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Statistics question

2021 · 20 Jul · Shift 1 · Q23
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Statistics question

2021 · 20 Jul · Shift 1 · Q23

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean of 6 distinct observations is 6.5 and their variance is 10.25. If 4 out of 6 observations are 2, 4, 5 and 7, then the remaining two observations are :
  1. A
    10, 11
  2. B
    3, 18
  3. C
    8, 13
  4. D
    1, 20
View written solutionFree

Correct answer: A

  1. Use the mean to find the sum of all 6 observations

Given mean =6.5=6.5=6.5 for 666 observations,

Total sum=6×6.5=39.\text{Total sum}=6\times 6.5=39.Total sum=6×6.5=39.

Four observations are 2,4,5,72,4,5,72,4,5,7, whose sum is

2+4+5+7=18.2+4+5+7=18.2+4+5+7=18.

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

x+y=39−18=21.(1)x+y=39-18=21. \qquad (1)x+y=39−18=21.(1)
  1. Use the variance to find the sum of squares

Variance is given by

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

Here,

10.25=∑xi26−(6.5)2.10.25=\frac{\sum x_i^2}{6}-(6.5)^2.10.25=6∑xi2​​−(6.5)2.

Since

(6.5)2=42.25,(6.5)^2=42.25,(6.5)2=42.25,

we get

∑xi26=10.25+42.25=52.5.\frac{\sum x_i^2}{6}=10.25+42.25=52.5.6∑xi2​​=10.25+42.25=52.5.

Thus,

∑xi2=6×52.5=315.\sum x_i^2=6\times 52.5=315.∑xi2​=6×52.5=315.

Now the sum of squares of the known four observations is

22+42+52+72=4+16+25+49=94.2^2+4^2+5^2+7^2=4+16+25+49=94.22+42+52+72=4+16+25+49=94.

Hence,

x2+y2=315−94=221.(2)x^2+y^2=315-94=221. \qquad (2)x2+y2=315−94=221.(2)
  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

From (1),

(x+y)2=212=441.(x+y)^2=21^2=441.(x+y)2=212=441.

So,

441=221+2xy441=221+2xy441=221+2xy 2xy=2202xy=2202xy=220 xy=110.xy=110.xy=110.
  1. Form the quadratic equation

The numbers xxx and yyy satisfy

t2−(x+y)t+xy=0.t^2-(x+y)t+xy=0.t2−(x+y)t+xy=0.

So,

t2−21t+110=0.t^2-21t+110=0.t2−21t+110=0.

Factorizing,

(t−10)(t−11)=0.(t-10)(t-11)=0.(t−10)(t−11)=0.

Hence,

t=10,11.t=10,11.t=10,11.

So the remaining two observations are

10,11.\boxed{10,11}.10,11​.
  1. Check options
  • A: 10,1110,1110,11 ✅
  • B: 3,183,183,18 gives sum 212121 but squares 9+324=333≠2219+324=333\neq2219+324=333=221 ❌
  • C: 8,138,138,13 gives sum 212121 but squares 64+169=233≠22164+169=233\neq22164+169=233=221 ❌
  • D: 1,201,201,20 gives sum 212121 but squares 1+400=401≠2211+400=401\neq2211+400=401=221 ❌

Therefore, the correct option is A.

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