Verified Solution • 2025–26 Session moderate

QJEEM19Maa8f2: JEE Main 2019 Mathematics - Class 11 JEEM19Maa8f2 Solution: If both the mean and the standard deviation of 50 observations…

If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2 , (x2 – 4)2 ,....., (x50 – 4)2 is :

Multiple Choice Options

  • A 400
  • B 480
  • C 380
  • D 525

Step-by-Step Verified Solution

Mean(μ)=∑xi50=16Mean(\mu ) = {{\sum {{x_i}} } \over {50}} = 16

∴\therefore ∑xi=16×50\sum {{x_i}} = 16 \times 50

S.D.(σ)=∑xi250−(μ)2=16S.D.\left( \sigma \right) = \sqrt {{{\sum {{x_i}^2} } \over {50}} - {{\left( \mu \right)}^2}} = 16

⇒∑xi250=256×2\Rightarrow {{\sum {{x_i}^2} } \over {50}} = 256 \times 2

Required mean = ∑(xi−4)250{{\sum {{{\left( {{x_i} - 4} \right)}^2}} } \over {50}}

⇒∑xi2+16×50−8∑xi50\Rightarrow {{\sum {{x_i}^2} + 16 \times 50 - 8\sum {{x_i}} } \over {50}}

⇒\Rightarrow 256 ×\times 2 + 16 - 8 ×\times 16

⇒\Rightarrow 400

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring