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 } = 0.35 \\) (type an integer or a decimal.)
\\( \sigma _ { w } = 0.572 \\) (round to three decimal places as needed.)
b. on average, how many breakdowns occur per day?
\\( 0.35 \\) (type an integer or a decimal.)
c. about how many breakdowns are expected during a 1 - year period, assuming 250 work days per year?
\\( \square \\) (type an integer or a decimal.)

Explanation:

Step1: Calculate the mean ($\mu_W$)

The formula for the mean of a discrete - probability distribution is $\mu_W=\sum_{w}w\times P(W = w)$.

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

The formula for the variance of a discrete - probability distribution is $\sigma_W^{2}=\sum_{w}(w-\mu_W)^{2}\times P(W = w)$.

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

The standard deviation is the square - root of the variance, $\sigma_W=\sqrt{\sigma_W^{2}}$.

$$ \sigma_W=\sqrt{0.3275}\approx0.572 $$

Step4: Calculate the expected number of breakdowns in a 1 - year period

If there are $n = 250$ workdays in a year and the mean number of breakdowns per day is $\mu_W=0.35$, then the expected number of breakdowns in a year is $n\times\mu_W$.

$$ 250\times0.35 = 87.5 $$

Answer:

a. $\mu_W = 0.35$, $\sigma_W\approx0.572$; b. $0.35$; c. $87.5$