JEE Main 2022 — Statistics Question with Solution

Given,

5 numbers are 3, 7, 12, a, 43 $$-$$ a

$$\therefore$$ Mean $$(\overline{x})=\frac{3+7+12+a+43-a}{5}$$

$$=\frac{65}{5}$$

$$=13$$

We know,

Variance $$({\sigma }^2)=\frac{\sum x_1^2}{n}-(\overline{x}{)}^2$$

$$=\frac{3^2+7^2+{12}^2+a^2+{(43-a)}^2}{5}-(13{)}^2$$

$$=\frac{9+49+144+a^2+1849+a^2-86a}{5}-169$$

$$=\frac{2a^2-86a+2051}{5}-169$$

$$=\frac{2a^2-86a+2051-845}{5}$$

$$=\frac{2a^2-86a+1206}{5}$$

$$=\frac{2}{5}(a^2-43a+603)$$

Variance will be natural number if $$a^2-43a+603$$ in multiple of 5.

$$\therefore$$ $$a^2-43a+603=5n$$

$$\Rightarrow a^2-43a+603-5n=0$$ ..... (1)

$$\therefore$$ $$a=\frac{43\pm \sqrt{D}}{2}$$

Now "a" will be natural number if

(1) D is a perfect square and

(2) $$43\pm \sqrt{D}$$ is multiple of 2

From equation (1),

$$D=(-43{)}^2-4.1.(603-5n)$$

$$=1849-2412+20n$$

$$=20n-563$$

For any value of n, unit digit of 20n is always 0. Then 20n $$-$$ 563 will give a number whose unit digit is 7.

For perfect square numbers ex : 1² = 1, 2² = 4, 3² = 9, 4² = 16, 5² = 25, 6² = 36, 7² = 49, 8² = 64, 9² = 81, 10² = 100

So, unit digit is either 1, 4, 6, 5, 6, 9, 0 it can't be 7. So D can't be perfect square.