Question

ten Student of the bcom class of a college have obtained the following marks in statistics out of 100 mark calculated the standard deviations by the as well as the direct method
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Answer 1

Step-by-step explanation:

Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.

Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.

Marks Number of students

0-10 3

10-20 5

20-30 7

30-40 10

40-50 12

50-60 15

60-70 12

70-80 6

80-90 2

90-100 8

Now, we compute mean arithmetic mean by taking 55 as the assumed mean.

Computative of Mean

Marks

(x

i

) Mid-value (f

i

) Frequency u

i

10

x

i

−55

f

i

u

i

0-10 5 3 -5 -15

10-20 15 5 -4 -20

20-30 25 7 -3 -21

30-40 35 10 -2 -20

40-50 45 12 -1 -20

50-60 55 15 0 0

60-70 65 12 1 12

70-80 75 6 2 12

80-90 85 2 3 6

90-100 95 8 4 32

Total

∑f

i

=80 ∑f

i

u

i

= -26

We have,

N= sumf

i

=80,∑f

i

u

i

=−26, A= 55 and h= 10

∴

X

=A+h[

N

1

∑f

i

u

i

]

⇒

X

=A+h[

N

1

∑f

i

u

i

]

⇒

X

=55+10×

80

−26

=55−3.25=51.75Marks.