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 \\( \mu _ { w } \\) and \\( \sigma _ { w } \\).
\\( \mu _ { w } = 35 \\) (type an integer or a decimal.)
\\( \sigma _ { w } = 572 \\) (round to three decimal places as needed.)
b. on average, how many breakdowns occur per day?
\\( \square \\) (type an integer or a decimal.)

Explanation:

Step1: Calculate the mean ($\mu_W$)

The formula for the mean of a discrete random variable 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 $\mu_W=0 + 0.25+0.10=0.35$.

Step2: Calculate the variance ($\sigma_{W}^{2}$)

The formula for the variance is $\sigma_{W}^{2}=\sum_{w}(w-\mu_W)^{2}\times P(W = w)$.
For $w = 0$: $(0 - 0.35)^{2}\times0.70=(0.1225)\times0.70 = 0.08575$.
For $w = 1$: $(1 - 0.35)^{2}\times0.25=(0.4225)\times0.25=0.105625$.
For $w = 2$: $(2 - 0.35)^{2}\times0.05=(2.7225)\times0.05 = 0.136125$.
$\sigma_{W}^{2}=0.08575+0.105625 + 0.136125=0.3275$.

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

The standard deviation is the square - root of the variance. $\sigma_W=\sqrt{\sigma_{W}^{2}}=\sqrt{0.3275}\approx0.572$.

Answer:

a. $\mu_W = 0.35$, $\sigma_W\approx0.572$
b. The mean $\mu_W$ represents the average number of breakdowns per day. So, on average, there are $0.35$ breakdowns per day.