Question

given the following relative - frequency table for the probability distribution of x, what are the expected value ($mu$) and the standard deviation ($sigma$)?
x p(x = x)
5 .10
10 .14
15 .29
20 .47
$mu = 15.65$, $sigma = 4.98$
$mu = 17.98$, $sigma = 15.94$
$mu = 15.65$, $sigma = 15.94$
$mu = 269.75$, $sigma = 4.98$

Explanation:

Step1: Calculate expected value $\mu$

$\mu=\sum_{i}x_iP(X = x_i)=5\times0.10 + 10\times0.14+15\times0.29 + 20\times0.47=0.5+1.4 + 4.35+9.4 = 15.65$

Step2: Calculate variance $\sigma^{2}$

$\sigma^{2}=\sum_{i}(x_i-\mu)^{2}P(X = x_i)=(5 - 15.65)^{2}\times0.10+(10 - 15.65)^{2}\times0.14+(15 - 15.65)^{2}\times0.29+(20 - 15.65)^{2}\times0.47$
$=(- 10.65)^{2}\times0.10+(-5.65)^{2}\times0.14+(-0.65)^{2}\times0.29+(4.35)^{2}\times0.47$
$=113.4225\times0.10 + 31.9225\times0.14+0.4225\times0.29 + 18.9225\times0.47$
$=11.34225+4.46915 + 0.122525+8.893575=24.8275$

Step3: Calculate standard - deviation $\sigma$

$\sigma=\sqrt{\sigma^{2}}=\sqrt{24.8275}\approx4.98$

Answer:

A. $\mu = 15.65, \sigma = 4.98$