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JEE Advanced 2024Paper 1
Exam
JEE Advanced
Year
2024
Paper
Paper 1
Subject
Mathematics
Chapter
Statistics

JEE Advanced 2024 · Paper 1 · Mathematics · Statistics

Let X be a random variable, and let P(X=x) denote the probability...

Let XX be a random variable, and let P(X=x)P(X=x) denote the probability that XX takes the value xx. Suppose that the points (x,P(X=x)),x=0,1,2,3,4(x, P(X=x)), x=0,1,2,3,4, lie on a fixed straight line in the xyx y-plane, and P(X=x)=0P(X=x)=0 for all x∈R−{0,1,2,3,4}x \in \mathbb{R}-\{0,1,2,3,4\}. If the mean of XX is 52\frac{5}{2}, and the variance of XX is α\alpha, then the value of 24α24 \alpha is ___

Correct answer

0

Step-by-step solution

Let equation of line is y = mx + c x0l234R−{0,1,2,3,4}P(x)cm+c2 m+c3 m+c4 m+c0\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{x} & 0 & \mathrm{l} & 2 & 3 & 4 & \mathrm{R}-\{0,1,2,3,4\} \\ \hline \mathrm{P}(\mathrm{x}) & \mathrm{c} & \mathrm{m}+\mathrm{c} & 2 \mathrm{~m}+\mathrm{c} & 3 \mathrm{~m}+\mathrm{c} & 4 \mathrm{~m}+\mathrm{c} & 0 \\ \hline \end{array} ∑x=04(mx+c)=1⇒10m+5c=1⇒2m+c=15...(1)\sum_{x=0}^4(m x+c)=1 \Rightarrow 10 m+5 c=1 \Rightarrow 2 m+c=\frac{1}{5}...(1)  mean =∑xiPi=∑i=04(mxi+c)⋅xi=30 m+10c=52∴3 m+c=14…(2)\begin{aligned} & \text{ mean }=\sum x_i P_i=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \cdot \mathrm{x}_{\mathrm{i}}=30 \mathrm{~m}+10 \mathrm{c}=\frac{5}{2} \\ & \therefore 3 \mathrm{~m}+\mathrm{c}=\frac{1}{4} \ldots(2)\end{aligned} ΣPixi2=∑i=04(mxi+c)x12=∑i=04(mxi3+cxi2)⇒100 m+30c (Now putting m and c ) ⇒ΣPixi2=5+3=8 Variance =ΣPixi2−(ΣPixi)2=8−(52)2=74∴24α=42\begin{aligned} & \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}+\mathrm{c}\right) \mathrm{x}_1^2 \\ & =\sum_{\mathrm{i}=0}^4\left(\mathrm{mx}_{\mathrm{i}}^3+\mathrm{cx}_{\mathrm{i}}^2\right) \Rightarrow 100 \mathrm{~m}+30 \mathrm{c} \text{ (Now putting } \mathrm{m} \text{ and } \mathrm{c} \text{ ) } \\ & \Rightarrow \Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=5+3=8 \\ & \text{ Variance }=\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2-\left(\Sigma \mathrm{P}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}\right)^2=8-\left(\frac{5}{2}\right)^2=\frac{7}{4} \\ & \therefore 24 \alpha=42\end{aligned}
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