Question

a factory manager collected data on the number of equipment breakdowns per day. from those data, she derived the probability distribution shown to the right, where w denotes the number of breakdowns on a given day. answer parts a through c.
a. determine μw and σw.
μw = □ (type an integer or a decimal.)

Explanation:

Step1: Calculate the mean ($\mu_W$)

The formula for the mean of a discrete probability distribution is $\mu_W=\sum_{w}w\times P(W = w)$.
For $w = 0$, $0\times0.70=0$.
For $w = 1$, $1\times0.25 = 0.25$.
For $w = 2$, $2\times0.05=0.10$.
Then $\mu_W=0 + 0.25+0.10$.

Step2: Calculate the variance ($\sigma_W^2$)

The formula for the variance of a discrete probability distribution is $\sigma_W^2=\sum_{w}(w-\mu_W)^2\times P(W = w)$.
First, $\mu_W = 0.35$.
For $w = 0$: $(0 - 0.35)^2\times0.70=( - 0.35)^2\times0.70=0.1225\times0.70 = 0.08575$.
For $w = 1$: $(1 - 0.35)^2\times0.25=(0.65)^2\times0.25=0.4225\times0.25 = 0.105625$.
For $w = 2$: $(2 - 0.35)^2\times0.05=(1.65)^2\times0.05=2.7225\times0.05=0.136125$.
Then $\sigma_W^2=0.08575 + 0.105625+0.136125$.

Step3: Calculate the standard deviation ($\sigma_W$)

The standard deviation is the square - root of the variance. $\sigma_W=\sqrt{\sigma_W^2}$.
Since $\sigma_W^2=0.08575 + 0.105625+0.136125 = 0.3275$, $\sigma_W=\sqrt{0.3275}\approx0.57$.

Answer:

$\mu_W = 0.35$, $\sigma_W\approx0.57$