Statistics & Linear ProgrammingmediumPYQ · JEE Main · 2 Apr 2026 · Shift 2 (Afternoon)Free

Mean and Variance Problem: Number of Observations n = 7 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
The mean and variance of nn observations are 88 and 1616 respectively. If the sum of the first (n−1)(n-1) observations is 4848 and the sum of squares of the first (n−1)(n-1) observations is 496496, then the value of nn is
A2121
B1616
C1313
D77correct
Solution
Step 1: Mean =8=8 means the total sum is 8n8n:
x1+x2+⋯+xn−1+xn=8n ⇒ 48+xn=8n ⇒ xn=8n−48.(1)x_1+x_2+\cdots+x_{n-1}+x_n=8n\ \Rightarrow\ 48+x_n=8n\ \Rightarrow\ x_n=8n-48.\quad(1)
 Step 2: Variance =16=16:
16=496+xn2n−(8)2 ⇒ 80n=496+xn2 ⇒ xn2=80n−496.(2)16=\frac{496+x_n^2}{n}-(8)^2\ \Rightarrow\ 80n=496+x_n^2\ \Rightarrow\ x_n^2=80n-496.\quad(2)
 Step 3: Substitute (1) into (2):
(8n−48)2=80n−496.(8n-48)^2=80n-496.
 Step 4: Simplify:
64(n−6)2=8(10n−62) ⇒ 8(n−6)2=10n−62 ⇒ 4(n−6)2=5n−31.64(n-6)^2=8(10n-62)\ \Rightarrow\ 8(n-6)^2=10n-62\ \Rightarrow\ 4(n-6)^2=5n-31.
 Step 5: Expand:
4(n2−12n+36)=5n−31 ⇒ 4n2−48n+144=5n−31 ⇒ 4n2−53n+175=0.4(n^2-12n+36)=5n-31\ \Rightarrow\ 4n^2-48n+144=5n-31\ \Rightarrow\ 4n^2-53n+175=0.
 Step 6: Factor:
4n2−28n−25n+175=0 ⇒ 4n(n−7)−25(n−7)=0 ⇒ (n−7)(4n−25)=0.4n^2-28n-25n+175=0\ \Rightarrow\ 4n(n-7)-25(n-7)=0\ \Rightarrow\ (n-7)(4n-25)=0.
 Since nn is a positive integer, n=7n=7. Correct answer: (4)
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