Question

Find the value of P for the following distribution whose mean is 10.

\[{{x}_{1}}\]	5	7	9	11	13	15	20
\[{{f}_{i}}\]	4	4	p	7	3	2	1

Answer 1

Hint: In this question, we will use the formula of the mean for uneven distribution, that is
\[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
where i = 1 to 7, \[{{f}_{i}}\] is the frequency of \[{{x}_{i}}\] and \[\sum \] represents the summation of all of the terms. By using this formula, we will find the value of p.

Complete Step-by-Step solution:
In this question, we are given a distribution table whose mean is 10 and we have to find the value of p. Now, we know that when we are given an uneven distribution table and we are asked to find the mean, we use the formula,
\[Mean=\dfrac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]
where \[{{f}_{i}}\] is the frequency for respective \[{{x}_{i}}\]. So, we will calculate \[{{f}_{i}}x\] for all i's and then we will solve.

\[{{x}_{i}}\]	\[{{f}_{i}}\]	\[{{x}_{i}}{{f}_{i}}\]
5	4	\[5\times 4=20\]
7	4	\[7\times 4=28\]
9	P	\[9\times p=9p\]
11	7	\[11\times 7=77\]
13	3	\[13\times 3=39\]
15	2	\[15\times 2=30\]
20	1	\[20\times 1=20\]

So, we can say that
\[\sum{{{x}_{i}}{{f}_{i}}=20+28+9p+77+39+30+20=214+9p}\]
And \[\sum{{{f}_{i}}=4+4+p+7+3+2+1=21+p}\]
Now, we will put the value of \[\sum{{{x}_{i}}{{f}_{i}}}\] and \[\sum{{{f}_{i}}}\] in the formula of the mean. So, we will get,
\[Mean=\dfrac{214+9p}{21+p}\]
And we have been given that mean = 10. Therefore, we can write
\[10=\dfrac{214+9p}{21+p}\]
Now, we will cross multiply the equation. So, we will get,
210 + 10p = 214 + 9p
Now, we will take all the terms with the variable p on the left-hand side and rest on the right side. So, we will get,
10p – 9p = 214 – 210
p = 4
Hence, we get the value of p as 4.

Note: In this question, the possible mistake one can make is by using the formula mean \[=\dfrac{\sum{{{f}_{i}}}}{n}\] where i = 1 to n. Obviously, the formula is not wrong but we cannot use this formula in this question.