Q 4. Calculate the Mean and Median of the following frequency distribution.

(10)

Marks No. of Students

0 - 10 3

10 - 20 6

20 - 30 5

30 - 40 8

40 - 50 15

50 - 60 6

60 - 70 10

Answer:

Mean (Calculated by long method)

Mean (M) = ∑fX/N

Mid point of each class interval (X) = Upper limit + Lower limit of class interval /2

fX = frequency of a class interval × corresponding mid point (X)

∑fX = Total sum of fX

N = Total number of observations or frequencies

Marks No. of Students (f) X fX

0-10 3 5 3 × 5 = 15

10-20 6 15 6 × 15 = 90

20-30 5 25 5 × 25 = 125

30-40 8 35 8 × 35 = 280

40-50 15 45 15 × 45 = 675

50-60 6 55 6 × 55 = 330

60-70 10 65 10 × 65 = 650

N = 53 ∑fX = 2165

N = 53

Efx = 2165

M = ∑ fX ÷ N

Mean = 2165 ÷ 53 = 40.85

Median

Median = l+{( [N /2 ]−F )÷ fm}×iN = Total number of frequencies

N/2 = One half of the total number of frequencies

l = Exact lower limit of the class interval within which the median lies

The class interval within which the median lies is found out by calculating the half of total

number of frequencies (N/2) followed by counting and adding the frequencies from first to last

interval and vice-versa, to locate the class-interval in which the calculated N/2 lies. This will lead

to the interval within which the median falls.

F = Sum of frequencies of all intervals from the lowest interval up to l

fm = Frequency of the interval within which the median falls

i = length of the interval

Marks No. of Students (f)

0-10 3

10-20 6

20-30 5

30-40 8

40-50 15

50-60 6

60-70 10

N = 53

N = 53

N/2 = 53/2 = 26.5

The class interval 40-50 should possibly have the median. Upon counting and adding

frequencies from the last to first interval and vice-versa, we find 26.5 lies in the interval 40 –

50.

l = 40

F = 3+6+5+8 = 22*

fm = 15

i = 10

Median = l+{( [N /2 ]−F )÷ fm}×i

Mdn = 40+ {( [53/2 ]−22 )÷15}×10Mdn = 40 + {(26.5−22 )÷15 }×10

Mdn = 40+ {4.5÷15 }×10

Mdn = 40+3

Mdn = 43

Therefore, Median is 43.

Q 5. Compute Standard Deviation of the following scores. (10)

IQ No. of Students

80 - 84 7

85 - 89 10

90 - 94 5

95 - 99 12

100 -104 20

105 -109 25

110 -114 32

115 -119 27

120 - 124 18

Answer:

Standard Deviation (Calculated by Long Method)

Standard Deviation SD =i×√∑ f d2∑N−(∑fd∑N )

2

N = Total number of frequencies

f = Frequency of a class interval

X = Mid point of a class interval

N = 156

Frequency = f

Assume Mean (AM) = 102

Interval(i)= 5

According to formula:

Class

intervals

of IQ

No. of

students

(f)

Mid

point

(X)

d=

X−AMi

d2

f . d2

f.d

80 - 84 7 82 -4 16 112 -28

85 – 89 10 87 -3 9 90 -30

90 – 94 5 92 -2 4 20 -10

95 -99 12 97 -1 1 12 -12

100 – 104 20 102 0 0 0 0

105 – 109 25 107 1 1 25 25

110 – 114 32 112 2 4 128 64

115 – 119 27 117 3 9 243 81

120 - 124 18 122 4 16 288 72

N = 156 918 162

SD =i×√∑ f d2∑N−(∑fd∑N )

2

SD = 5×√ 918156−( 162156 )2

SD = 5 ×√5.8−(1.03 )2

SD = 5 × √5.8−1.06SD = 5 × √4.74SD = 5 × 2.1

SD = 10.5

Therefore, Standard deviation of the IQ scores of the students in 10.5

Q 7. In a Psychology examination the marks obtained by two groups of students

are given below. Examine the significance of difference between the means of

the marks obtained by the students of two groups (10)

First Group : 50 49 18 20 34 22 61 60 56

Second Group : 35 30 29 28 46 15 55 54 47

Answer:

N = Total number of subjects

∑X = Sum total of subjects

Median (M) = ∑X/N

t value of two groups : t = M1−M 2

SED

Standard Error Difference (SED) between 2 groups SED = SD √ N1+N2N1×N2

Standard Deviation (x) = X – M

Standard Deviation SD = √∑ x12+∑ x2

2

N 1+N 2−2

Degree of freedom (df) = (N₁+N₂ - 2)

First group

(X₁)

SD (x₁) = X₁ -

M₁(x1 )2 Second group

(X₂)

SD (x₂) = X₂ -

M₂(x2 )2

50 8.9 79.21 35 -2.6 6.76

49 -7.9 62.41 30 -7.6 57.76

18 -23.1 533.61 29 -8.6 73.96

20 -21.1 445.21 28 -9.6 92.16

34 -7.1 50.41 46 8.4 70.56

22 -19.1 364.81 15 -22.6 510.76

61 19.9 396.01 55 17.4 302.76

60 18.9 357.21 54 16.4 268.96

56 14.9 222.01 47 9.4 88.36

∑X₁ = 370 ∑ ( x1 )2=¿¿

2510.89 ∑X₂ = 339

∑ ( x2 )2=¿

1472.04

First Group

∑X₁ = 370

N₁ = 9

Median of first group (M₁) = ∑X₁/N₁ = 370/9 = 41.1

Second Group

∑X₂ = 339

N₂ = 9

Median of second group (M₂) = ∑X₂/N₂ = 339/9 = 37.6

Formula for ‘t’ test is:

t = M1−M 2

SED---1

SED = SD √ N1+N2N1×N2---2

SD = √∑ x12+∑ x2

2

N 1+N 2−2---3

From -3

SD = √ 2510.89+1472.049+9−2

SD =√ 3982.9316

SD =√248.933125SD = 15.77

From-2

SED = 15.77×√ 9+99×9

SED = 15.77×√ 1881SED = 15.77×√0.2

SED = 15.77×0.4

SED = 6.3

From-1

t value = (M 1−M 2 )SED

t value = (41.1 –37.6 )

6.3

t value = 3.56.3

t value = 0.55

Degree of freedom (df):

df = (N₁+N₂ - 2)

df = (9+9 - 2)

df = 18 – 2

df = 16

Upon entering the value on 16 df, we get 2.12 at the 0.5 and 2.91 at the 0.1 level.

t value (0.55) is less than 2.12 and 2.91

Therefore, we will say that the significance of difference between the means of the marks

obtained by the students of two groups is “Non Significant”.

Q 9. Find Pearson product moment correlation coefficient from the following data (10)Marks in Psychology: 50 75 35 30 70 30 85 35Marks in Statistics : 60 80 40 60 65 40 80 25

Answer:

Subject Marks in Psychology (X)

Marks in Statistics (Y)

SD of X (x=X-M1)

SD of Y (y=Y-M2)

x2 y2 xy

1 50 60 -1.25 3.75 1.5625 14.0625 -4.68752 75 80 23.75 23.75 564.0625 564.0625 564.062

53 35 40 -16.25 -16.25 264.0625 264.0625 264.062

54 30 60 -21.25 3.75 451.5625 14.0625 -79.68755 70 65 18.75 8.75 351.5625 76.5625 164.062

56 30 40 -21.25 -16.25 451.5625 264.0625 345.312

57 85 80 33.75 23.75 1139.0625 564.0625 801.562

58 35 25 -16.25 -31.25 264.0625 976.5625 507.812

5N = 8 ∑X =410 ∑Y = 450 ∑x2 ∑y2 ∑xy

=3487.5 =2737.5 =2562.5M1 = ∑X/N = 51.25

M2= ∑Y/N= 56.25

Mean = ∑f/N∑X = Sum total of marks of psychology∑Y = Sum total of marks of statisticsN (Total no. of subjects) = 8M1 (Mean of X) = 51.25M2 (Mean of Y) = 56.25x = Standard deviation of X = X – M1

y = Standard deviation of Y = Y – M2

x2 = Square of SD of Xy2 = Square of SD of Y∑x2 (Sum total of squares of SD of X) = 3487.5∑y2 (Sum total of squares of SD of Y)= 2737.5xy = Multiplication of standard deviation of X with standard deviation of y∑xy (Sum total of SD of X multiplied by SD of Y) = 2562.5

Pearson Product Moment Correlation Formula (r):

r = ∑xy

N (σ x .σ y )

σx = √∑ x2/Nσx = √3487.5/8σx = √435.9σx = 20.8

σy = √∑ y2/Nσy = √2737.5/8σy = √342.18σy = 18.4

r (Pearson’s Product Moment Correlation Coefficient) = ∑ xy /N ×σx×σyr = 2562.5 /8×20.8×18.4r = 2562.5 /3061.76r = 2562.5 /3061.76r = 0.83r = +0.83

Since r = +0.83 which is closer to +1, the marks of psychology and marks of statistics have a stronger positive correlation.