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Statistics question

2019 · 8 Apr · Shift 1 · Q33
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Statistics question

2019 · 8 Apr · Shift 1 · Q33

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :
  1. A
    40
  2. B
    48
  3. C
    49
  4. D
    45
View written solutionFree

Correct answer: B

  1. Use the mean to find the sum of all 7 observations

Given mean =8=8=8 for 777 observations,

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

The 5 known 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.

So if the remaining two observations are xxx and yyy, then

x+y=56−42=14.x+y=56-42=14.x+y=56−42=14.
  1. Use the variance to find the sum of squares of all 7 observations

For variance,

σ2=∑xi2n−xˉ2.\sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^2.σ2=n∑xi2​​−xˉ2.

Given variance =16=16=16 and mean =8=8=8,

16=∑xi27−8216=\frac{\sum x_i^2}{7}-8^216=7∑xi2​​−82 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, the sum of squares of the 5 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.

So,

x2+y2=560−460=100.x^2+y^2=560-460=100.x2+y2=560−460=100.
  1. Use the identity to find xyxyxy

We know

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

Substitute x+y=14x+y=14x+y=14 and x2+y2=100x^2+y^2=100x2+y2=100:

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. Match with the options

Thus, the product of the remaining two observations is

48.\boxed{48}.48​.

So the correct option is B.

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