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Statistics question

2020 · 5 Sep · Shift 1 · Q32
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  5. /2020 · 5 Sep · Shift 1 · Q32

Statistics question

2020 · 5 Sep · Shift 1 · Q32

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and variance of 7 observations are 8 and 16, respectively. If five observations are 2, 4, 10, 12, 14, then the absolute difference of the remaining two observations is :
  1. A
    2
  2. B
    3
  3. C
    1
  4. D
    4
View written solutionFree

Correct answer: A

  1. Use the mean to get the total sum

For 777 observations with mean 888,

sum of observations=7×8=56.\text{sum of observations} = 7 \times 8 = 56.sum of observations=7×8=56.

The five given observations are 2,4,10,12,142,4,10,12,142,4,10,12,14, whose sum is

2+4+10+12+14=42.2+4+10+12+14=42.2+4+10+12+14=42.

Let the remaining two observations be xxx and yyy. Then

x+y=56−42=14.x+y = 56-42 = 14.x+y=56−42=14.
  1. Use the variance to get the sum of squares

Variance is given by

σ2=∑xi2n−(xˉ)2.\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2.σ2=n∑xi2​​−(xˉ)2.

Here,

16=∑xi27−82.16 = \frac{\sum x_i^2}{7} - 8^2.16=7∑xi2​​−82.

So,

16=∑xi27−6416 = \frac{\sum x_i^2}{7} - 6416=7∑xi2​​−64 ∑xi27=80\frac{\sum x_i^2}{7} = 807∑xi2​​=80 ∑xi2=560.\sum x_i^2 = 560.∑xi2​=560.

Now, sum of squares of the five known observations is

22+42+102+122+142=4+16+100+144+196=460.2^2+4^2+10^2+12^2+14^2 = 4+16+100+144+196 = 460.22+42+102+122+142=4+16+100+144+196=460.

Hence,

x2+y2=560−460=100.x^2+y^2 = 560-460 = 100.x2+y2=560−460=100.
  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.

Substitute the values:

142=100+2xy14^2 = 100 + 2xy142=100+2xy 196=100+2xy196 = 100 + 2xy196=100+2xy 2xy=962xy = 962xy=96 xy=48.xy = 48.xy=48.
  1. Find ∣x−y∣|x-y|∣x−y∣

Use

(x−y)2=(x+y)2−4xy.(x-y)^2 = (x+y)^2 - 4xy.(x−y)2=(x+y)2−4xy.

So,

(x−y)2=142−4(48)=196−192=4.(x-y)^2 = 14^2 - 4(48) = 196 - 192 = 4.(x−y)2=142−4(48)=196−192=4.

Thus,

∣x−y∣=2.|x-y| = 2.∣x−y∣=2.
  1. Match with the options

The absolute difference of the remaining two observations is

2.\boxed{2}.2​.

So the correct option is A.

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