Mathematics · Probability
JEE Main 2026 — 2 April, Evening Shift — Question 31
The mean and variance of n observations are 8 and 16, respectively. If the sum of the first (n - 1) observations is 48 and the sum of squares of the first (n - 1) observations is 496, then the value of n is:
- Option A:
21
- Option B:
16
- Option C:
13
- Option D:Correct
7
Answer: D
Step-by-step solution
\Rightarrow \mathrm{x}_{\mathrm{n}}=8 \mathrm{n}-48 \end{gathered}$$ $\Rightarrow 16=\frac{496+x_{n}^{2}}{n}-(8)^{2}$ $\Rightarrow 80 \mathrm{n}=496+\mathrm{x}_{\mathrm{n}}^{2}$ $$\begin{gathered} \Rightarrow \mathrm{x}_{\mathrm{n}}^{2}=80 \mathrm{n}-496 \end{gathered}$$ $\Rightarrow(8 \mathrm{n}-48)^{2}=80 \mathrm{n}-496$ $\Rightarrow 64(\mathrm{n}-6)^{2}=8(10 \mathrm{n}-62)$ $\Rightarrow 8(\mathrm{n}-6)^{2}=10 \mathrm{n}-62$ $\Rightarrow 4(\mathrm{n}-6)^{2}=5 \mathrm{n}-31$ $\Rightarrow 4\left(\mathrm{n}^{2}-36-12 \mathrm{n}\right)=5 \mathrm{n}-31$ $\Rightarrow 4 n^{2}+144-48 n=5 n-31$ $\Rightarrow 4 \mathrm{n}^{2}-53 \mathrm{n}+175=0$ $\Rightarrow 4 \mathrm{n}^{2}-28 \mathrm{n}-25 \mathrm{n}+175=0$ $\Rightarrow 4 \mathrm{n}(\mathrm{n}-7)-25(\mathrm{n}-7)=0$ $\mathrm{n}=7$
Answer key and solution verified before publishing.
Practise Probability
Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.
- Exam
- JEE Main 2026
- Subject
- Mathematics
- Chapter
- Probability
- Topic
- Mean, variance, expected values of distributions