Question

The mean of the following distribution is 23.4. Then find the value of p:

Class-interval	0-8	8-16	16-24	24-32	32-40	40-48
Frequency	5	3	10	p	4	2

Answer 1

Hint:In this, we will find the value of p by finding the mean of the given distribution. We will find the mean of the distribution by the direct method indirect method we first find the class of each class interval i.e. mean of each interval then we will find the sum of multiplication class and frequency. After that we will find the mean by the formula given below:
$\overline{X}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}$
Where ${{x}_{i}}$ is the class of each interval and ${{f}_{i}}$ is the frequency of each interval.

Complete step by step answer:
Now we will construct a table.

Class interval	Class (${{x}_{i}}$)	Frequency (${{f}_{i}}$)	Class x frequency.
0-8	4	5	20
8-16	12	3	36
16-24	20	10	200
24-32	33	P	33p
32-40	36	4	144
40-48	44	2	88

Now we will find the sum of the class x frequency of intervals
$\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}}={{x}_{1}}{{f}_{1}}+{{x}_{2}}{{f}_{2}}+{{x}_{3}}{{f}_{3}}+{{x}_{4}}{{f}_{4}}+{{x}_{5}}{{f}_{5}}+{{x}_{6}}{{f}_{6}}$

By substituting the value of class frequency for each interval, we get
$\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}}=20+36+200+33p+144+88$
$\sum\limits_{i=1}^{6}{{{x}_{i}}{{f}_{i}}}=488+33p$
Now we will find the sum of the frequency of intervals
$\sum\limits_{i=1}^{6}{{{f}_{i}}}={{f}_{1}}+{{f}_{2}}+{{f}_{3}}+{{f}_{4}}+{{f}_{5}}+{{f}_{6}}$
By substituting the value of frequency for each interval, we get
 $\sum\limits_{i=1}^{6}{{{f}_{i}}}=5+3+10+p+4+2$
$\sum\limits_{i=1}^{6}{{{f}_{i}}}=24+p$
Now we will use the formula of mean which is given in hint
$\overline{X}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}{{f}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}}$
$\overline{X}=\dfrac{488+33p}{24+p}$
Since the mean of the distribution is given that mean = 23.4. Then,
$23.4=\dfrac{488+33p}{24+p}$
By cross multiplication we get
$23.4\left( 24+p \right)=488+33p$
561.6 + 23.4p = 488 + 33p
561.6 – 488 = 33p - 23.4p
73.6 = 9.6p
By dividing both sides by 9.6, we get
p = 7.67
Hence the value of p is 7.67.
Note:
 Alternative method, Assumed mean method.In the assumed mean method we first consider assumed mean A and the find the deviation for each interval by ${{d}_{i}}={{x}_{i}}-A$. After that, we find the product of frequency and deviation of each interval. Then we find the mean of product of frequency and deviation of each interval.
To find the mean of distribution we add assumed mean and mean of product of frequency and deviation of each interval.