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Question: The mean of five numbers is \(18\). If one number is excluded, then their mean is \(16\), the exclud...

The mean of five numbers is 1818. If one number is excluded, then their mean is 1616, the excluded number is
A) 2424
B) 2626
C) 2828
D) 2525

Explanation

Solution

In this problem we are going to find an excluded number with the use of the mean values. Here it is given that the mean of five numbers and one number will be excluded so there must be some difference in mean. Also we are given the mean value after the exclusion of one number. Now students aim is to find the excluded number.

Formula used: Mean value = sum of observationnumber of observation\dfrac{{{\text{sum of observation}}}}{{{\text{number of observation}}}}
That is X\overline X =1n(Xi)\dfrac{1}{n}\left( {\sum {{X_i}} } \right) (1) - - - - - \left( 1 \right)

Complete step-by-step solution:
It is given that, the mean of 55 numbers is 1818
Here the number of observations is 55 and mean value is 1818.
That is, n=5n = 5 and X=18\overline X = 18
Now students aim to find the sum of observation.
That is to claim  Xi{\text{ }}\sum {{X_i}}
And Xi=X1+X2+X3+X4+X5\sum {{X_i} = {X_1} + {X_2} + {X_3} + {X_4} + {X_5}}
Now substitute the given values in equation (1)\left( 1 \right)
We get 18=15Xi18 = \dfrac{1}{5}\sum {{X_i}}
Cross multiply the equation,
\Rightarrow 1818×55 =Xi= \sum {{X_i}}
\Rightarrow 90=Xi90 = \sum {{X_i}}
\Rightarrow X1{X_1}+X2{X_2}+X3{X_3}+X4{X_4}+X5{X_5}=9090 (2) - - - - - \left( 2 \right)
That is, sum of observation =90 = 90
Let the number has been excluded be X5{X_5}
After the exclusion of one number, then their mean is 1616
Here, n=4n = 4 and X=16\overline X = 16
\therefore Equation (2)\left( 2 \right)becomes, X1{X_1}+X2{X_2}+X3{X_3}+X4{X_4}=9090X5{X_5}
Now the mean value of 44 numbers is equal to 90X54\dfrac{{90 - {X_5}}}{4}
If one number is excluded, then their mean is 1616 is given
Therefore, 16=90X5416 = \dfrac{{90 - {X_5}}}{4}
Cross multiply the equation, we get
\Rightarrow 1616×44=9090X5{X_5}
\Rightarrow 6464=9090X5{X_5}
\Rightarrow X5{X_5}=90906464
\Rightarrow X5{X_5}=2626
Therefore the excluded number is 2626

The answer is option (B)\left( B \right)

Additional information: There are several kinds of mean in mathematics, especially in statistics. For a data set, the arithmetic mean, also called the expected value or average, is the central value of a discrete set of numbers, specifically the sum of the values divided by the number of values.

Note: The simple way to find mean value is just like an average.
Alternative method: Mean of 55 numbers =1818
\Rightarrow Sum of those 55numbers =18×518 \times 5=9090
Let x be the excluded number
New mean = 16=16 = 90x4\dfrac{{90 - x}}{4} () - - - - - \left( * \right)
Solving equation ()\left( * \right), we get
90x=6490 - x = 64
 x = 26\therefore {\text{ x = 26}}