Question

5 students of a class have an average height 150 cm and a variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of the six students is,
A. 22
B. 20
C. 16
D. 18

Answer 1

Hint: First assume the 5 students. Now substitute the value of mean and number of students in the mean formula $\dfrac{{\sum {{x_i}} }}{n}$ to get the total sum of the height of 5 students. After that substitute the value of the mean, number of students, and variance in the variance formula to get $\sum {{x_i}^2} $. Now add the height of the new student to the total height of students to get a total height of 6 students and find the mean by $\dfrac{{\sum {{x_i}} }}{n}$. Then, square the height of the new student and add it to $\sum {{x_i}^2} $ to get a new value. After that find the new variance by the formula $\dfrac{{\sum {{x_i}^2} }}{n} - {\left( {\bar x} \right)^2}$.

Complete step by step solution:
Let the 5 students be ${x_1},{x_2},{x_3},{x_4},{x_5}$.
The formula of mean is,
Mean $ = \dfrac{{\sum {{x_i}} }}{5}$
Substitute the values,
$ \Rightarrow 150 = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5}$
Cross-multiply the terms,
$ \Rightarrow {x_1} + {x_2} + {x_3} + {x_4} + {x_5} = 750$ ….. (1)
The formula for the variance is,
${\sigma ^2} = \dfrac{{\sum {{x_i}^2} }}{n} - {\left( {\bar x} \right)^2}$
Substitute the values in the above formula,
$ \Rightarrow 18 = \dfrac{{\sum {{x_i}^2} }}{5} - {\left( {150} \right)^2}$
Move the constant term on one side and simplify,
$ \Rightarrow \dfrac{{\sum {{x_i}^2} }}{5} = 22518$
Cross multiply the terms,
$ \Rightarrow \sum {{x_i}^2} = 112590$ ….. (2)
The height of the student (${x_6}$) is 156 cm.
Then the sum of the height of 6 students will be,
$ \Rightarrow \sum {{x_i}} = {x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}$
Substitute the values,
$ \Rightarrow \sum {{x_i}} = 750 + 156$
Add the terms,
$ \Rightarrow \sum {{x_i}} = 906$
Substitute the values in the mean formula,
$ \Rightarrow $ Mean $ = \dfrac{{906}}{6}$
Divide numerator by the denominator,
$ \Rightarrow $ Mean $ = 51$
Now find the sum of the square of the height of students,
$ \Rightarrow \sum {{x_i}^2} = {x_1}^2 + {x_2}^2 + {x_3}^2 + {x_4}^2 + {x_5}^2 + {x_6}^2$
Substitute the values from equation (2),
$ \Rightarrow \sum {{x_i}^2} = 112590 + {\left( {156} \right)^2}$
Square the term,
$ \Rightarrow \sum {{x_i}^2} = 112590 + 24336$
Add the terms,
$ \Rightarrow \sum {{x_i}^2} = 136926$
Now substitute the values in variance formula,
$ \Rightarrow {\sigma ^2} = \dfrac{{136926}}{6} - {\left( {151} \right)^2}$
Simplify the terms,
$ \Rightarrow {\sigma ^2} = 22821 - 22801$
Subtract the terms,
$\therefore {\sigma ^2} = 20$

So, the correct answer is “Option B”.

Note: Mean is the arithmetic average of the data given. Mean is also used to calculate the variance and standard deviation of the data in statistics. The mean is affected by extremely high or low values, called outliers.
In statistics, the difference between the mean and its data points is known as the variance. In simpler terms, it is the measure of how far a set of data points are far from their mean value.