Find the value of p, if the mean of the following distribution is 20.
x: | 15 | 17 | 19 | 20+p | 23 |
y: | 2 | 3 | 4 | 5p | 6 |
X | y | yx |
15 | 2 | 30 |
17 | 3 | 51 |
19 | 4 | 76 |
20 + p | 5p | 100p + 5p2 |
23 | 6 | 138 |
N = 5p + 15 | ∑yx = 295 + 100p + 5p2 |
Given,
Mean = 20
= 20
= 20
295 + 100p + 5p2 = 100p + 300
295 + 5p2 = 300
5p2 = 300 – 295
5p2 – 5 = 0
5 (p2 – 1) = 0
p2 – 1 = 0
(p + 1) (p – 1) = 0
p = � 1
If p + 1 = 0, p = - 1 (Reject)
Or p – 1 = 0, p = 1