1.

Find the value of p, if the mean of the following distribution is 20.x:15171920+p23y:2345p6

Answer»
Xyyx
15230
17351
19476
20 + p5p100p + 5p2
236138
N = 5p + 15\(\sum\)yx = 295 + 100 + 5p2

Given,

Mean = 20

\(\frac{\sum yx}{N}=20\)

\(\frac{295+100p+5}{5p+15}=20\)

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



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