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Statistics question

2019 · 12 Jan · Shift 2 · Q34
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Statistics question

2019 · 12 Jan · Shift 2 · Q34

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :
  1. A
    1
  2. B
    7
  3. C
    3
  4. D
    5
View written solutionFree

Correct answer: B

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

Let the five observations be 3,4,4,x,y3, 4, 4, x, y3,4,4,x,y.

Given mean =4=4=4 for 555 observations, so 3+4+4+x+y5=4\frac{3+4+4+x+y}{5}=453+4+4+x+y​=4 11+x+y5=4\frac{11+x+y}{5}=4511+x+y​=4 11+x+y=2011+x+y=2011+x+y=20 x+y=9x+y=9x+y=9

  1. Use the variance formula

Given variance =5.20=5.20=5.20.

For n=5n=5n=5, variance is σ2=∑xi25−(∑xi5)2\sigma^2=\frac{\sum x_i^2}{5}-\left(\frac{\sum x_i}{5}\right)^2σ2=5∑xi2​​−(5∑xi​​)2

Since mean =4=4=4, 5.20=32+42+42+x2+y25−425.20=\frac{3^2+4^2+4^2+x^2+y^2}{5}-4^25.20=532+42+42+x2+y2​−42 5.20=9+16+16+x2+y25−165.20=\frac{9+16+16+x^2+y^2}{5}-165.20=59+16+16+x2+y2​−16 5.20=41+x2+y25−165.20=\frac{41+x^2+y^2}{5}-165.20=541+x2+y2​−16

So, 41+x2+y25=21.20\frac{41+x^2+y^2}{5}=21.20541+x2+y2​=21.20 41+x2+y2=10641+x^2+y^2=10641+x2+y2=106 x2+y2=65x^2+y^2=65x2+y2=65

  1. Use (x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy(x+y)2=x2+y2+2xy to find xyxyxy

We know x+y=9,x2+y2=65x+y=9, \quad x^2+y^2=65x+y=9,x2+y2=65

Now, (x+y)2=x2+y2+2xy(x+y)^2=x^2+y^2+2xy(x+y)2=x2+y2+2xy 92=65+2xy9^2=65+2xy92=65+2xy 81=65+2xy81=65+2xy81=65+2xy 2xy=162xy=162xy=16 xy=8xy=8xy=8

  1. Find the two numbers

So xxx and yyy satisfy t2−9t+8=0t^2-9t+8=0t2−9t+8=0 Factoring, (t−1)(t−8)=0(t-1)(t-8)=0(t−1)(t−8)=0 Thus, the two observations are 111 and 888.

  1. Find the absolute difference

∣x−y∣=∣8−1∣=7|x-y|=|8-1|=7∣x−y∣=∣8−1∣=7

  1. Check options

The correct option is B: 7.

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