Question

The mean of 5 observations is 9. If the mean of
the first 3 observations is 6 and the mean of the last 3 observations is 11, then the third observation is

Answer 1

Given :-

• Mean of 5 observation is 9 .

• Mean of first three observation is 6.

• Mean of last three observations is 11 .

To find :-

• The value of third observation .

Solution :-

Let's assume that the 5 numbers are :- $\sf{\purple{ a \:, b\:,c\:,d\:,e}}$

${\boxed{\sf{\implies Formula\: = \: \frac{ Sum\: of \:all\: observations}{number\: of \:observations} }}}$

${\sf{\implies 9\: = \: \frac{ a+b+c+d+e}{5} }}$

$\sf{\implies 45 = a+b+c+d+e }$

Now average of first three observation

${\sf{\implies 6\: = \: \frac{ a+b+c}{3} }}$

$\sf{\implies 18 =\: a+b+c }$

Now average of last three observations.

${\sf{\implies 11\: = \: \frac{c+d+e}{3} }}$1

$\sf{\implies 33 = c+d+e }$

Now adding sum of first three observation and sum of last three observations.

$\sf{\purple{\implies a+b+c+c+d+e= 18 + 33 }}$

${\underline{\sf{\purple{\implies a+b+2c+d= 51 \:\:\:eq\:1st}}}}$

${\underline{\sf{\purple{\implies 45 = a+b+c+d+e \:\:\:\: eq\:2nd}}}}$

Sub equation equation second from first

→ a+b+2c+d+e - a-b-c-d-e = 51-45

→ c = 6.

$\boxed{\sf{\purple{Third\:term\:is\:6}}}$

Answer 2

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$\huge\tt{GIVEN:}$

• The mean of 5 observations is 9.
• The mean of the first 3 observations is 6
• The mean of the last 3 observations is 11

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$\huge\tt{TO~FIND:}$

• THE THIRD OBSERVATION

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$\huge\tt{SOLUTION:}$

Let the 5 numbers be a,b,c,d,e

⇝Mean = (sum of observations/number of observations)

The average of all the 5 numbers :-

⇝9 = (a+b+c+d+e/5)

⇝9×5 = (a+b+c+d+e)

⇝ 45 = a+b+c+d+e

The average of first three observations :-

⇝ 6 = (a+b+c/5)

⇝6×5 = (a+b+c)

⇝18 = a+b+c

Average of last three observations

⇝11 = (c+d+e/3)

⇝ 11×3 = c+d+e

⇝ 33 = c+d+e

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Now, let's add all of them....

⇝a+b+c+c+d+e = 18 + 33

⇝ a + b +2c + d = 51 ____(EQ.1)

⇝45 =a+b+c+d+e ____(EQ.2)

Then, if we compare both of the equations,

➩a+b+2c+d+e-a-b-c-d-e = 51-45

➩c = 6

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