Question

15. The mean of the following frequency table is 50. But the frequencies f, and f,
classes 20-40 and 60-80 are missing. The missing frequencies are
20-40 40-50 60 - 80 80 - 100 Total
Frequency
32
19
120
a=28
D) 1. = 28
c) si = 24
d) f. = 24​

Answer 1

Step-by-step explanation:

R

Class f

1

Frequency x

i

Mid-values u

i

=

h

x

i

−A

f

i

u

i

0−20 17 10 −2 −34

20−40 f

1

30 −1 −f

1

40−60 32 50 0 0

60−80 f

2

70 1 −f

2

80−100 19 90 2 38

N= ∑f

i

=63+f

1

+f

2

∑f

i

u

i

=4−f

1

+f

2

We have,

N=∑f

i

=120 [Given]

⇒68+f

1

+f

2

=120

⇒f

1

+f

2

=52

Now,

Mean = 50

⇒A+h[

N

1

∑f

i

u

i

]=50

⇒50+20×[

120

4−f

1

+f

2

]=50

⇒

6

4−f

1

+f

2

=0

⇒4−f

1

+f

2

=0

⇒f

1

−f

2

=4

Solving equations (i) and (ii), we get f

1

= 28 and f

2

=24.

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