Verified Solution • 2025–26 Session moderate

QJEEM19Ma1219: JEE Main 2019 Mathematics - Class 11 JEEM19Ma1219 Solution: 5 students of a class have an average height 150 cm and…

5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :

Multiple Choice Options

  • A 16
  • B 22
  • C 20
  • D 18

Step-by-Step Verified Solution

Average height of 5 students,

x‾=x1+x2+x3+x4+x55=150\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150

⇒   ∑i=15xi=750\Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750

We know,

Variance (σ)=∑xi25−(x‾)2\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}

given that,

∑xi25−(150)2=18{{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18

⇒   ∑xi2=112590\Rightarrow \,\,\,\sum {x_i^2} = 112590

Height of new student, x6 == 156 cm

New average height   (x‾new)=750+1566=151\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151

New variance    = ∑i=16xi26−(x‾new)2= {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}

=112590+(156)26−(151)2= {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}

=22821−22801= 22821 - 22801

=20= 20

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