Question

Equipment breakdowns: A factory manager collected data on the number of equipment breakdown per day. From those data she derived the probability distribution shown in the following table, Where W denote the number of breakdown on a given day.

Part (a) Determine ${\mu }_{w}and{\sigma }_w$, Round your answer for the standard deviation to three decimal places.

Part (b) On average, how many breakdown occur per day?

Part (c) About how many breakdown are expected during a 1 year period assuming 250 work days per year?

Short answer

Part (a) ${\mu }_w=0.25and{\sigma }_w=0.526$

Part (b) 0.25

Part (c) 62.5

Step-by-step solution

Part (a) Step 1. Given information. 

Consider the following data table:

w	0	1	2
P(W = w)	0.80	0.15	0.05

Part (a) Step 2.μ and  σ are the values.

The random variable W's mean is:

$$\begin{gathered} {\mu }_w=\sum _{}w·P(W=w) \\ {\mu }_w=(0\times 0.080)+(1\times 0.15)+(2\times 0.05) \\ {\mu }_w=0+0.15+0.10 \\ {\mu }_w=0.25 \end{gathered}$$

The random variable W's standard deviation is;

$$\begin{gathered} {\sigma }_w=\sqrt{\sum _{}w·P(W=w)-{\mu }^2} \\ {\sigma }_w=\sqrt{(0^2\times 0.080)+(1^2\times 0.15)+(2^2\times 0.05)-0.{25}^2} \\ {\sigma }_w=\sqrt{0+0.15+0.20-0.0625} \\ {\sigma }_w=0.536 \end{gathered}$$

Part (b) Step 1. On average, there are a number of breakdowns every day.

The mean value =0.25 comes from section a.

The average number of breakdowns per day is equivalent to the mean value.

As a results, The average number of breakdowns per day is 0.25.

Part (b) Step 1.The anticipated number of breakdowns over a one-year period

The number of working days in a calendar year = 240

The expected number of breakdowns;

$$\begin{gathered} {\mu }_W\times 250=0.25\left(250\right) \\ =62.5 \end{gathered}$$

As a result, number of breakdowns over a one-year period = 62.5