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Statistics question

2023 · 1 Feb · Shift 1 · Q29
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Statistics question

2023 · 1 Feb · Shift 1 · Q29

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and variance of 5 observations are 5 and 8 respectively. If 3 observations are 1, 3, 5, then the sum of cubes of the remaining two observations is :
  1. A
    1792
  2. B
    1216
  3. C
    1456
  4. D
    1072
View written solutionFree

Correct answer: D

  1. Use the mean to find the sum of all 5 observations

Let the 5 observations be 1,3,5,x,y1, 3, 5, x, y1,3,5,x,y.

Given mean =5=5=5, so

1+3+5+x+y5=5\frac{1+3+5+x+y}{5}=551+3+5+x+y​=5 1+3+5+x+y=251+3+5+x+y=251+3+5+x+y=25 x+y=25−9=16x+y=25-9=16x+y=25−9=16

So,

x+y=16x+y=16x+y=16
  1. Use the variance to find x2+y2x^2+y^2x2+y2

For observations a1,a2,…,ana_1,a_2,\dots,a_na1​,a2​,…,an​, variance is

σ2=∑xi2n−(∑xin)2\sigma^2=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2σ2=n∑xi2​​−(n∑xi​​)2

Given variance =8=8=8 and mean =5=5=5, hence

8=∑xi25−258=\frac{\sum x_i^2}{5}-258=5∑xi2​​−25 ∑xi25=33\frac{\sum x_i^2}{5}=335∑xi2​​=33 ∑xi2=165\sum x_i^2=165∑xi2​=165

Now,

12+32+52+x2+y2=1651^2+3^2+5^2+x^2+y^2=16512+32+52+x2+y2=165 1+9+25+x2+y2=1651+9+25+x^2+y^2=1651+9+25+x2+y2=165 35+x2+y2=16535+x^2+y^2=16535+x2+y2=165 x2+y2=130x^2+y^2=130x2+y2=130
  1. Find xyxyxy using (x+y)2(x+y)^2(x+y)2

We know

(x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy(x+y)2=x2+y2+2xy

So,

162=130+2xy16^2=130+2xy162=130+2xy 256=130+2xy256=130+2xy256=130+2xy 2xy=1262xy=1262xy=126 xy=63xy=63xy=63
  1. Find x3+y3x^3+y^3x3+y3

Use the identity

x3+y3=(x+y)3−3xy(x+y)x^3+y^3=(x+y)^3-3xy(x+y)x3+y3=(x+y)3−3xy(x+y)

Substitute x+y=16x+y=16x+y=16 and xy=63xy=63xy=63:

x3+y3=163−3(63)(16)x^3+y^3=16^3-3(63)(16)x3+y3=163−3(63)(16) =4096−3024=4096-3024=4096−3024 =1072=1072=1072
  1. Match with the options

Thus, the sum of cubes of the remaining two observations is

1072\boxed{1072}1072​

So the correct option is D.

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