Let the Mean and Variance of five observations x_1=1, x_2=3, x_3=a, x_4=7 and x_5=b, a\gt b , be 5 and 10 respectively. Then the Variance of the observations n+x_n, n=1,2, \ldots, 5 is

Statistics PYQ — JEE Main 2025 · 3 Apr · Shift 2 · Q37 with Solution

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Correct answer: B

The five observations are $x_1=1,\; x_2=3,\; x_3=a,\; x_4=7,\; x_5=b$ with mean $5$ and variance $10$ .

We need the variance of the new observations $n+x_n,\quad n=1,2,3,4,5.$ So the new set is $1+x_1,\; 2+x_2,\; 3+x_3,\; 4+x_4,\; 5+x_5.$

Since mean is $5$ for 5 observations, $\frac{1+3+a+7+b}{5}=5$ $1+3+7+a+b=25$ $11+a+b=25$ $a+b=14.$

For observations with mean $5$ and variance $10$ , $\frac{1}{5}\sum_{i=1}^5 (x_i-5)^2=10.$ Hence, $\sum_{i=1}^5 (x_i-5)^2=50.$

Now,

\quad (3-5)^2=4, \quad (7-5)^2=4. $$ So, $$16+4+4+(a-5)^2+(b-5)^2=50$$ $$24+(a-5)^2+(b-5)^2=50$$ $$ (a-5)^2+(b-5)^2=26. $$ Using $a+b=14$, $$[(a-5)+(b-5)] = a+b-10 = 4.$$ So if we let $$u=a-5,\quad v=b-5,$$ then $$u+v=4,\quad u^2+v^2=26.$$ Now, $$(u+v)^2=u^2+v^2+2uv$$ $$16=26+2uv$$ $$2uv=-10$$ $$uv=-5.$$ Thus $u,v$ are roots of $$t^2-4t-5=0,$$ which gives $$t=5,-1.$$ Hence, $$a-5=5,\; b-5=-1$$ (since $a>b$), so $$a=10,\quad b=4.$$ --- 4. **Form the new observations** The new observations are $$1+x_1=2, \quad 2+x_2=5, \quad 3+x_3=13, \quad 4+x_4=11, \quad 5+x_5=9.$$ So the set is $$2,5,13,11,9.$$ --- 5. **Find the mean of the new observations** $$\bar y=\frac{2+5+13+11+9}{5}=\frac{40}{5}=8.$$ --- 6. **Find the variance of the new observations** Variance is $$\frac{1}{5}\left[(2-8)^2+(5-8)^2+(13-8)^2+(11-8)^2+(9-8)^2\right].$$ Compute squares: $$(-6)^2=36, \quad (-3)^2=9, \quad 5^2=25, \quad 3^2=9, \quad 1^2=1.$$ Sum: $$36+9+25+9+1=80.$$ Therefore, $$\text{Variance}=\frac{80}{5}=16.$$ --- 7. **Check the options** The correct option is $$\boxed{16}$$ which is **Option B**. --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.

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Correct answer: B

The five observations are $x_1=1,\; x_2=3,\; x_3=a,\; x_4=7,\; x_5=b$ with mean $5$ and variance $10$ .

We need the variance of the new observations $n+x_n,\quad n=1,2,3,4,5.$ So the new set is $1+x_1,\; 2+x_2,\; 3+x_3,\; 4+x_4,\; 5+x_5.$

Since mean is $5$ for 5 observations, $\frac{1+3+a+7+b}{5}=5$ $1+3+7+a+b=25$ $11+a+b=25$ $a+b=14.$

For observations with mean $5$ and variance $10$ , $\frac{1}{5}\sum_{i=1}^5 (x_i-5)^2=10.$ Hence, $\sum_{i=1}^5 (x_i-5)^2=50.$

Now,

\quad (3-5)^2=4, \quad (7-5)^2=4. $$ So, $$16+4+4+(a-5)^2+(b-5)^2=50$$ $$24+(a-5)^2+(b-5)^2=50$$ $$ (a-5)^2+(b-5)^2=26. $$ Using $a+b=14$, $$[(a-5)+(b-5)] = a+b-10 = 4.$$ So if we let $$u=a-5,\quad v=b-5,$$ then $$u+v=4,\quad u^2+v^2=26.$$ Now, $$(u+v)^2=u^2+v^2+2uv$$ $$16=26+2uv$$ $$2uv=-10$$ $$uv=-5.$$ Thus $u,v$ are roots of $$t^2-4t-5=0,$$ which gives $$t=5,-1.$$ Hence, $$a-5=5,\; b-5=-1$$ (since $a>b$), so $$a=10,\quad b=4.$$ --- 4. **Form the new observations** The new observations are $$1+x_1=2, \quad 2+x_2=5, \quad 3+x_3=13, \quad 4+x_4=11, \quad 5+x_5=9.$$ So the set is $$2,5,13,11,9.$$ --- 5. **Find the mean of the new observations** $$\bar y=\frac{2+5+13+11+9}{5}=\frac{40}{5}=8.$$ --- 6. **Find the variance of the new observations** Variance is $$\frac{1}{5}\left[(2-8)^2+(5-8)^2+(13-8)^2+(11-8)^2+(9-8)^2\right].$$ Compute squares: $$(-6)^2=36, \quad (-3)^2=9, \quad 5^2=25, \quad 3^2=9, \quad 1^2=1.$$ Sum: $$36+9+25+9+1=80.$$ Therefore, $$\text{Variance}=\frac{80}{5}=16.$$ --- 7. **Check the options** The correct option is $$\boxed{16}$$ which is **Option B**. --- 8. **Comparison with stored answer** Stored correct answer: **B** Our derived answer: **B** So they agree.