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Statistics – Level 2
C.S.VEERARAGAVAN
The mid value of the class
27.5 – 37.5 is
32
32.5
33
33.5
0
4
The mid value of the class 27.5 – 37.5 is
32
32.5
33
33.5
Mid value =
Lower limit+Upper Limit
2
0
4
The mid value of the class 27.5 – 37.5 is
32
32.5
33
33.5
Mid value =
Lower limit+Upper Limit
2
Mid value =
27.5 + 37.5
2
0
4
If the mid value of an inclusive
class of size 7 is 9, Then the
class interval is
5 – 13
6 – 12
8 – 10
None of these
0
3
If the mid value of an inclusive
class of size 7 is 9, Then the
class interval is
5 – 13
6 – 12
8 – 10
None of these
Lower limit is 9 –
7−1
2
= 9 – 3 =
6
0
3
If the mid value of an
inclusive class of size 7 is 9,
Then the class interval is
5 – 13
6 – 12
8 – 10
None of these
Lower limit is 9 –
7−1
2
= 9 – 3 =
6
Upper limit is 9 +
7−1
2
= 9 + 3 = 12
0
3
The size of the exclusive class
interval 24 – 34 is
9
11
10
24
0
1
The difference between the lower (
or upper) limits of two successive
classes is the
Lower bound
Upper bound
Mid value of the class
Size of the class, for a continous
distribution
0
2
The arithmetic mean of the
series 2,5,8,11,14
8
6
9
7
0
5
The arithmetic mean of the
series 2,5,8,11,14
8
6
9
7
Mean of A.P =
First Number + Last number
2
0
5
Mean deviation of 8 and 17 is
4
3.5
4.5
5.5
0
6
Mean deviation of 8 and 17 is
4
3.5
4.5
5.5
Mean deviation =
17 −8
2
= 4.5
0
6
Mode of 3, 1 , 2 , 3 , 2 , 1, x , 3 , 4 , 3, 6
3
2
x
Cannot be determined
0
7
The upper boundary of an
inclusive type class 10 – 14 is
14
10
14.5
9.5
0
8
The upper boundary of an
inclusive type class 10 – 14 is
14
10
14.5
9.5
Boundaries of a class are obtained by
Subtracting 0.5 from Lower limit and
Adding 0.5 to Upper limit.
0
8
The range of the values
7, 8, 12, 9, 6, 13, 15, 21, 19, 5 is
15
13
14
16
0
9
The range of the values
7, 8, 12, 9, 6, 13, 15, 21, 19, 5 is
15
13
14
16
Range = 21 – 5 = 16
0
9
When a constant ‘c’ is subtracted from
every observation of given individual data
then the standard deviation of the data is
Increases by c
Decreases by c
Unchanged
Cannot be determined
1
0
The sum of the deviations about
mean of an individual data is equal
to
0
its arithmetic mean
its mean deviation
its range
11
The sum of deviations is least
when taken about
Mean
Median
Mode
All of the above
1
2
If the variance of x1, x2,x3…xn is p, then the s.d of
2x1 + 3, 2x2 + 3, …2xn + 3 is
𝑝
2 𝑝 + 3
2p + 3
2 𝑝
3
0
When 10 < x < 15, then the median
of the data 6, 18 , 21, 9 , 23, 5 and
x is
9
21
x
Cannot be determined
1
3
The A.M and the sum of
observations of individual data is 9
and 108 resp. The no. of
observations = ?12
10
11
5
1
4
The A.M and the sum of
observations of individual data is 9
and 108 resp. The no. of
observations = ?12
10
11
5
A.M =
Sum of observations
No of observations
1
4
For a symmetric distribution, the
mode is 24. The A.M of the
distribution is
22
26
24
Cannot be distributed
1
5
For a moderately symmetric distribution,
Mode – Median = ?
Median – Mean
Mode – Mean
3(Median – Mean)
2(Median – mean)
1
6
For a moderately symmetric distribution,
Mode – Median = ?
Median – Mean
Mode – Mean
3(Median – Mean)
2(Median – mean)
For a moderately
symmetric distribution
Mode = 3 median – 2
mean
1
6
The arithmetic mean of
the first n natural
numbers is
n n+1
2
n
2
n+1
2
n+1
2n
1
7
The A.M of the series x1, x2,x3… is 𝑥 then
the A.M of x1 – a , x2 – a , x3 – a , … xn – a is
𝑥
𝑥 – a
𝑥 – a
a 𝑥
2
9
Median of 8, 12, 13, 17 and 19 is
12.5
13
13.5
6.5
1
8
Median of the data
6, 15, 21, 28, 32 and 40
is24.5
24
21.5
28
1
9
The median of the first five prime
numbers is
11
5
7
2
2
7
The median of five
observations is the third
observation.
12-06-2015VEERARAGAVAN C S
veeraa1729@gmail.com 9894834264
34
The median of five
observations is the third
observation.
The third prime no is 5.
12-06-2015VEERARAGAVAN C S
veeraa1729@gmail.com 9894834264
35
In some individual data consisting of 20
observations, the observation a0 occurs
for the greatest number of times. The
mode isa0
a0
2
2a0
Cannot determine
2
0
The G.M of the data 1, 3, 12 is
36
6
3
36
3
2
1
If A, G and H are A.M, G.M & H.M of
2 +ve nos. a and b, then which is
true? A
G
=
H
A
G
H
=
H
A
A
G
=
G
A
A
G
=
G
H
2
2
If each observation is increased by
5, then the range of the data
Increases by 5
Decreases by 5
Does not change
May or may not change
2
3
If the range and the minimum value of
the observations are 17 and 88 resp.,
then the maximum value of the data
is
100
105
71
110
2
4
The first quartile (Q1) of the observations
4, 8, 10, 15, 17, 29 and 32 is
8
16
29
53
2
5
The first quartile (Q1) of the observations
4, 8, 10, 15, 17, 29 and 32 is
53
29
16
8
If the data is in ascending order,
then
Q1 =
n+1
4
𝑡ℎ
data.
2
5
The third quartile ( Q3) of the data 16,
21, 23, 25, 29, 32, 46, 48, 51, 53
, 54
51
48
29
53
2
6
The third quartile ( Q3) of the data 16,
21, 23, 25, 29, 32, 46, 48, 51, 53
, 54
51
48
29
53
2
6
The third quartile is the
3
4
Q3 is
3
4
n+1 th data = 9th data

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Statistics – level 2

  • 1. Statistics – Level 2 C.S.VEERARAGAVAN
  • 2. The mid value of the class 27.5 – 37.5 is 32 32.5 33 33.5 0 4
  • 3. The mid value of the class 27.5 – 37.5 is 32 32.5 33 33.5 Mid value = Lower limit+Upper Limit 2 0 4
  • 4. The mid value of the class 27.5 – 37.5 is 32 32.5 33 33.5 Mid value = Lower limit+Upper Limit 2 Mid value = 27.5 + 37.5 2 0 4
  • 5. If the mid value of an inclusive class of size 7 is 9, Then the class interval is 5 – 13 6 – 12 8 – 10 None of these 0 3
  • 6. If the mid value of an inclusive class of size 7 is 9, Then the class interval is 5 – 13 6 – 12 8 – 10 None of these Lower limit is 9 – 7−1 2 = 9 – 3 = 6 0 3
  • 7. If the mid value of an inclusive class of size 7 is 9, Then the class interval is 5 – 13 6 – 12 8 – 10 None of these Lower limit is 9 – 7−1 2 = 9 – 3 = 6 Upper limit is 9 + 7−1 2 = 9 + 3 = 12 0 3
  • 8. The size of the exclusive class interval 24 – 34 is 9 11 10 24 0 1
  • 9. The difference between the lower ( or upper) limits of two successive classes is the Lower bound Upper bound Mid value of the class Size of the class, for a continous distribution 0 2
  • 10. The arithmetic mean of the series 2,5,8,11,14 8 6 9 7 0 5
  • 11. The arithmetic mean of the series 2,5,8,11,14 8 6 9 7 Mean of A.P = First Number + Last number 2 0 5
  • 12. Mean deviation of 8 and 17 is 4 3.5 4.5 5.5 0 6
  • 13. Mean deviation of 8 and 17 is 4 3.5 4.5 5.5 Mean deviation = 17 −8 2 = 4.5 0 6
  • 14. Mode of 3, 1 , 2 , 3 , 2 , 1, x , 3 , 4 , 3, 6 3 2 x Cannot be determined 0 7
  • 15. The upper boundary of an inclusive type class 10 – 14 is 14 10 14.5 9.5 0 8
  • 16. The upper boundary of an inclusive type class 10 – 14 is 14 10 14.5 9.5 Boundaries of a class are obtained by Subtracting 0.5 from Lower limit and Adding 0.5 to Upper limit. 0 8
  • 17. The range of the values 7, 8, 12, 9, 6, 13, 15, 21, 19, 5 is 15 13 14 16 0 9
  • 18. The range of the values 7, 8, 12, 9, 6, 13, 15, 21, 19, 5 is 15 13 14 16 Range = 21 – 5 = 16 0 9
  • 19. When a constant ‘c’ is subtracted from every observation of given individual data then the standard deviation of the data is Increases by c Decreases by c Unchanged Cannot be determined 1 0
  • 20. The sum of the deviations about mean of an individual data is equal to 0 its arithmetic mean its mean deviation its range 11
  • 21. The sum of deviations is least when taken about Mean Median Mode All of the above 1 2
  • 22. If the variance of x1, x2,x3…xn is p, then the s.d of 2x1 + 3, 2x2 + 3, …2xn + 3 is 𝑝 2 𝑝 + 3 2p + 3 2 𝑝 3 0
  • 23. When 10 < x < 15, then the median of the data 6, 18 , 21, 9 , 23, 5 and x is 9 21 x Cannot be determined 1 3
  • 24. The A.M and the sum of observations of individual data is 9 and 108 resp. The no. of observations = ?12 10 11 5 1 4
  • 25. The A.M and the sum of observations of individual data is 9 and 108 resp. The no. of observations = ?12 10 11 5 A.M = Sum of observations No of observations 1 4
  • 26. For a symmetric distribution, the mode is 24. The A.M of the distribution is 22 26 24 Cannot be distributed 1 5
  • 27. For a moderately symmetric distribution, Mode – Median = ? Median – Mean Mode – Mean 3(Median – Mean) 2(Median – mean) 1 6
  • 28. For a moderately symmetric distribution, Mode – Median = ? Median – Mean Mode – Mean 3(Median – Mean) 2(Median – mean) For a moderately symmetric distribution Mode = 3 median – 2 mean 1 6
  • 29. The arithmetic mean of the first n natural numbers is n n+1 2 n 2 n+1 2 n+1 2n 1 7
  • 30. The A.M of the series x1, x2,x3… is 𝑥 then the A.M of x1 – a , x2 – a , x3 – a , … xn – a is 𝑥 𝑥 – a 𝑥 – a a 𝑥 2 9
  • 31. Median of 8, 12, 13, 17 and 19 is 12.5 13 13.5 6.5 1 8
  • 32. Median of the data 6, 15, 21, 28, 32 and 40 is24.5 24 21.5 28 1 9
  • 33. The median of the first five prime numbers is 11 5 7 2 2 7
  • 34. The median of five observations is the third observation. 12-06-2015VEERARAGAVAN C S veeraa1729@gmail.com 9894834264 34
  • 35. The median of five observations is the third observation. The third prime no is 5. 12-06-2015VEERARAGAVAN C S veeraa1729@gmail.com 9894834264 35
  • 36. In some individual data consisting of 20 observations, the observation a0 occurs for the greatest number of times. The mode isa0 a0 2 2a0 Cannot determine 2 0
  • 37. The G.M of the data 1, 3, 12 is 36 6 3 36 3 2 1
  • 38. If A, G and H are A.M, G.M & H.M of 2 +ve nos. a and b, then which is true? A G = H A G H = H A A G = G A A G = G H 2 2
  • 39. If each observation is increased by 5, then the range of the data Increases by 5 Decreases by 5 Does not change May or may not change 2 3
  • 40. If the range and the minimum value of the observations are 17 and 88 resp., then the maximum value of the data is 100 105 71 110 2 4
  • 41. The first quartile (Q1) of the observations 4, 8, 10, 15, 17, 29 and 32 is 8 16 29 53 2 5
  • 42. The first quartile (Q1) of the observations 4, 8, 10, 15, 17, 29 and 32 is 53 29 16 8 If the data is in ascending order, then Q1 = n+1 4 𝑡ℎ data. 2 5
  • 43. The third quartile ( Q3) of the data 16, 21, 23, 25, 29, 32, 46, 48, 51, 53 , 54 51 48 29 53 2 6
  • 44. The third quartile ( Q3) of the data 16, 21, 23, 25, 29, 32, 46, 48, 51, 53 , 54 51 48 29 53 2 6 The third quartile is the 3 4 Q3 is 3 4 n+1 th data = 9th data