Jee Main Math Class 12 Statistics JEE Main Physics: Authentic Shift PYQs (2021-2026) with 2.5-Min Shortcut Tricks & Trap Defense

Correct Answer: (b)

Explanation: Concept: Use properties of probability distribution: sum of probabilities equals 1, and $E(X^2) = \sigma^2 + \mu^2 = 2$ . Explanation: Let $p(0)=\frac{2a+1}{30}$ , $p(1)=\frac{8a-1}{30}$ , $p(2)=\frac{4a+1}{30}$ , $p(3)=b$ . Since $\sum p = 1$ , we have $\frac{2a+1}{30}+\frac{8a-1}{30}+\frac{4a+1}{30}+b=1$ . Combine numerators: $\frac{14a+1}{30}+b=1 \Rightarrow 14a+1+30b=30 \Rightarrow 14a+30b-29=0$ ...(1) Given $\sigma^2+\mu^2=2$ and $\sigma^2=E(X^2)-\mu^2$ , so $E(X^2)=2$ . $E(X^2)=\sum x^2 p(x) = 0^2\cdot p(0)+1^2\cdot p(1)+2^2\cdot p(2)+3^2\cdot p(3)$ . $= \frac{8a-1}{30}+4\times\frac{4a+1}{30}+9b = \frac{8a-1+16a+4}{30}+9b = \frac{24a+3}{30}+9b$ . Set equal to 2: $\frac{24a+3}{30}+9b=2$ . Multiply by 30: $24a+3+270b=60 \Rightarrow 24a+270b=57$ . Divide by 3: $8a+90b=19 \Rightarrow 8a+90b-19=0$ ...(2) Solve (1) and (2): Multiply (1) by 3: $42a+90b-87=0$ . Subtract (2): $(42a+90b-87)-(8a+90b-19)=0 \Rightarrow 34a-68=0 \Rightarrow a=2$ . Substitute $a=2$ in (1): $14(2)+30b-29=0 \Rightarrow 28+30b-29=0 \Rightarrow 30b=1 \Rightarrow b=\frac{1}{30}$ . Thus $\frac{a}{b}=2 \div \frac{1}{30}=60$ . Answer: $\boxed{60}$ (Option B)

Correct Answer: (b)

Explanation: $\text{ Mean }\;=\;\frac{4+5+6+6+7+8+x+y}{8}=6$ $\therefore x+y=12 . . . . \;\text{ (i) }$ And variance $=\;\frac{2^2+1^2+0^2+0^2+1^2+2^2+(x-6)^2+(y-6)^2}{8}$ $=\;\frac{9}{4}$ $\therefore (x-6)^2+(y-6)^2=8 . . . . \;\text{ (ii) }$ From (i) and (ii) $x=4$ and $y=8$ $\therefore x^4+y^2=320$

Correct Answer: (c)

Explanation: Mean $(\overline{x})=10$ $\Rightarrow\;\frac{\Sigma x_i}{20}=10$ $\Sigma x_i =10 \times 20=200$ If 8 is replaced by 12 , then $\Sigma x_1=200-8+12=204$ $\therefore\;\text{ Correct mean }(\overline{x})=\;\frac{\Sigma x_i}{20}$ $=\;\frac{204}{20}=10.2$ $\because\;\text{ Standard deviation }\;=2$ $\therefore\;\text{ Variance }\;=(\;\text{ S.D. }\;)^2=2^2=4$ $\Rightarrow\;\frac{\Sigma x_i^2}{20}- \left(\frac{\Sigma x_i}{20}\right)^2=4$ $\Rightarrow\;\frac{\Sigma x_i^2}{20}-(10)^2=4$ $\Rightarrow\;\frac{\Sigma x_i^2}{20}=104$ $\Rightarrow\;\Sigma x_i^2=2080$ Now, replaced ' 8 ' observations by ' 12 ' Then, $\Sigma x_i^2=2080-8^2+12^2=2160$ $\therefore$ Variance of removing observations $\Rightarrow\;\frac{\Sigma x_i^2}{20}- \left(\frac{\Sigma x_i}{20}\right)^2$ $\Rightarrow\;\frac{2160}{20}-(10.2)^2$ $\Rightarrow\;108-104.04$ $\Rightarrow\;3.96$ Correct standard deviation $=\sqrt{3.96}$

Correct Answer: (b)

Explanation: $\sigma^2 = \left(\frac{\Sigma x_i^2}{n}\right) - \overset{-}{x}^2$ $\overset{-}{x} = \frac{\sum\limits_{r=1}^{50} 2r}{50} = 51$ $\sigma^2 = \frac{\sum\limits_{r=1}^{50} 4r^2}{50} - (51)^2 = 833$

Correct Answer: (c)

Explanation: (a) is true as mode can be computed from histogram (b) is true as median will be changed if data's are changed (c) is false as variance depends on change of scale and independent of change of origin.