Question

6. Find the value of p in each of the following frequency distributions.
(i)
Observation (x;)
5
15
25
35
Frequency (f;)
20
16
р
2
if the mean is 14.2.​

Answer 1

★ Given :-

Mean=14.2

★ Frequency distribution table :-

$\large\boxed{\begin{array}{c| c| c }\bf Observation (X_i) &\bf Frequency (F_i)&\bf F_iX_i&&&&\sf 5&\sf 20&\sf 100&\sf 15&\sf 16&\sf 240&\sf 25&\sf P&\sf 25P&\sf35&\sf2&\sf70&\cline{1-3}\bf Total&\sf \sum F_i=38+P&\sf \sum F_iX_i=410+25P\end{array}}$

We have Formula for mean::

$\large\boxed{\displaystyle{\sf Mean=\dfrac{\sum F_iX_i}{\sum F_i}}}$

By substituting values::

$:\implies\sf 14.2=\dfrac{ 410+25P}{38+P}$

By cross multiplication::

$:\implies\sf (14.2)(38+P)={ 410+25P}$

By solving it::

$:\implies\sf 539.6+14.2P =410+25P$

By rearranging it::

$:\implies\sf 539.6-410 =25P-14.2P$

By solving it, we get::

$:\implies\sf 129.6=10.8 P$

By rearranging::

$:\implies\sf \dfrac{129.6}{10.8}= P$

By dividing it, we get::

$:\implies\sf 12= P$

So the required value of P is 12.

Answer 2

Answer:

★ Given :-

Mean=14.2

★ Frequency distribution table :-

\large\boxed{\begin{array}{c| c| c }\bf Observation (X_i) &\bf Frequency (F_i)&\bf F_iX_i&&&&\sf 5&\sf 20&\sf 100&\sf 15&\sf 16&\sf 240&\sf 25&\sf P&\sf 25P&\sf35&\sf2&\sf70&\cline{1-3}\bf Total&\sf \sum F_i=38+P&\sf \sum F_iX_i=410+25P\end{array}}

Observation(X

i

)

Frequency(F

i

)

F

i

X

i

5

20

100

15

16

240

25

P

25P

35

2

70

\cline1−3Total

∑F

i

=38+P

∑F

i

X

i

=410+25P

We have Formula for mean::

\large\boxed{\displaystyle{\sf Mean=\dfrac{\sum F_iX_i}{\sum F_i}}}

Mean=

∑F

i

∑F

i

X

i

By substituting values::

:\implies\sf 14.2=\dfrac{ 410+25P}{38+P}:⟹14.2=

38+P

410+25P

By cross multiplication::

:\implies\sf (14.2)(38+P)={ 410+25P}:⟹(14.2)(38+P)=410+25P

By solving it::

:\implies\sf 539.6+14.2P =410+25P:⟹539.6+14.2P=410+25P

By rearranging it::

:\implies\sf 539.6-410 =25P-14.2P:⟹539.6−410=25P−14.2P

By solving it, we get::

:\implies\sf 129.6=10.8 P:⟹129.6=10.8P

By rearranging::

:\implies\sf \dfrac{129.6}{10.8}= P:⟹

10.8

129.6

=P

By dividing it, we get::

:\implies\sf 12= P:⟹12=P

So the required value of P is 12.