Question

The mean of a set of 30 observations is 75. If each other observation is multiplied by a non-zero number $\lambda $ and then each of them is decreased by 25, their mean remains the same. The $\lambda $ is equal to
a) $\dfrac{{10}}{3}$
b) $\dfrac{4}{3}$
c) $\dfrac{1}{3}$
d) $\dfrac{2}{3}$

Answer 1

Hint: In this question find the sum of observations using the mean formula,
${\rm{mean = }}\dfrac{{{\rm{sum \space of \space observation}}}}{{{\rm{total \space observation}}}}$. Now, multiply each observation by ‘$\lambda $’, and subtract by 25 and apply the same formula to find ‘$\lambda $’ value.

Complete step-by-step answer:
Mean of a set of observations is given by,
${\rm{mean = }}\dfrac{{{\rm{sum \space of \space observation}}}}{{{\rm{total \space observation}}}}$
i.e.,
$a = \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_{\rm{n}}}}}{{\rm{n}}}$
It is given that mean, $a = 75$
Total observation, $n = 30$
Then equation (1) becomes
$ \Rightarrow 75 = \dfrac{{{a_1} + {a_2} + {a_3} + ... + {a_{30}}}}{{30}}$
$ \Rightarrow {a_1} + {a_2} + {a_3} + ... + {a_{30}} = 75 \times 30 = 2250$
Now, we need to multiply by $\lambda $ and subsequently subtract them with 25 for each observation then we get,
$ \Rightarrow \dfrac{{({a_1}\lambda - 25) + ({a_2}\lambda - 25) + ({a_3}\lambda - 25) + ... + ({a_{30}}\lambda - 25)}}{{30}} = 75$
Here, the above equation has the same mean i.e., 75 as per the question.
In equation (3), 25 is repeated 30 times so equation (3) becomes,
$\Rightarrow \dfrac{{{a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda - (25 \times 30)}}{{30}} = 75$
$ \Rightarrow \dfrac{{{a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda - 750}}{{30}} = 75$
On simplification,
$ \Rightarrow {a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda - 750 = 75 \times 30$
$\Rightarrow {a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda - 750 = 2250$
$ \Rightarrow {a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda = 2250 + 750$
$\Rightarrow {a_1}\lambda + {a_2}\lambda + {a_3}\lambda + ... + {a_{30}}\lambda = 3000$
Taking $\lambda $ common on LHS we get,
$ \Rightarrow \lambda ({a_1} + {a_2} + {a_3} + ... + {a_{30}}) = 3000$
Substitute equation (2), we get,
$ \Rightarrow \lambda (2250) = 3000$
$ \Rightarrow \lambda = \dfrac{{3000}}{{2250}}$
$ \Rightarrow \lambda = \dfrac{4}{3}$
Thus, option (b) is the correct answer.

Note: Whenever we face such types of problems the key is to use mean or average concepts. This concept will help you to find out the solution easily.