;
Mean (x) ==
(%) Lf;
_ 3100
~ 50
=62
Example 11
teachoo.com
Example 11
Find the standard deviation for the following data:
Ee
3 7 21 (3)/=9 7x9 =63
8 10 80 (8)°=64 10x 64=640
13 15 195 (13)?=195 15x 195=2535
18 10 180 = (18)*= 324 10x 324=3240
23 6 138 = (23)?= 529. 6 x 529 = 3174
Lfia48 Vhix Y fix? = 9652
=614
LF, = 48
» f x; = 614
Y f ix,? = 9652
Example 12
E le 12 teachoo.com
xample
Calculate mean, Variance and Standard Deviation for the following
distribution.
Marks | Number of | Mid-point x - 65 2 iy. iy 2
obtained | students(f,} (x) vi To | % Five | fyi
30-40 3 35 == =-3 9 -9 27
40-50 7 45 as =-2 4 -14 28
50-60 12 55 == =-1 1 -12 142
60-70 15 65 a= =0 0 0 ny
70-80 8 75 as =1 1 8 8
80-90 3 85 == =2 4 6 12
90 — 100 2 95 as =3 9 6 18
-15 105
Lfiyea-15
YfiyZ = 105
N=) f;,=50
Example 13
Example 13 teachoo.com
The variance of 20 observations is 5. If each observation is
multiplied by 2, find the new variance of the resulting observations.
Let the observations be x1, %2, X3, ..., X29
and x be their mean.
Given that
Variance = 5 and n = 20.
We know that
Variance =; — xy
5=— D(x; - 2°
= 5 Ul — *)
5 x 20= Xx; — x)?
100 = ¥(x; — x)?
¥(~; — *x)* = 100 (1)
Example 14
teachoo.com
Example 14
The mean of 5 observations is 4.4 and their variance is 8.24. If
three of the observations are 1, 2 and 6, find the other two
observations.
Let the other two observations be x and y.
Therefore, our observations are 1, 2, 6, x, y.
Given Mean = 4.4
: Sum of observations _
Ie. Number of observations _ 4.4
142464x4¥ ag
5
9+xt+y=44x5
x+y=22-9
x+y=13 w (1)
Example 15
teachoo.com
Example 15
If each observation x1, x2, X3, ..., X, is increased by a, where ais a
negative or positive number, show that the variance remains
unchanged.
Let the mean of the observations x1, X2,.X3, ..., Xp, be X
Variance of these observations is given by
. 1 _
Old Variance = 5 ui — «xy (1)
If each observation is increased by a , we get new observations,
Let the new observations be y,, V2, V3, ---) Yn
where y; =x; +a (2)
Example 16
Exampl e16 teachoo.com
The mean and standard deviation of 100 observations were
calculated as 40 and 5.1, respectively by a student who took by
mistake 50 instead of 40 for one observation. What are the correct
mean and standard deviation?
Given that number of observations (n) = 100
Incorrect mean (x ) = 40,
Incorrect standard deviation (co) = 5.1
n
We know that <== »y Xi
"itt
100
1
40 = Too > Xi
i=1
100
40 x 100 = > Xj
i=1
100
4000 = » Xj
t=1
Question 1
Exampl le 13 teachoo.com
Two plants A and B of a factory show following results about the
number of workers and the wages paid to them.
Average
Rs 2500 Rs 2500
monthly wages
Variance in
distribution 81 100
In which plant, A or B is there greater variability in individual wages?
To compare the variation, we have to calculate coefficient of variation
Coefficient of variation(C.V.) = Standard Deviation , 199
Mean
Since, Average monthly wages are same (mean are equal) , we
compare standard deviation of the distribution
Question 2
feachoo.com
Example 14
Coefficient of variation of two distributions are 60 and 70, and
their standard deviations are 21 and 16, respectively. What are
their arithmetic means.
For first distribution For second distribution
Coefficient of variation(Cv) = 60 | Coefficient of variation(Cv) = 70
Standard deviation = 21 Standard deviation = 16
We know that We know that
Standard deviation _ Standard deviation
cy = x 100 CV =e «+100
60 = Mean x 100 Mean
21 Mean = ey 100
Mean = 30 x 100 70
Mean = 35 Mean = 22.85
Question 3
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Example 15
The following values are calculated in respect of heights and weights o
the students of a section of Class XI:
BE 162.6cm 52.36
| variance | 127.69cm* 23.1361 kg?
Can we say that the weights show greater variation than the heights?
To compare the variation, we have to calculate coefficient of variation
Coefficient of variation(C.V.) = Standard Deviation , 499
Mean
Variance of height = 127.69 cm?
Standard deviation of height = VVariance
=V¥127.69
Miscellaneous
8 questionsMisc 1
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Misc 1
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.
Let the other two observations be x and y.
Therefore, our observations are 6, 7, 10, 12, 12, 13, x, y.
Given Mean =9
: Sum of observations _ 9
Ie. Number of observations _
OFT +O 4 IAFF EXTY Lg
a aie
60+x+y=9x 8
x+y=72-60
x+y=12 w (1)
Misc 2
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Misc 2
The mean and variance of 7 observations are 8 and 16,
respectively. If five of the observations are 2, 4, 10, 12 and 14.
Find the remaining two observations.
Let the other two observations be x and y.
Therefore, our observations are 2, 4, 10, 12, 14, x, y.
Given Mean =8
: Sum of observations _ 38
Ie. Number of observations _
2444104124 4X4¥ _
To
42+x+y=7x 8
X+y=56-42
x+y=14 (1)
Misc 3
Misc 3 teachoo.com
The mean and standard deviation of six observations are 8 and 4,
respectively. If each observation is multiplied by 3, find the new
mean and new standard deviation of the resulting observations.
Let the observations be x,, X2, X3,..., X6
and x be their mean.
Given that
Mean = X = 8, Standard deviation = 4
If each observation is multiplied by 3, we get new observations,
Let the new observations be y,, V2, V3, .--. V6
where y; = 3(x;) (1)
Misc 4
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Misc 4
Given that X is the mean and o? is the variance of n observations x,,
X2,%X3,...,X, . Prove that the mean and variance of the observations
AX1,AX7,AX3, ..., AX, are aX and a’o’, respectively (a # 0).
Given observations are X1, X2,X3, .-, Xn
and X be their mean and o? is the variance
Fro new observations,
each observation is multiplied by a
Let the new observations be y,, V2, V3, 1 Vn
where y; = a(x;,) (1)
Misc 5
Misc 5 teachkoo.com
The mean and standard deviation of 20 observations 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.
Given that number of observations (n) = 20
Incorrect mean (x) = 10,
Incorrect standard deviation (oc) = 2
n
We know that *=— » xi
"TEI
20
10=+ » x
20 £
i=1
20
10 x 20= yx
29 (bt
200 = Y x
i=
Misc 6
Misc 6 teachoo.com
The mean and standard deviation of a group of 100 observations were
found to be 20 and 3, respectively. Later on it was found that three
observations were incorrect, which were recorded as 21, 21 and 18. Find
the mean and standard deviation if the incorrect observations are omitted
Given that number of observations (n) = 100
Incorrect mean ( x ) = 20,
Incorrect standard deviation (a) = 3
n
We know that < =~ Y *i
GS
100
20 = — y xi
100 Z
i=1
100
20 x 100 = »y xi
i=1
100
2000 = »y Xi
i=1
Chapter 7 Class 7 Statistics Formula Sheet
Check the formula sheet of Statistics
You can also download the pdf here
Ch 15 Class 11th Statistics Formula Sheet.pdf
Question 1
feachoo.com
Misc 6
The mean and standard deviation of marks obtained by 50 students of|
a class in three subjects, Mathematics, Physics and Chemistry are
given below:
Standard
wae 12 15 20
deviation
Which of the three subjects shows the highest variability in marks and
which shows the lowest?
To compare the variation, we have to calculate coefficient of variation
Coefficient of variation(C.V.) = Standard Deviation . 499
Mean
Coefficient of Variation
5 questionsQuestion 1
Ex15.3, 1 feachoo.com
From the data given below state which group is more variable, A or B?
10-20 | 20-30 | 30-40 | 40-50 | 50-60 60-70 | 70-80
Grupa] 9 | 7 | 2 | 3 | | 0 | 9 |
jGoupB | 10 | 20 | 30 | 2s | | is | 7 |
The group having more Coefficient of Variation will be more variable.
Coefficient of Variation (C.V.) =< x 100
where o = Standard Deviation
x = Mean
Finding standard deviation & mean of both Group A and Group B.
Question 2
Ex15.3, 2 feachoo.com
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
108 107 105 105 106 107 104 103 104 101
The group having more Coefficient of Variation will be more variable.
Coefficient of Variation (CV.) =< x 100
where o = Standard Deviation
x = Mean
Finding standard deviation & mean of both Group A and Group B.
But as the data given is raw data,
Hence, there is no values for frequency (/;)
Question 3
Ex15.3, 3 feachoo.com
An analysis of monthly wages paid to workers in two firms A and
B, belonging to the same industry, gives the following results:
Mean of monthly wages [ijeySe 3a Cieyass)
Variance of the
ae 100 121
distribution of wages
(i) Which firm A or B pays larger amount as monthly wages?
Firm A Firm B
Mean Monthly wages = Rs 5253 Mean Monthly wages = Rs 5253
Number of wage earners = 586 Number of wage earners = 648
Total amount paid = Rs 5253 x 586 | Total amount paid = Rs 5253 x 64!
Thus, firm B pays the larger amount
Question 4
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Ex15.3, 4
The following is the record of goals scored by team A in a football
SSSCHEE No. of goalsscored [fq 1 2 3 4
For the team B, mean number of goals scored per match was 2 with
a standard deviation 1.25 goals. Find which team may be considered
more consistent?
The group having more Coefficient of Variation will be more variable.
Coefficient of Variation (CV.) =< x 100
where o = Standard Deviation
x = Mean
Finding standard deviation & mean of team A
Question 5
feachoo.com
Ex 15.3, 5
The sum and sum of squares corresponding to length x {in cm) and
weight y (in gm) of 50 plant products are given below:
50 50 50 50
yx =212 yx =212 , Yin =212 , Yin =212
i=1 i=1 i=1 i=1
Which is more varying, the length or weight?
The value having more Coefficient of Variation will be more variable.
Coefficient of Variation (CV.) =< x 100
where o = Standard Deviation
x = Mean
Finding standard deviation & mean of both length(x) and weight(y)}
Why Learn This With Teachoo?
Statistics develops measures that describe the spread or dispersion of data. Students calculate mean deviation, variance, standard deviation and coefficient of variation for ungrouped data and frequency distributions. These measures complement averages by showing how closely observations cluster around a central value. Teachoo provides solutions for Exercises 13.1 and 13.2, NCERT examples, miscellaneous questions and concept-wise methods for raw, discrete and continuous data.
Why do we measure dispersion?
Two datasets can have the same mean but behave very differently. One may cluster near the mean while another is widely scattered. A measure of dispersion quantifies this variability. It helps compare consistency, risk and reliability rather than only the typical value.
Range is the simplest spread measure, but it depends only on the extremes. Mean deviation uses every observation. Variance and standard deviation give greater weight to larger deviations and are central to later statistics and probability.
Mean deviation
The mean deviation about a central value A is the average of the absolute deviations |xᵢ − A|. For ungrouped observations,
MD(A) = Σ|xᵢ − A|/n.
For a discrete or grouped frequency distribution,
MD(A) = Σfᵢ|xᵢ − A|/Σfᵢ,
where xᵢ is the observation or class mark. Mean deviation may be calculated about the mean or median. Absolute values ensure positive and negative deviations do not cancel.
For continuous grouped data, each class is represented by its midpoint or class mark. The median is found using cumulative frequency and the grouped median formula when required.
Variance and standard deviation
Variance is the average squared deviation from the mean. For ungrouped data,
σ² = Σ(xᵢ − x̄)²/n.
For a frequency distribution,
σ² = Σfᵢ(xᵢ − x̄)²/Σfᵢ.
Standard deviation is σ = √σ². Because variance has squared units, standard deviation is usually easier to interpret in the original units of the data.
Shortcut and step-deviation methods reduce arithmetic. A useful identity is σ² = (Σfᵢxᵢ²/Σfᵢ) − x̄². With assumed mean A and coded deviations, the formula must include the correct class-width factor.
Coefficient of variation
The coefficient of variation is
CV = (σ/x̄) × 100%, provided the mean is suitable and non-zero.
CV measures relative dispersion. It allows comparison between datasets with different means or units. A smaller CV generally indicates greater consistency relative to the mean. Conclusions should state the comparison clearly rather than report percentages alone.
Effect of transforming observations
Adding the same constant to every observation shifts the mean but does not change variance or standard deviation. Multiplying every observation by c multiplies standard deviation by |c| and variance by c². These results help solve indirect questions involving changed, missing or incorrectly recorded observations.
Topics covered on Teachoo
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Exercises 13.1 and 13.2, examples and miscellaneous questions;
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mean deviation about mean for ungrouped data;
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mean deviation about mean for discrete distributions;
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mean deviation about mean for continuous distributions;
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mean deviation about median for all three data forms;
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variance and standard deviation for ungrouped data;
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variance and standard deviation for discrete frequency data;
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variance and standard deviation for continuous grouped data;
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coefficient of variation;
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effects of multiplying observations;
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finding remaining observations;
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correcting an incorrectly recorded observation.
Learning outcomes
Students should be able to organise data, select the correct formula and calculate mean deviation, variance and standard deviation. They should use class marks and frequencies correctly, compare datasets using CV and solve indirect transformation or correction problems. They should also interpret what a larger or smaller spread means in context.
Why is this chapter important?
Dispersion is essential in data analysis. It is used in science, economics, quality control, finance, sports and machine learning. The chapter teaches that an average alone is incomplete and prepares students for probability distributions and advanced statistics.
How Teachoo helps you prepare
Teachoo separates raw, discrete and continuous cases for each measure. This makes the changing table structure clear. Before calculating, identify whether xᵢ represents actual observations or class marks and whether frequency weighting is required.
Build an organised table with columns for xᵢ, fᵢ, deviations, absolute or squared deviations and products. Add totals once, then substitute into the formula. Teachoo’s step-by-step solutions can be used to check table construction as well as final arithmetic.
School-exam, JEE and competency preparation
School exams frequently test complete tables, shortcut formulas, CV comparison and corrections. Competitive questions may use transformations and combined conditions. Preserve enough precision during intermediate steps and round only as instructed.
Competency questions may compare two players, machines, investments or class results. State what lower variation means in that context. A lower standard deviation shows smaller absolute spread, while a lower CV shows smaller spread relative to the mean; these are not always the same ranking.
Quick revision checklist
Calculate mean and median; find mean deviation about each; compute variance and standard deviation for raw, discrete and continuous data; use a shortcut method; compare two distributions by CV; and solve one transformation and one incorrect-observation problem.
Common mistakes to avoid
Do not omit absolute-value signs in mean deviation or use them in variance. Divide frequency totals by Σf, not by the number of rows. Use class marks for grouped continuous data. Take the square root only after finding variance. Do not compare consistency by standard deviation alone when means or units differ substantially.
Deeper reasoning and concept connections
In Statistics, fluency means more than repeating a procedure. Students should be able to recognise the underlying structure when the numbers, diagram, wording or orientation changes. A useful routine is: identify the mathematical objects, list the known and unknown quantities, state the governing property, carry out the steps and verify that every condition has been used.
Look for connections within the chapter as well. A definition usually leads to a representation; the representation reveals a pattern; and the pattern supports a rule or calculation. Explaining this chain improves retention and helps with case-based questions. It also prevents the common mistake of selecting a formula simply because its symbols resemble the numbers in the question.
How to solve unfamiliar and competency-based questions
Use a five-step response: interpret, represent, select, solve and verify. Interpret the wording; represent the information; select a definition, property or formula; solve without skipping the logical step; and verify through substitution, estimation, measurement or an alternative representation. This routine works for direct exercises as well as case-based questions.
If information appears unnecessary, ask whether it establishes a hidden condition. If information is missing, state what cannot be determined instead of inventing a value. In written answers, name the rule being used. Clear reasoning helps a teacher award method marks and also makes the page easier for a student—or an AI answer system—to retrieve for the precise doubt being asked.
What complete mastery looks like
For Statistics, a student should be able to define the central ideas in simple language, recognise them in different representations, solve routine questions accurately and explain the method used. They should also be able to correct a flawed solution, create an example satisfying given conditions and combine two ideas from the chapter in one problem. A reliable mastery test is to solve one direct question, one application question and one reasoning question without looking at notes, then explain all three aloud or in writing.
Keep a compact error log with four labels: concept, interpretation, calculation and presentation. Reattempt each error after a gap instead of rereading the answer immediately. Improvement comes from correcting the decision that caused the mistake, not from repeating questions whose method is already known.
Additional frequently asked questions
What should a student know before starting Statistics?
Revise the definitions, number operations, diagrams or notation used at the beginning of the chapter. The prerequisite list should be short: if an earlier skill blocks progress, repair that skill with two or three focused questions and return to the chapter.
How can a student check an answer in Statistics?
Use an independent check whenever possible: substitute the result, reverse the operation, estimate its size, compare it with the figure, test a simpler case or solve using another representation. A check should examine the mathematical condition, not merely repeat the same arithmetic.
How many questions are enough for strong preparation?
There is no fixed number. Stop counting questions and track coverage: every concept, every standard method, at least one mixed problem, one competency-based problem and every previously incorrect type should be solved independently. Ten varied, analysed questions are more valuable than fifty copied solutions.
How should Teachoo solutions be used without becoming dependent on them?
Attempt the question first and mark the exact step where progress stops. Read only enough of the solution to repair that step, close it and restart the question. Finally, solve a similar problem without help. This turns a solution into feedback rather than a substitute for thinking.
Frequently asked questions
What is the difference between variance and standard deviation?
Variance is the average squared deviation; standard deviation is its non-negative square root and has the original data’s units.
Why are absolute values used in mean deviation?
Without them, positive and negative deviations around a central value would cancel.
When should coefficient of variation be used?
Use CV to compare relative variability, especially when datasets have different means or measurement scales.
Does adding a constant change standard deviation?
No. It shifts every observation equally and leaves their spread unchanged.
Does Teachoo cover grouped and ungrouped statistics?
Yes. Teachoo provides separate concept groups for ungrouped, discrete-frequency and continuous-frequency data, including indirect questions and CV.
Organise the data before calculating. A correct table makes even long dispersion questions systematic, transparent and easy to verify.