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Statistics question

2021 · 27 Jul · Shift 2 · Q28
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  5. /2021 · 27 Jul · Shift 2 · Q28

Statistics question

2021 · 27 Jul · Shift 2 · Q28

JEE MainMathematicsStatisticsMCQ+4 / −1
Let the mean and variance of the frequency distribution x:x1=2x2=6x3=8x4=9f:44αβ\begin{matrix} {x:} & {{x_1} = 2} & {{x_2} = 6} & {{x_3} = 8} & {{x_4} = 9} \\ {f:} & 4 & 4 & \alpha & \beta \\ \end{matrix}x:f:​x1​=24​x2​=64​x3​=8α​x4​=9β​ be 6 and 6.8 respectively. If x3 is changed from 8 to 7, then the mean for the new data will be :
  1. A
    4
  2. B
    5
  3. C
    173{{17} \over 3}317​
  4. D
    163{{16} \over 3}316​
View written solutionFree

Correct answer: C

  1. Given distribution
x2689f44αβ\begin{array}{c|cccc} x & 2 & 6 & 8 & 9 \\ f & 4 & 4 & \alpha & \beta \end{array}xf​24​64​8α​9β​

Mean =6=6=6 and variance =6.8=6.8=6.8.

We use:

xˉ=∑fx∑f,σ2=∑fx2∑f−xˉ2\bar x=\frac{\sum f x}{\sum f}, \qquad \sigma^2=\frac{\sum f x^2}{\sum f}-\bar x^2xˉ=∑f∑fx​,σ2=∑f∑fx2​−xˉ2
  1. Form equation using mean

Total frequency:

N=4+4+α+β=8+α+βN=4+4+\alpha+\beta=8+\alpha+\betaN=4+4+α+β=8+α+β

Also,

∑fx=4(2)+4(6)+8α+9β=32+8α+9β\sum fx=4(2)+4(6)+8\alpha+9\beta=32+8\alpha+9\beta∑fx=4(2)+4(6)+8α+9β=32+8α+9β

Since mean is 666,

32+8α+9β8+α+β=6\frac{32+8\alpha+9\beta}{8+\alpha+\beta}=68+α+β32+8α+9β​=6

So,

32+8α+9β=48+6α+6β32+8\alpha+9\beta=48+6\alpha+6\beta32+8α+9β=48+6α+6β 2α+3β=16...(1)2\alpha+3\beta=16 \quad ...(1)2α+3β=16...(1)
  1. Form equation using variance

Given variance =6.8=6.8=6.8, so

∑fx2N−62=6.8\frac{\sum f x^2}{N}-6^2=6.8N∑fx2​−62=6.8

Thus,

∑fx2N=42.8\frac{\sum f x^2}{N}=42.8N∑fx2​=42.8

Now,

∑fx2=4(22)+4(62)+α(82)+β(92)\sum f x^2=4(2^2)+4(6^2)+\alpha(8^2)+\beta(9^2)∑fx2=4(22)+4(62)+α(82)+β(92) =16+144+64α+81β=160+64α+81β=16+144+64\alpha+81\beta=160+64\alpha+81\beta=16+144+64α+81β=160+64α+81β

Hence,

160+64α+81β8+α+β=42.8\frac{160+64\alpha+81\beta}{8+\alpha+\beta}=42.88+α+β160+64α+81β​=42.8 160+64α+81β=42.8(8+α+β)160+64\alpha+81\beta=42.8(8+\alpha+\beta)160+64α+81β=42.8(8+α+β) 160+64α+81β=342.4+42.8α+42.8β160+64\alpha+81\beta=342.4+42.8\alpha+42.8\beta160+64α+81β=342.4+42.8α+42.8β 21.2α+38.2β=182.421.2\alpha+38.2\beta=182.421.2α+38.2β=182.4

Multiplying by 555:

106α+191β=912...(2)106\alpha+191\beta=912 \quad ...(2)106α+191β=912...(2)
  1. Solve (1) and (2)

From (1):

2α+3β=162\alpha+3\beta=162α+3β=16

Multiply by 535353:

106α+159β=848106\alpha+159\beta=848106α+159β=848

Subtract from (2):

(106α+191β)−(106α+159β)=912−848(106\alpha+191\beta)-(106\alpha+159\beta)=912-848(106α+191β)−(106α+159β)=912−848 32β=6432\beta=6432β=64 β=2\beta=2β=2

Now from (1):

2α+3(2)=162\alpha+3(2)=162α+3(2)=16 2α=10⇒α=52\alpha=10 \Rightarrow \alpha=52α=10⇒α=5
  1. Construct the new distribution

Now x3x_3x3​ changes from 888 to 777.

So new values are:

x2679f4452\begin{array}{c|cccc} x & 2 & 6 & 7 & 9 \\ f & 4 & 4 & 5 & 2 \end{array}xf​24​64​75​92​

Total frequency:

N=4+4+5+2=15N=4+4+5+2=15N=4+4+5+2=15

New sum:

∑fx=4(2)+4(6)+5(7)+2(9)\sum fx=4(2)+4(6)+5(7)+2(9)∑fx=4(2)+4(6)+5(7)+2(9) =8+24+35+18=85=8+24+35+18=85=8+24+35+18=85

Therefore new mean is

xˉnew=8515=173\bar x_{\text{new}}=\frac{85}{15}=\frac{17}{3}xˉnew​=1585​=317​
  1. Final answer
173\boxed{\frac{17}{3}}317​​

So the correct option is C.

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