
The mean of six numbers is \[42\]. If one number is excluded, the mean of the remaining numbers is \[45\]. Find the excluded number.
Answer
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Hint: In this question, we need to follow up two steps very carefully. In the first step, we need to calculate the total of these six numbers. Also, we will calculate the total of five numbers by considering that one number is excluded from the data set. Finally, we will subtract the smaller number from the bigger number to find the excluded number.
Formula used: Average of \[{x_1},{x_2},{x_3},......,{x_n} = \dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}\] .
Complete step by step solution:
Let us consider six numbers are \[{x_1},{x_2},{x_3},{x_4},{x_5},{x_6}\] and we have excluded \[{x_6}\] from the data set.
We know that the formula for mean of numbers is equal to the sum of all the numbers divided by number of numbers.
According to the question, it is stated as, \[\dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 42\].
So, total of all the numbers\[ = \]\[{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = (6 \times 42) = 252.\]
Now, if we exclude \[{x_6}\] from the data set then the mean of the rest of the numbers would be \[45\].
So, according to the question it is stated as, \[\dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 45.\]
So, total of the rest of the five numbers is
Now, we can say that,
\[{x_6} = ({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}) - ({x_1} + {x_2} + {x_3} + {x_4} + {x_5})\].
so, the required excluded number is \[{x_6} = (252 - 225)\]
Let us subtract the terms and we get
\[ = 27.\]
\[\therefore \] The excluded number is \[27\].
Note: Average (mean) of numbers means the sum of all the numbers divided by the number of numbers present in the given data set.
We have to remember that if any number is excluded from the given data set, then the sum of those numbers would become less than the original sum of all the numbers.
Sometimes we can observe that the average of numbers can be reduced by some number after the exclusion of a number from the data set, but the total of these numbers cannot be exceeded by the total of the numbers of the original data set.
Formula used: Average of \[{x_1},{x_2},{x_3},......,{x_n} = \dfrac{{{x_1} + {x_2} + {x_3} + ..... + {x_n}}}{n}\] .
Complete step by step solution:
Let us consider six numbers are \[{x_1},{x_2},{x_3},{x_4},{x_5},{x_6}\] and we have excluded \[{x_6}\] from the data set.
We know that the formula for mean of numbers is equal to the sum of all the numbers divided by number of numbers.
According to the question, it is stated as, \[\dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 42\].
So, total of all the numbers\[ = \]\[{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = (6 \times 42) = 252.\]
Now, if we exclude \[{x_6}\] from the data set then the mean of the rest of the numbers would be \[45\].
So, according to the question it is stated as, \[\dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}}}{5} = 45.\]
So, total of the rest of the five numbers is
Now, we can say that,
\[{x_6} = ({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}) - ({x_1} + {x_2} + {x_3} + {x_4} + {x_5})\].
so, the required excluded number is \[{x_6} = (252 - 225)\]
Let us subtract the terms and we get
\[ = 27.\]
\[\therefore \] The excluded number is \[27\].
Note: Average (mean) of numbers means the sum of all the numbers divided by the number of numbers present in the given data set.
We have to remember that if any number is excluded from the given data set, then the sum of those numbers would become less than the original sum of all the numbers.
Sometimes we can observe that the average of numbers can be reduced by some number after the exclusion of a number from the data set, but the total of these numbers cannot be exceeded by the total of the numbers of the original data set.
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