JEE MAIN - Mathematics (2015 (Offline) - No. 2)
The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is deleted
and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant data, is :
15.8
14.0
16.8
16.0
Explanation
Initially we have $$16$$ observations and among them one is $$16.$$
So, we have $$15$$ unknowns. Let those are $${a_1},a{}_2,{a_3}.....{a_{15}}$$
$$\therefore\,\,\,$$ Mean of $$16$$ datal set
$$ = {{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}}$$
According to the question,
$${{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}} = 16$$
$$ \Rightarrow {a_1} + {a_2} + ...... + {a_{15}} = 256 - 16 = 240$$
Now we deleted $$16$$ and replaced by there new numbers $$3,4,$$ and $$5.$$
So, new mean
$$ = {{{a_1} + {a_2} + .......{a_{15}} + \left( {3 + 4 + 5} \right)} \over {18}}$$
$$ = {{240 + \left( {3 + 4 + 5} \right)} \over {18}}$$
$$ = {{240 + 12} \over {18}}$$
$$=14$$
So, we have $$15$$ unknowns. Let those are $${a_1},a{}_2,{a_3}.....{a_{15}}$$
$$\therefore\,\,\,$$ Mean of $$16$$ datal set
$$ = {{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}}$$
According to the question,
$${{{a_1} + {a_2} + .....{a_{15}} + 16} \over {16}} = 16$$
$$ \Rightarrow {a_1} + {a_2} + ...... + {a_{15}} = 256 - 16 = 240$$
Now we deleted $$16$$ and replaced by there new numbers $$3,4,$$ and $$5.$$
So, new mean
$$ = {{{a_1} + {a_2} + .......{a_{15}} + \left( {3 + 4 + 5} \right)} \over {18}}$$
$$ = {{240 + \left( {3 + 4 + 5} \right)} \over {18}}$$
$$ = {{240 + 12} \over {18}}$$
$$=14$$
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