Question

If the mean of the following frequency distribution is 7.2. Find the value of p and k when total frequency is 5.
X:2 4 6 8 10 12
F:4 p 10 16 k 3

Answer 1

Question:

If the mean of the following frequency distribution is 7.2. Find the value of p and k when total frequency is 5.

$\begin{array}{|c|c|c|c|c|c|c|}\cline{1-7}X_i & 2 & 4 & 6 & 8 & 10 & 12 \\ \cline{1-7} F_i & 4 & p & 10 & 16 & k & 3 \\ \cline{1-7}\end{array}$

Solution:

$\begin{tabular}{|c|c|c|}\cline{1-3} X_i & F_i & F_iX_i \\ \cline{1-3}2 & 4 & 8 \\ \cline{1-3}4 & p & 4p \\ \cline{1-3}6 & 10 & 60 \\ \cline{1-3}8 & 16 & 128 \\ \cline{1-3}10 & k & 10k \\ \cline{1-3}12 & 3 & 36 \\ \cline{1-3}Total & \Sigma\:F_i = 33 + p + k = 5& \Sigma X_iF_i = 232 + 4p + 10k \\ \cline{1-3} \end{tabular}$

In question we have given that mean is 7.2

And we know that -

$\sf{Mean\:=\:\frac{Sum\:of\:all\:observations}{Total\:number\:of\:observations}}$

$\sf{Mean\:=\:\dfrac{\Sigma\:X_iF_i}{\Sigma\:F_i}}$

From above $\sf{\Sigma\:X_iF_i}$ = 232 + 4p + 10k

$\sf{\Sigma\:F_i}$ = 33 + p + k

$\implies\:\sf{7.2\:=\:\dfrac{232\:+\:4p\:+\:10k}{33\:+\:p\:+\:k}}$

$\implies\:\sf{7.2(33\:+\:p\:+\:k)\:=\:232\:+\:4p\:+\:10k}$

$\implies\:\sf{237.6\:+\:7.2p\:+\:7.2k\:=\:232\:+\:4p\:+\:10k}$

$\implies\:\sf{237.6\:-\:232\:=\:4p\:-\:7.2p\:+\:10k\:-\:7.2k}$

$\implies\:\sf{5.6\:=\:-\:3.2p\:+\:2.8k}$ --- [1]

Total frequency is 5.

So,

$\implies\:\sf{33\:+\:p\:+\:k\:=\:5}$

$\implies\:\sf{p\:+\:k\:=\:-28}$

$\implies\:\sf{p\:=\:-28\:-\:k}$

Substitute value of p = - 28 - k in equation (1)

$\implies\:\sf{5.6\:=\:-\:3.2(-28-k)\:+\:2.8k}$

$\implies\:\sf{5.6\:=\:+89.6\:+\:3.2k\:+\:2.8k}$

$\implies\:\sf{-84\:=\:6k}$

$\implies\:\sf{k\:=\:-14}$

Substitute value of k = -14 in p

$\implies\:\sf{p\:=\:-28\:-\:(-14)}$

$\implies\:\sf{p\:=\:-28\:+\:14}$

$\implies\:\sf{p\:=\:-14}$

Answer 2

Given :-----

• x = 2, 4 ,6 ,8 , 10, 12
• F = 4, p , 10, 16, k , 3

Question :-----

• we have to Find value of p and k ..

Formula used :------

• Mean of ungrouped date is given by sum of fixi , divided by sum of fi ...

Calculation :-----

From image we can see that ,,,

→ sum of fi we get = 33 + p + k

→ sum of fixi we get = 232+4p+10k

so, above told formula we get,

→ 7.2 = (232+4p+10k)/(33+p+k)

cross - multiplying we get,,,,

→ 33×7.2 + 7.2k + 7.2p = 232 + 4p + 10k

→ 7.2k - 10k + 7.2p - 4p = 232 - 237.6

→ (-2.8)k + 3.2p = (- 5.6) ------ Equation (1)

Now since it is given ,

sum if frequency is 5 .

→ 33 + p + k = 5

→ p + k = - 28

→ p = (-28-k) --------------- Equation (2)

Putting value of this in Equation (1) we get,,

→(-2.8)k + 3.2(-28-k) = ( -5.6)

→ -2.8k - 89.6 -3.2k = -5.6

→ (-6k) = (-5.6) + 89.6

→ (-6k) = 84

dividing both side by (-6) we get,

→ k = (-14)

Now how is value of frequency in negative yar ,

kuch bhe matlab ,,,

saaf saaf dikh raha question galat hai, but u need answer with given date , ok ... fine .....

so,

putting value of k in Equation (2) we get,,,

→ p = [(-28) - (-14)]

→ p = [ -28 + 14]

→ p = (-14) ....

Very nice Question ,

so,

value of p is (-14) and value of k is also (-14) ....

Given data is only possible when both have same values (-14)..

Thank you so much for providing us such a great Question ...

___________________________

→ mean ( or simply average ) of simple data = sum of observation / No. of observation

→ mean of grouped data is same as mean of un-grouped data ...