Let the mean and variance of the frequency distribution x: $\mathrm{x}_{1}=2$, $\mathrm{x}_{2}=6$, $\mathrm{x}_{3}=8$, $\mathrm{x}_{4}=9$ f: 4, 4, $\alpha$, $\beta$ be 6 and 6.8 respectively. If $x_{3}$ is changed from 8 to 7, then the mean for the new data will be :

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Statistics

Let the mean and variance of the frequency distribution

x:

x_{1} = 2

x_{2} = 6

x_{3} = 8

x_{4} = 9

f: 4, 4,

α

β

be 6 and 6.8 respectively. If

x_{3}

is changed from 8 to 7, then the mean for the new data will be :

Answer: (C)

Let

N = 8 + α + β

. The mean condition gives

\frac{2 \cdot 4 + 6 \cdot 4 + 8 α + 9 β}{N} = 6 ⟹ 32 + 8 α + 9 β = 48 + 6 α + 6 β ⟹ 2 α + 3 β = 16

Measuring the variance about the mean

6

\frac{4 ( 2 - 6 ) ^{2} + 4 ( 6 - 6 ) ^{2} + α ( 8 - 6 ) ^{2} + β ( 9 - 6 ) ^{2}}{N} = 6.8 ⟹ 64 + 4 α + 9 β = 6.8 (8 + α + β)

which reduces to

14 α - 11 β = 48

. Solving this with

2 α + 3 β = 16

gives

α = 5

and

β = 2

, so

N = 15

.
Changing

x_{3}

from

8

7

alters only that term of the sum of observations:

new mean = \frac{2 \cdot 4 + 6 \cdot 4 + 7 \cdot 5 + 9 \cdot 2}{15} = \frac{85}{15} = \frac{17}{3}

So the mean for the new data is

\frac{17}{3}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option