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Question
The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is
Options
98
99
100
101
Solution
Given that the mean of 7 numbers is 81. Let us denote the numbers by`X_1,...,X_7`.
If`bar(X) `be the mean of the n observations`X_1,...,X_n, `then we have
`bar(X) = 1/n sum_(i=1)^n X_i`
`⇒ sum_(i=1)^n X_i = nbar(X)`
Hence the sum of 7 numbers is
`sum_(i=1)^7 X_i = 7 xx 81`
=567
If one number is discarded then the mean becomes 78 and the total numbers becomes 6.
Let the number discarded is x.
After discarding one number the sum becomes 567-x and then the mean is
`(567-x)/6`
But it is given that after discarding one number the mean becomes 78.
Hence we have
`(567-x)/6 =78`
⇒567 - x = 468
⇒567 = x + 468
⇒ x+ 468 = 567
⇒ x = 567 -468
⇒ x = 99
Thus the excluded number is 99 .
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