Question

If the mean of the observations a, a+6, a+2 , a+8 and a+4 is 11, find (i) the value of 'a' (ii) the median

Answer 1

Value of a is 7 and median is 11 .

Given :

Observations a, a+6, a+2 , a+8 and a+4 .

Mean , M = 11 .

We know ,

$11=\dfrac{a+(a+6)+(a+2)+(a+8)+(a+4)}{5}\\\\11=\dfrac{5a+20}{5}\\\\5a+20=55\\\\a=7$

Now , we need to find median .

So , arranging the observation in ascending order we get ,

7 , 9 , 11 , 13 , 15

Median is the center term :

So , median is 11 .

Hence , this is the required solution .

Learn More :

Mean , Mode and Median

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Answer 2

$Mean \: of \: the \: observations \: a, a+6, a+2, \\a+8 \:and \:a+4 \:is \:11. \: (given)$

$\boxed { \pink { Mean = \frac{Sum \:of \:the \: observations }{ Number \:of \: observations } }}$

$Mean = 11 \:(given)$

$\implies \frac{ a+(a+6)+(a+2)+(a+8)+(a+4)}{5} = 11$

$\implies \frac{ 5a + 20 }{5} = 11$

$\implies \frac{5(a+4)}{5} = 11$

$\implies a + 4 = 11$

$\implies a = 11 - 4$

$\implies a = 7$

Therefore.,

$substitute \: a= 7 \: in \:the \:data , we \:get$

$5 \: observations \:are \: 7, 13, 9, 15\:and 11$

$Arranging \:the \:observations \: in \\ ascending\:order \:we \:get$

$7, 9 , 11, 13 \:and \: 15$

$Number \:of \: observations (n) = 5 \:( Odd )$

$Median = \big( \frac{n+1}{2}\big)^{th} \: observation\\= \big( \frac{5+1}{2}\big)^{th} \: observation \\= \big( \frac{6}{2}\big)^{th} \: observation \\= 3^{rd} \: observation \\= 11$

Therefore.,

$\red { Value \:of \: a } \green {= 7 }$

$\red { Median \:of \:the \:data }\green { = 11 }$

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