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Question
Find the value of p, if the mean of the following distribution is 20.
| x: | 15 | 17 | 19 | 20+p | 23 |
| f: | 2 | 3 | 4 | 5p | 6 |
Solution
The given data in tabulated form is
| x | 15 | 17 | 19 | 20+p | 23 |
| f | 2 | 3 | 4 | 5p | 6 |
We have to find the value of p using the information that the mean of the data is 20.
Prepare the following frequency table of which the first column consists of the values of the variate and the second column the corresponding frequencies. Multiply the frequency of each row with the corresponding values of variable to obtain the third column containing `f_ix_i`.
| `x_i` | `f_i` | `f_ix_i` |
| 15 | 2 | 30 |
| 17 | 3 | 51 |
| 19 | 4 | 76 |
| 20+p | 5p | 5p(20+p) |
| 23 | 6 | 138 |
Find the sum of all entries in the second and third column to obtain N and `sum f_ix_I` respectively. Therefore,
` N = sum_(i=1)^5 f_i`
= 2+3+4+5p+6
=15+5p
` sum_(i=1)^5 f_ix`
=30+51+76+5p(20+p)+138
=5p2 + 100p +295
The mean is
`bar(X) = (sum_(i=1)^5 f_ix_i)/N`
`=(5p^2 +100p+295)/(15+5p)`
Hence, we have
`=(5p^2 +100p+295)/(15+5p)=20`
⇒5p2 + 100p +295=300+100p
⇒5p2 = 300-295
⇒ 5p2 = 5
⇒ p2 =`5/5`
⇒ p2 = 1
`⇒ p =+- 1`
If p is negative then the 4th frequency becomes negative. But frequency can’t be negative. Hence the possible value of p is 1
p = 1
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