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