Statistics — JEE Main practice

1. Since $X$ takes values $0,1,2,3$ with the given probabilities, compute the mean:

$$\mu=0\cdot\frac{2a+1}{30}+1\cdot\frac{8a-1}{30}+2\cdot\frac{4a+1}{30}+3b=\frac{16a+90b+1}{30}$$

2. Note that $\sigma^2+\mu^2=E(X^2)$, since $\sigma^2=E(X^2)-\mu^2$. So the given condition $\sigma^2+\mu^2=2$ directly means:

$$E(X^2)=2$$

3. Computing $E(X^2)=\sum X_i^2P(X_i)$:

$$0+1\cdot\frac{8a-1}{30}+4\cdot\frac{4a+1}{30}+9b=2$$

4. Simplifying:

$$8a+90b-19=0 \quad(i)$$

5. Since the total probability must sum to 1:

$$\frac{2a+1}{30}+\frac{8a-1}{30}+\frac{4a+1}{30}+b=1 \;\Rightarrow\; 14a+1+30b=30$$

6. So:

$$14a+30b-29=0 \quad(ii)$$

7. Solving equations (i) and (ii) simultaneously:

$$a=2,\qquad b=\frac{1}{30}$$

8. Therefore:

$$\frac{a}{b}=\frac{2}{1/30}=60$$

Hence, the answer is Option D.

1. For the numbers $k, 2k, \ldots, 1000k$, the median (average of the 500th and 501st terms) is:

$$\text{Median}=\frac{500k+501k}{2}=\frac{1001k}{2}$$

2. The mean deviation about the median is $\frac{\sum|X_i-X_M|}{n}$. By symmetry, the deviations pair up around the median, and the sum can be written as:

$$\text{M.D.}=\frac{2\left(\frac{k}{2}+\frac{3k}{2}+\frac{5k}{2}+\cdots\ 500\ \text{terms}\right)}{1000}$$

3. The sum inside is an arithmetic series of 500 odd multiples of $\frac{k}{2}$, whose sum is $\frac{k}{2}\times500^2$. So:

$$\text{M.D.}=\frac{2\cdot\frac{k}{2}(500)^2}{1000}=\frac{500k}{2}$$

4. Setting this equal to the given value of 500:

$$\frac{500k}{2}=500 \;\Rightarrow\; k=2$$

5. Therefore:

$$k^2=4$$

Hence, the answer is Option B.

1. Using the mean condition for the 8 numbers:

$$\frac{-10-7-1+x+y+16+2+9}{8}=\frac72 \;\Rightarrow\; x+y+9=28 \;\Rightarrow\; x+y=19$$

2. Using the variance condition:

$$\frac{100+49+1+x^2+y^2+256+4+81}{8}-\left(\frac72\right)^2=\frac{293}{4}$$

3. Simplifying:

$$\frac{491+x^2+y^2}{8}=\frac{342}{4} \;\Rightarrow\; x^2+y^2=193$$

4. Solving $x+y=19$ and $x^2+y^2=193$ together gives:

$$(x,y)=(7,12)\ \text{or}\ (12,7)$$

5. Since the numbers $x, y$ are interchangeable in this context, $|x-y|=5$ either way. The mean of the 4 numbers $x, y, x+y+1, |x-y|$ is:

$$\text{Mean}=\frac{x+y+(x+y+1)+|x-y|}{4}=\frac{19+20+5}{4}$$

6. Evaluating:

$$\text{Mean}=\frac{44}{4}=11$$

Hence, the answer is Option D.