written 6.2 years ago by | • modified 3.9 years ago |
If the mean of the following distribution is 16 ,find m,n and variance
X: | 8 | 12 | 16 | 20 | 24 |
---|---|---|---|---|---|
P(X=x): | 1/8 | m | n | 1/4 | 1/12 |
written 6.2 years ago by | • modified 3.9 years ago |
If the mean of the following distribution is 16 ,find m,n and variance
X: | 8 | 12 | 16 | 20 | 24 |
---|---|---|---|---|---|
P(X=x): | 1/8 | m | n | 1/4 | 1/12 |
written 6.1 years ago by | • modified 5.9 years ago |
We know that, $ ∑p_i $ =1, we have
$ \frac{1}{8} + m + n + \frac{1}{4} + \frac{1}{12} $ = 1
∴ m+n = $\frac{13}{24}$ ……………(1)
since mean =16, $∑p_i x_i $ = 1
∴ $1 + 12m+ 16n +5 +2 =16$
∴ $12m +16n = 8 or 3m +4n =2$ ……………(2)
solving (1) & (2) we get m = $\frac{1}{6}$ & n = $\frac{3}{8}$
$$ \begin{array}{|c|c|c|c|c|c|} \hline X & 8 & 12 & 16 & 20 & 24 \\ \hline P(X=x) & \frac{1}{8} & \frac{1}{6} & \frac{3}{8} & \frac{1}{4} & \frac{1}{12} \\ \hline \end{array} $$
$Var(X) =E(X^2) – [E(X)]^2$
$E(X) =16$
$E[X^2]= ∑p_i x_i^2 = \frac{1}{8} (64)+\frac{1}{6} (144)+\frac{3}{8} (256)+\frac{1}{4} (400)+\frac{1}{12}(576) $ = 276
$Var(X) =E(X$^2$) – [E(X)]$^2$ = 276 – 256 = 20$