statistics for exam like cgl railway or any exam.pdf

for competitive exam

1. | STATISTICS

à.
iz A N)
v . N
y . “hs
>
y x
Luz? % y 8

EE :
Class pot. by Aditya Ranjan Si Sir
fom + YER y

2. OND UWP

F Mey
CHAPTER INCLUDES

. MEAN (HT) 2. MEDIAN (aie)

MODE (3317) 4. RANGE (URI)

+ VARIANCE (HU)

. STANDARD DEVIATION (AS far)
. MEAN DEVIATION (area aaa)

. COEFICIENT VARRIANCE (HRRUT Turis)

7 SS

3. 12434+4s +So+2u

5

= 168” 33
=

The arithmetic mean of a given data is the sum
of all observations divided by the number of
observations. For example, a cricketer's scores
in five ODI matches are as follows: 12, 34, 45,
50, 24. To find his average score we je calculate
the arithmetic mean of data using the mean
formula:

Fr feu a Set ar ares Bert ma cat eo at
wen à far wet a una gate aro fa, wise
vetada teat à um Bret a HAN 12, 34, 45, 50, 24
$ sam alte eer ae AA A few EU oes a at
Hera À Set AT BAA AE A TA I

4. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

Sum of all observations / ait deu al ame
Number of observations / NAT #1 HET
Mean/WE#A = (12 + 34 + 45 + 50 + 24)/5
Mean/W = 165/5 = 33
Mean is denoted by x (pronounced as x bar).

Wet at x gra aid ara El

5. Tf X,, Xe, RR ‚x, are n values of a
variable X, then the arithmetic mean or simply
mean of these values is denoted by X and is
defined as:

6. ag: Stlotis = 100
3

7. COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

COMPLETE

1. The arithmetic mean of the following data is

= 1243444S+S0t 24 ——
mans III 12, 34, 45, 50, 24

5
_ 33 Fate Wen at Gana aa u
7 a am
12, 34, 45, 50, 24
(a) 30 (b) 36

(c) 33 (d) 25

8. =
Average (ar)
Berge 2 (Deviatien Method)

— .
40, 14,42, 13, 14 14212, 11212 (219) 11215, 11216
id 1 . À simi

© oy tovisiasistiy (De 2 - O er pro
Ss

Case 1

= Ons= PTA
FO

9. Average (atta)
9 (Deviation Method)

er
14212, (1213) 11214, 1215, 11216 14219, 11213, 11214, 11215, 11216
Pe: (==)
(A= 11200)
» LL © gr +2 43
Ds +2, +13, HY 418 416
V9 deviation = +3 - a:
ZO Z-O
On -
Ong Ira “= 100414

E Uy u

10. ©. 4332, 4328,4349) 4335, aus
A
D 8 -12 0 +s 48
OvgD = -28 4 =
=
5

Ans.
TE 4336

12. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

Q3, 11,60) 19, 21
ag. jo zu =o

Ong = 20

The arithmetic mean of the following data is

23, 17,20,19,21
o eT
23, 17,20,19,21

\9¥20 (b) 19

(e) 23 (a) 21

13. ey ADITYA RANJAN SIR

3. If the mean of six observations 5, 7, 9, a, 11

Fe, and 12 is 9 then the value of a is :
q añ we tau 5, 7, 9, a, 1197 12 51 909%, À
aa um À:
mt Stt9+at412 = q (a) 10 (b) 15
6 (e) 22 (d) 25
> «Hy=-sy
> x= 10

16. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

Le) 98+26422+H1+13+X =204 If the mean of the data 28, 26, 22, 11, 13, x
6 is 20, then find the value of 'x'.

3 1004X, =20 af 28, 26, 22, 11, 13, x langt a mara 20 À, À

6 ‘x! AT aa PAU ı
4 ER) ay 20 (b) 30
(e) 25 (d) 28
m
=< 38, 26, 22, 11,13 x
O
4 20
8, +6, 42 -Q a

x:g

18. 103

120

ut 108

+8 |-8

X= 10-8 = |

iS

I

A

19. FOR DISCRETE FREQUENCY DISTRIBUTION

(xi) avg 40 20 30 If X takes values x,, x,, x,........….
i) 6 corresponding frequencies f,, f,, f,, . à
Ei data 4 5 7 respectively, then arithmetic mean of these

Basic values is given by:
= av > Sum AÑ X Um x, x, x,
2 3

AA AA
ene wa dm. ines |

S_ 5x +£x,+£,x, +

a X= z -
AS 3 e f,+f,+f, +

20. Ma ©. (xi) Avg > 20 30 49

(fi) Data ?

Qns-

10 4 3

10X20+ 4X 30 +3 x40
SALOF 1X30+3x40_

104143

S3
= 26.
$ 265

-10 © +0
OA > 20 40
Œipatar 10 4 3

Ovg D2 =l00+0+30 =-18
So 28
ne

Ong- 30-3.5

F6,

21. E <Q -S
© (1) Avg 123 1122 Gs 1120
(Hide 4 3 8 10

Basic 1I23xu 4 1122 FX + Woxto

22. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

quevage
MA = UXSHIOXGFIOXQENXIO+8NIS 5. Find the mean)of the following distribution:
Stl0+l0+7 +8 Fafafada fer at area ara AT:
= 368 q Avg cto | 4 | 6 9 10 15
Kia Dada et | 5 | 10 10 7 8
A 9 (b) 10

(ce) 12 (d) 8

23. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR
a vage
m 5. Find the cano the following distribution:
Frafafaa feu ar area ara ae:
—5 _-3 Oo +f +6
Avg ét | 4 [| 6 10 | 15
Dada et | 5 | 10 10 7 8
\ T9 (b) 10
(ce) 12 (d) 8
Daya - -2
97 28-3044 4y
O SS
4o SStss =0

24. COMPLETE

ONS 26 424422104116 14S

COURSE (For all govt. exams ) By ADITYA RANJAN SIR

843

loo

The following table shows the number of
commercial clerks at 100 stations in a electric
department:

Frerferfern af à ur faa few à 100 Bun
= age act at den aad M à:

Number of
Commercial | (x)
Clerks
Number of
stations

Find the mean from the above.
sugar 9 ma ara PAU!
(a) 2.50 \by2.73

(ce) 2.33 (d) 2.58

25. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

7. Find the mean of the following distribution:

frafatad fear ar area ara afar:

(a) 8.325 (b) 9.125
sy 7.025 (d) 5.225
Ang -
° SONG oe rara
= So
= 8
Yo

org

26. COMPLETE

I
AA
> 6+8+I8H0+2p+io = 66
3

S242p=66

> PRAY

or all govt. exams ) By ADITYA RANJAN SIR

If the mean of the following distribution is 6,

a (6)
q ang = 3x2+2X4+3%6 + IXO+2 o. find the value of p.

añ frafataa faro at area 6 à, at pa m
aa PAU

ect | 2 4 6 10 | p+5
Daaet | 8 | 2 | 3 1 2
DT ro qe
xBr7 (b) 8

(c) 9 (a) 10

27. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

9. Ifthe mean of the following data is 15, then
find the value of k.

afa frafafaa Set ar ma 15 à, at k AT om am
CAL

aval (x) PS 10 5 |} 20 25
GE Bete je
(a) 7 Nbys

(c) 6 (d) 10

28. COMPLETE

COURSE (For all govt. exams ) By ADITYA RANJAN SIR
(gasió)

10. Find the mean of the following frequency
avg» 5 18 25 35 us distribution:
Dada 4 jo 1S 8 10

ass - à Fer ar AA aa PU:
tis iS 2s Ss us
and 0-10 | 10-20 | 20-30 | 30-40 | 40-50
- interval:
ag: MXS+10XIS-NSX25+ BX 35+10X4S) Noor

workers 7 10 15 8 10
(0:

(a) 25.8 (b) 24.8
(c) 25.9 (d) 24.9

29. COMPLETE

By ADITYA RANJAN SIR

-20 -I0 +10 49010. Find the mean of the following frequency

ovg > 5 (Ss 3S ys distribution:

Dao 4 Jo 1S 8 10 AUGE: ¡IA

is 25 35 us
Dag = yo =100 +0 +80 +200 avg Ki 0-10 | 10-20 | 20-30 | 30-40 | 40-50
So No.of |
Data | workers 7 10 15 8 10
= 4% -08 i
Se (a) 25.8 (b) 24.8

Quy = 9408: 25. (c) 25.9 (d) 24.9

30. ey ADITYA RANJAN SIR

11. Arithmetic Mean (AM) of the following data is-
Fafafaa Stet at Wait (AM) ura à-

Class- 8 12 T6 DO TZ
interval | 6-10 10-14 14-18 18-22 22-26 |
Frequency 5 | 12 7 5 1 |
(a) 10 (b) 12
yoria o «(is
95 8 1G CS ty +8
S 1
Og = —
TO MBtorrotg | an.
>. 30 16-2 <Q
TEN

31. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

12. The mean of the following distribution is 26,
then what is the value of k?

añ Prof den at Ta 26 à, at k at I M à?
9) s Ss as ys

L
{Class 0-10 | 10-20 | 20-30 | 30-40 | 40-50
| Frequency 8 10 | K 6 12
(a) 8 (b) 1
Basic usrá (a) 10

MY S 18 28 38 us

Doig 8 lo kK 6 lo
D CEE LE CE
36+

Worse ane

3 = Br

32. ey ADITYA RANJAN SIR

12. The mean of the following distribution is @6,)
then what is the value of k?
añ Preafeiad seat or ara 26 à, at aa ae À?
€ gt et ne ge

1 EX
[ Class 0-10 10-20 20-30 30-40 40-50

| Frequency 8 10 | K 6 12
(a) 8 (b) 1
Deviation, \sr# (d) 10

21 A 1 +4 19
MW S 18 28 38 us

Dag 8 lo kK 6

33. COMPLETE For all govt. exams ) By ADITYA RANJAN SIR

15. Find the arithmetic mean of the following
frequency distribution by the assumed mean
method:

aaa ma fafa gro Rafa mann sex aT
HATAT AeA ATA 2 Fu © +20 +80 +100
Ovg_[ Wages (in Rs.): | 800 Ent PMC 920 | 980 | 1000
No. | No. of Workers: | 7 14 | 19 | 25 | 20 | 10 5
ler Rs. 891.2 (b) Rs. 890.2

(c) Rs. 895.6 (d) Rs. 898.6
Da

2-1
9 00-1120 -N6o +04 4004 8004 800

ng.

34. COMPLETE

For all govt. exams )

By ADITYA RANJAN SIR

14. Find the arithmetic mean of the following
frequency distribution by the assumed mean

-200 -l00 © +100 +200 :
i bo don en) 400 soo method:
8 ia la fear area fafa grt Frerfefen mann dea ar
Die 16 10 22 JS Auer wea ara re:
100 _200 _ 300 4oo__soo_
Day. = ~3200-1000 +1S0042400 ng y | En 50-150 | 150-250 | 250-350 | 350-450 | 450-550
6 id 1S = = [Frequency | 16 10 | 22 15 12 |
(a) 290 (b) 296
Quy = 300-4 = 99 E (e) 285 (d) 250

35. The value of the middlemost observation,
obtained after arranging the data in ascending
or descending order, is called the median of the
data.

Sar at ame wa À ahaa ae à are wera Derr
wal HT Hed ZI

For example, consider the data: 4, 4, 6, 3, 2.
Let's arrange this data in ascending order: 2, 3,
4, 4, 6. There are 5 observations. Thus, median
= middle value i.e. 4.

ZEN à feu AT 4, 4, 6, 3, 2 aig Ser 34 waa
Weel ame HA 2, 3, 4, 4, 6 À aaftaa ara dla 5
dart #1 ga fea arf = qa aa aria 4 $1

36. O 1% 10, 64, 41,18

viele

Median 7)

& Fr

Qns
==

37. © 12 10, 8,4, 9, 11,18
Y

ye

Median 7)

de $.1,901,m:8 es << Oe es”
Se) ow
2 y
qe y = Utero

04 =19

38. © 13, 16, 10,6
Median?

u Ah

Ow =_10H12 = D
2

39. © 13, 16, 10,6
Median: ?

da Qe lé
Amel (2), (89% term
7 _ E 3 (4+)

= =, 3rd

“sou <p

40. COMPLETE

STEP I:

STEP II:

COURSE (For all govt. exams )

Arrange the observations x,, x,,
x, in ascending or descending
order of magnitude.

RUT x,, x,,...
am ar pe aa aaa at
Determine the total number of
observations, say, n

Want a ae den aia at, ara cif, n

By ADITYA RANJAN SIR

STEP II:

If n is odd, then median is the value of

n+1)\"

oa

af n fawn à, a eal

aq a AA $1
If n is even, then median is the AM of
un un
n n
the values of E) and E)
observations.
.

aña mt, (2) x

“
(4) dei & ari ar AM À

41. ey ADITYA RANJAN SIR

15. The following are the marks of 9 students in
a class. Find the median.

PARA

363 2% 48, ME dat ae
(c) 38 + 21

PRK
N$= 32 GS

42. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

16. Find the median of the daily wages of ten
workers from the following data:

frafatad ist à aa ae at afr oat ar

ara aa PAU:

re. IHM (IMD
Aa 16 (b) 18

(e) 20 (d) 22

AAA oa
Qng >

43. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

17. The median of the following data will be

32, 25,33,27, 35, 29 and 30
32, 25,33,27, 35, 29 ait 30

(a) 32 (b) 27

er 30° (d) 29
OR

dus. 30

44. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

18. The median of a set of observations 15, 16, 18,
22, ER Dig. t+ 26, 27, 30 arranged in
ascending order 24) then find the value of x.
aná wa sahen Wau à um Ae 15, 16, 18,
22,x+2,x+3,26, 27, 30 at wma 24 À,
a x a aM E

(a) 26 (b) 25
(c) 20 gr 22

EOS AD y, 296
X42=2y

A

45. COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

COMPLETE

3
19. The median of observations Ea k + 2,
1 1
k-1,k+4,k+ 7,k-3,k+4>) is .
, 3 1
avi, k-7,k +2, k-1,k+4,k+ 7,k-3

1 à
k+4, St ae ara AO
1
2
(ce) k- 1 (d) k+2

(a) k- (b) k+

DIo

46. => -2

TA Se 6 | à 3 Ye -

47. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

FIP OPS

3
19. The median of observations Ea k + 2,

1 1
k-1,k+4,k+>,k-3,k+4> is __
1-5 z0-5
Ro, +2, K +4, $8 Y
MS

INTPC 01/04/2021 (Shift-03

3 b)/
la) -; (e)
1

D ple

(4) +

48. ey ADITYA RANJAN SIR

20. What is the difference between mean and
medain of the given data.

feu ma ats à area ae aa À aot aa El

BST IL VOB BI
(a) 4 (b) 2

5 (d) 1.5

Mean- $4 84941412 4164649413 Fig,
ng VA phy
x Mion
O U:

49. MEO

4, lo
2

9S
“ Cr you

50. 230519

Pore

51. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

21. The median of a set of 11 distinct observations
is 73.2. If each of the largest five observations

ST of the set is increased by 3, then the median
asec Ss mess of the new set:
+L +I+[-[-| 11 fart Ago & wR yea aa 73.2 $

añ ge A wae as we Wait À à wes À 3
Aa at art à, at ga AU anes at a :
(a) Is 3 times that of the original set
We ae ar 3 mE
(b) Is increased by 3/3 Hi af a à
«jala the same as that of the original set
ya de HOT na
(d) Is decreased by 3/3 À wer &

52. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

22. The median of a set of 7 distinct observation
is 21.5. If each of the largest 3 observations
of the set is increased by 4, then the median
of the new sets-

= = — 7 fa Dev a ayers ait aaa 21.5 $1 ale
Lid ayaa & wat ag 3 daw à à were Ha at ate
wet at aU, at AU aera ait aaa -
(a) Will decrease by 4/4 a grit
(b) Will be four times the original median
wet an WT Et
(c) Will remain the same as that of the original
set/ YA Wye A WAM À wet
(d) Will increase by 4/4 así

53. MEDIAN OF DISCRETE FREQUENCY

DISTRIBUTION
WAI INARA Fea EAT

54. COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

STEPI: Find the cumulative frequencies (c.f.)

N A
STEP II: Find >, where N= Ih
&

N > .
— am at, NY,
2 ia
STEP III: See the cumulative frequency (c.f.) just
N
greater than > and determine the
corresponding value of the variable.
#
AA y
2 à ote afer dat anata (AL, ) à

ait at am ana a aia ati

STEP IV: The value obtained in step III is the
median.

ru a are at hen 71

55. COMPLETE

COURSE (For all govt. exams ) By ADITYA RANJAN SIR

23. Obtain the median for the following frequency
distribution:

Pratetad safe fer à fer aaa wre al:

x|1/2 3 4 5 6 7 89
f|8]ı0/11/16/20/25/15/9|6
(a) 3 (b
(e) 7 (d) 10
CFL
“EBs =60

56. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

24. Find the medain of the following data.

e R
= E E Fafafada atest at aaa ara A
4 275 Term (x) s|7|9]11]13]16
a 4,9 Frequency |3|2|4|/6|3]|5

We 6 ==
13 fe (a) 15 (b) 12
ls Ss e (c) 10.5 on

57. Medion

=?

58. MEDIAN OF A GROUPED OR
CONTINUOUS FREQUENCY

DISTRIBUTION
Add ALARA dea Ht

59. By ADITYA RANJAN SIR

COMPLETE

COURSE (For all govt. exams )

STEP I: Obtain the frequency distribution. STEP V: Use the following formula:
aña faaror ura at,
STEP Il: Prepare the cumulative frequency
column and obtain N = If,
tect angfa ein dar at sit N= Df wa
Fr
N y
STEP HI: Find 2 am at NP
M 2
STEPIV: See the cumulative frequency just fort « L+ | — xh

N
greater than > and determine the

corresponding class. This class is
known as the median class.

2a dha fe woh st a af

a aa ati ga at at ofa ot à
ay à AAT STAT $

where, 1= lower limit of the median class
Wel, L = aaa ai at a dear

f = frequency of the median class

f = af at at angi

h = (size) of the median class

60. Lit (a tox(g-<)

61. COMPLETE

ws F |

540 5 Dc

DE DO
bi at GD

20-25 10 36

28-30 Sel

DI 4 41

35-40 9 = 4s

Uo-us 9 =

=N

COURSE (For all govt. exams )

By ADITYA RANJAN SIR

25. Calculate the median from the following

distribution:

Free fur aaa at ort at:

class _|5-10]10-15]15-20]20-25/25-30|30-35|35-40|40-45,
frequency) 5 | 6 | 15 | 10 5 | 4 | 2 | 2
(a) 17.4 (b) 18.4
(c) 14.2 (919.5

= IS + Ls 1) re)

FIs
_ wes
us

Sg.
ng

62. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

26. Find the median of the following data.

Clow my Cf frafafaa atest at aaa ara AO
Ss yl
= 2 a O: Class |10-15]15-20]25-30/30-35]35-40
us oe ap" le $ O nn Frequency| 7 9 11 8 18
30-35 en (a) 20: .1
35-40 18 <Se N (c) .35
A = 83-26.
2 3 <S Midi (+ (4-4) (a
em =)
= acy F
TA] * ss.)
2 _ sx log
SH SS

63. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

(a) 38 (b) 40 27. Find the median of the following data.
(1:85 La fafafaa siterst at aaa ara O
<= 835 Class (CI) Frequency (F)| Cf
mee 0-10 8 8
Omg = b+ (dr (g-<) 10-20 3 u
+ ? 20-30 7 18
= Uo+ 18 x = 30-40 4 22 €
= + ls ° =) 4 40-502 10 ED
ES 50-60 1 33
~ Abs 60-70 3 36
70-80 5 yy
80-90 2 Ya
90-100 4
Da

64. The mode or modal value of a distribution is
that value of the variable for which the
frequency is maximum.

ae aM fau nu Ser À aaa af oe feed à à
art sean rate are Ser MER aaa El

FAR

65. ey ADITYA RANJAN SIR

28. Find the mode of the following data:
Frorfefera Ser ar ae ama at:
25, 16, 19 48, 19; 20, 34, 15, 19, 20; 21, 24,
19,16, 22, 16, 18, 20, 16,19
(a) 16 Yai NULS sr
(c) 20 3 AR (d) 22 40

66. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

29. What will be the mode of the following data?
Prafetaa atest ar agen aa er?
13, 15, 31, 12,27) 13,27)60)27)@8)and 16
(a) 28 147% (b) 31 Acie
(e) 30 IM (727 8 MS

67. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

30. What is the mode of the given data?
fa mm Ser AGE A a
(6)7, 9,7,87,6)7, 8, 6, 7
DS AM b)6 AR

(ce) 592m (d) 3 Laie

68. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

31. Find the mode and median of 8, 7, 3, 7, 9, 4,

(a) 9,8 Nr, 7
(c) 8,6 d) 7,8

69. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

32. Find the sum of the mean, median and mode
of the given data.
fau m Set 0 ma, aaa ait age aT am

Sum = 22425425 = Pz,

z 0 935,90, 96,95, 38, 25

INTPC 30/01/2021 (Shift-01)|
(a) 50 (b) 4
(c) 75 ay

FE 224
Mean: ere PT
: Mode = Qs

= Sy

ae

Median 288

70. MBM A
Mean ($) Mode = 3Median-2 Mean

8 = 3x3-
he a x3 e
Median = 3 Ye =
N
g

71. ey ADITYA RANJAN SIR

Relation between mean, median and mode:

aa, afta ait agen à a dae:

Mode/SE th
= 3(Median/ATÈAAT) - 2(Mean/ATeA)

(mode = 3Median - Amıan

72. COMPLETE

COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

=60 33. Mean of a sample data = 60 and median = 48.
Mean =6'

a edion= 48 Find the mode of this distribution.

mods = ? TR TAT atest FT AA = 60 ait aa = 48 $1

ga der AT agers aa AU

SSC CGL TIER- II 06/03/2023}

(a) 36 (b) 18
24 (4) 48
Mode = 3Median- a mean
= 3xug- So
OS

= ey

73. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

34. Find the mode if mean and median are 4 and
5 respectively.

afa urea ait af mam: 4 ait 5 À À age

ara PAU
TPC 05/02/2021 (Shift-01)|
(a) 9 b) 7
(e) 11 (d) 5
Mean =y
Medion = s

Mody = IM edian - Mean
a 3XS-2xy

74. COMPUTATION OF MODE FORA
CONTINUOUS FREQUENCY
DISTRIBUTION
Add INARA ICH AGA

—

75. LeL- 23,44 4/44, 3,9, 96,6.

02

49,

7

x
q
on (8)
6
lo

SSsess 86,6,6,6, 10
“hay = g

F

2

ul
5

PE

Mode = Y

03 = 5 7
Q
80-40 2 5 te
40- so

Mod = IH 4- |

76. COMPLETE

STEPI:

STEP II:

Obtain the continuous frequency
distribution.

Waa arate far ura ati

Determine the class of maximum
frequency either by inspection or by
grouping method. This class is called
the modal class.

Fran ar gérer fahr grt aña arate
ar at fratfta ati ga at at aa at mer
ara #1

COURSE (For all govt. exams )

STEP Ill:

STEP Iv:

By ADITYA RANJAN SIR

Obtain the values of the following from
the frequency distributio:
arafe far à farafataa $ ur ura at:
1= lower limit of the modal class,

asa at at fraeft diem,

f = frequency of the modal class

tea ai at anata

h = width of the modal class,

Hise ot at a,

f, = frequency of the class preceding the
modal class,

tea at à weet are at at ama

f, = frequency of the class following the
modal class.

Wea wi a ara ai at angie
Substitute the values obtained in step
II in the following formula:

au à wa at at afan qa à

va:

f~f,

wae

Mode/ AR = 1+

77. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

35. Given below is the data of the age of the
various children.
An fates aeat a ar Ser fear mar +
What is the difference between the mean and
mode of the ages?
ay à ma ait agen À aa stat 2?

Age (years) Number of
(x) Children (F)

2.6 (b) 2.5
(c) 3.5 (d) 3.6
Mode = S
= TPs ase
+2
= agg BS ES
Mg

78. ey ADITYA RANJAN SIR

f:2 2-2 36. For the following grouped frequency
$,=10 h=3 distribution, find the mode:
452 Frorferfar ef amt recor © TL, MER ar
ang: EST xh di à
2441-1, [ Class: E 6-9 | 9-12 15 | 15-18 | 18-21 | 21-24 |
Frequency: | 2 | 5 | 10 + 21 12 | 3
= "+ sa (2) 13.6 1 (b) 15.6 ©
ve) (c) 15.4 (d) 14.6
ef
a Ae y
= p 3
t96

e

79. COMPLETE

COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

37. Find the mode for the given distribution

(rounded off to two decimal places).

volveras u alia
sen)

Class
5 LE 15/15-20 20-25 30 ss
Interval

Frequency | 8 7 6 ¡HOJ De =
ISSC CGL TIER - II 02/03/2023]

(a) 35.25 (b) 40.25
(c) 30.33 (4) 48.33

80. COMPLETE

COURSE (For all govt. exams ) By ADITYA RANJAN SIR

a 38. The mode for the above grouped frequency
modız À EL xh distribution is-
OF Fife

sated aga arcana dea fea age à-

= 20+/ 10 x lo Category Frequency
Y-10-5

= 204 2ywxwa
SEs
Y

81. COMPLETE COURSE (For all govt. exams ) By ADITYA RANJAN SIR

Mod = A+ 4 1 Xh 39. Given below is the distribution of Y 8students
! 2$-f;-f2

present in the class on the basis of their
attendance (days): Find mock.
* … ara aman À aaa YQ maña ar fur saat
36-13-8

9

ES i ages =?

> M4 8 xy Number of days of Attendance | 6.10 [10.14 ($18 | 18-22 | 22-26
[ar Number of Students
= a fern an wm ? a
* 1583 Es
=e (a) 15.29 (b) 15.33

(c) 15.60 ) 16.50

82. . Range e

ee oe ye
a
Median LT © Standant Deviation A

. eR
Mode. uns | Mis Manus.
en

83. Difference between highest and lowest
numbers, is called Range.
sere ait Pen dent a à siat at aa
wed El
How to find the Range:
We Ha qa ml
(i) Put the numbers in ascending order.
FEN wt wae vest al wa À aña al
(ii) Subtract the lowest value from the largest.
waa as à Aaa Be Hm a vere!
Range = largest value - smallest value

84. ey ADITYA RANJAN SIR

40. What is the range of the following data?

frafefiaa à Stet ar gf aa à? i
Data/ZIzI: 35, 40,25) 27, 38, 45, 50,65)
(a) 44 (b) 45
(c) 38 \ (91-40
Vane = les
= 68-25

~ 4o

85. ey ADITYA RANJAN SIR

41. Find the range of 12, 22, 7, 1, 5, 27, 30, 43.
12, 22, 7,() 5, 27, 30, Det a ara Rf
(a) 28 (b) 48

(c) 35 sr 42

86. COMPLETE

For all govt. exams ) By ADITYA RANJAN SIR

42. Calculate the range for the given frequency
distribution.

saat

(a) 50
(e) 60

Set ar ura (Ya) wm à?

Class Interval | Frequency

L 40,20 2
20-30 3

30-40 14
40-50 8
50-60 3
60-70 8
70-80) 4 | 2

70

(d) 55

87. VARIANCE

Variance is a measure of variability in statistics. It
assesses the average squared difference between data
values and the mean.

warn - fa aaa à aftadagitera ar ua ug $1 aE Set
unit ait ma & ata sited at stat at en AT 21

It is denoted by (0)/3% (of fea frat art 21
variance /faarut

88. © 2,5,6,% 10 - Find Variance =?

fos Step t- X = Siseórito = €)
5
Step 2:- CHE -4,-1,0 ,1,4
Step 3: (xi-k) = 6, 4 0
Si 2 (278, 416
te
== => l6+1+0
A ete

74

89. (For all govt. exams ) By ADITYA RANJAN SIR

How to compute variance and standard deviation?

Warr ait are fe ait worn à wt?

Step 1 - Compute the simple mean X.

aro 1- Aer area XX at UT AI

Step 2 - Calculate the difference of x,- x, for each

value in the data set.

caro 2- er te eee a ah fo x SC TOT ar

Step 3 - Calculate the squared difference (x,- x),

for each value in the data set.

arm 3 - Sar te trees Ta a SAT (x - x), BOT A

Step 4- Sum of Differences of the the squares Y”, (xx P.

wa een Dil:

Step 5 - Divide the sum of squared differences with n,
a 2

variance pers alos)

as - ai sit & am at nd fera at

far (gs) 2 E

90. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

43. Calculate the variance from the following data.

E lias

INTPC 02/03/2021 (Shift-03)

(a) 3 2

(e) 2.2 ae

X= ag +e

ms CO Py CÉOETCS
= “so °

"©

91. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

44. Calculate the variance for the following data:
Frafataa Ser & far fear at wort at:

2, 5, 6, 8,9
NTPC 19/03/2021 (Shift-03)
(a) 3 (b) 4
6 (d) 5
X= 20 6
os
- (-ur
Mes (Gp, (OY )*4 (3
TEN
sl,
ER

a 6

92. By ADITYA RANJAN SIR

45. The variance of the seven observations 6, 7,
10, 12, 13, 8, 14 is:

ara Hero 6, 7, 10, 12, 13, 8, 14 wT HATUT E:
INTPC 19/03/2021 (Shift-01)

COURSE (For all govt. exams )

COMPLETE

(a) 9 (b) 9.25
(c) 8.50 Ad) 8.29
X: 38 10

E

- (yn
Cu, (Py, ES

BUCH 7
ALTER

I HG "Sa.

are *

93. Yariauce

1,2, 3 4,8, 12
Y= 14243 = X= 44842 = 8
A O 3

2
a= CO os q tugs (+ CoN (y?
2 4 4

> Be
3

94. Note
O Numbers 5 DE add] Subrad —> No Change in Variouce

© Aumbers x" ima Variance X (7 (x)?

by x
Ex: am 812,16, uy
= lo sz

=l0xy?
=]
Lo

95. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

46. The variance of 20 observations is 5. If each
observation is multiplied by 2, then the
variance of the resulting observation will be:
20 Rant wt wart 5 1 ata were ter A 2 À
rom fear wre, at attend eur at war ET:

TPC 10/02/2021 (Shift-03

(a) 5 (b) 2x 5
‚Jsr2? x 5 (d) 2x 5?
Va
New Variante > 51%

96. Standard Deviation

2
Yariane = & If o? is the variance, then o, is called the
§-D=V Variance standard deviation./1% o fur à, À o ums
ez” fa rara à

eet Standard Deviation (0) = variance

a en"
are fren (0) = ma = [Zul = 2°

n-1

Varance =o

97. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

47. If the variance of 5 value is 0.81, then what is
its standard deviation?

afa 5 Wri at URRU 0,81 à, at SUN A free aa E?

ISSC MTS 26/10/2021 (Shift-01)

(a) 0.09 ‚br6.9
(c) 2.7 (d) 0.27
S = 081
S.
D= {ogy

nu

0.
0.9

98. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

Nariance 48. Calculate the standad deviation for the
e following data.
x5 frafattad Ser à fea ame face at TOT atl

Yorione= Cr C+ 0% P42? > 4,8) 6,7
5 INTPC 14/03/2021 (Shift-01)|
"+ me (b) V3
qe ¿Dña (a) 2
So

99. 49. Find the standared deviation of the following

Variance data (rounded off = two decimal nn
ES Camas tara a quite )i
4 4 5,3, 4,7
aid
(58) ( DE?) 1.48 (b) 3.21
y (e) 4.12 (d) 2.45
= y
16 + +4 + ET
nu

4

6
Spe
nes = $x LS

100. COMPLETE COURSE (For all govt. exams ) BY ADITYA RANJAN SIR

Yationce 50. Calculate the standard deviation for the
3. q following data.
XL Frefefiaa Ser & fea oe fa at wort at
A 7 4, 7, 9, 10, 15
s? = +20 6)
L s (a) 2.733 (b) 3.133

- 88 (c) 3.533 __(d) 3:633

SD= S6XS=
Sys” cs eB ns.