Question

Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skateboards and a standard deviation of 55 skateboards.

Answer 1

Answer:

a) 84.13%

b) 2.28%

c) 15.86%

Step-by-step explanation:

Given:

the numbers of skateboards produced per day at a certain factory were normally distributed

mean = 20, 500

standard deviation = 55

To find:

a) On what percent of the day did the factories produced 20,555 or fewer?

b) On what percent of the day did the factories produced 20,610 or fewer?

c) On what percent of the day did the factories produced 20445 or fewer?

To determine the answers, we will use the z-score formula and then use the standard normal table to get the equivalence of the z-score

The formula of score is given as:

`\begin{gathered} z=(X-μ)/(σ) \\ \mu\text{ = mean} \\ σ\text{ = standard deviation} \\ =\text{ value we want to find} \end{gathered}`
`\begin{gathered} a)\text{ X}=\text{ 20555} \\ z\text{ = }\frac{20555\text{ - 20500}}{55}\text{ } \\ z\text{ = }(55)/(55)\text{ = 1} \\ on\text{ the standard normal table, z = 1 gives 0.84134} \\ Percent\text{ that they produced 20555 or fewer = 84.13\%} \end{gathered}`
`\begin{gathered} b)\text{ X}=\text{ 20610} \\ z\text{ = }\frac{20610\text{ - 20500}}{55} \\ z\text{ = 2} \\ On\text{ the standard normal table, z = 2 corresponds to 0.97725} \\ \\ In\text{ this case, we were asked for the percent that produce 20610 or more} \\ To\text{ get ths percent, we will subtract 0.97725 from 1} \\ =\text{ 1 - 0.97725 = 0.02275 } \\ percent\text{ that produced 20610 or more = 2.28\%} \end{gathered}`
`\begin{gathered} c)\text{ X = 20445} \\ z\text{ = }\frac{20445\text{ - 20500}}{55} \\ z\text{ = -1} \\ This\text{ translate to 0.1586} \\ percent\text{ that produced 20445 or fewer = 15.86\%} \end{gathered}`