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Statistics question

2019 · 9 Jan · Shift 1 · Q36
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Statistics question

2019 · 9 Jan · Shift 1 · Q36

JEE MainMathematicsStatisticsMCQ+4 / −1
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :
  1. A
    16
  2. B
    22
  3. C
    20
  4. D
    18
View written solutionFree

Correct answer: C

  1. Given data for 5 students
  • Number of students: n=5n=5n=5
  • Mean height: xˉ=150\bar{x}=150xˉ=150 cm
  • Variance: σ2=18\sigma^2=18σ2=18 cm2^22

Using the formula

σ2=1n∑xi2−xˉ2\sigma^2=\frac{1}{n}\sum x_i^2-\bar{x}^2σ2=n1​∑xi2​−xˉ2

we get

18=15∑xi2−150218=\frac{1}{5}\sum x_i^2-150^218=51​∑xi2​−1502

So,

15∑xi2=18+22500=22518\frac{1}{5}\sum x_i^2=18+22500=2251851​∑xi2​=18+22500=22518 ∑xi2=5×22518=112590\sum x_i^2=5\times 22518=112590∑xi2​=5×22518=112590

Also, the sum of heights of the 5 students is

∑xi=5×150=750\sum x_i = 5\times 150=750∑xi​=5×150=750
  1. Add the new student

New student's height = 156156156 cm.

So for 6 students:

New sum

∑xi=750+156=906\sum x_i = 750+156=906∑xi​=750+156=906

New mean

xˉnew=9066=151\bar{x}_{new}=\frac{906}{6}=151xˉnew​=6906​=151

New sum of squares

∑xi2=112590+1562\sum x_i^2 = 112590+156^2∑xi2​=112590+1562 1562=24336156^2=243361562=24336

Thus,

∑xi2=112590+24336=136926\sum x_i^2=112590+24336=136926∑xi2​=112590+24336=136926
  1. Compute new variance

Again using

σnew2=16∑xi2−xˉnew2\sigma_{new}^2=\frac{1}{6}\sum x_i^2-\bar{x}_{new}^2σnew2​=61​∑xi2​−xˉnew2​

we get

σnew2=1369266−1512\sigma_{new}^2=\frac{136926}{6}-151^2σnew2​=6136926​−1512 1369266=22821\frac{136926}{6}=228216136926​=22821

and

1512=22801151^2=228011512=22801

Therefore,

σnew2=22821−22801=20\sigma_{new}^2=22821-22801=20σnew2​=22821−22801=20
  1. Check options
  • A: 161616
  • B: 222222
  • C: 202020 ✅
  • D: 181818

So the correct option is C.

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