Question

During 1 week an overnight delivery company found that the weights of its parcels were normally distributed, with a mean of 20 ounces and a standard deviation of 5 ounces. A). What percent of the parcels weighed between 10 ounces and 25 ounces? (Round your answer to one decimal place.) B). What percent of the parcels weighed more than 35 ounces? (Round your answer to two decimal places.)

Answer 1

Step 1: Problem

Normal distribution

Step 2: Concept

$\begin{gathered} z\text{ = }\frac{x\text{ - }\mu}{\sigma} \\ \mu\text{ = mean} \\ \sigma\text{ = standard deviation} \end{gathered}$

Step 3: Method

A) Pr(10 <= x <= 25)

$\begin{gathered} \mu\text{ = 20} \\ \sigma\text{ = 5} \\ z1\text{ = }\frac{x\text{ - }\mu}{\sigma}\text{ = }\frac{10\text{ - 20}}{5}\text{ = }\frac{-10}{5}\text{ = -2 z value of -2 from normal distributio = }0.4772 \\ z2\text{ = }\frac{x\text{ - }\mu}{\sigma}\text{ = }\frac{25\text{ - 20}}{5}\text{ = }\frac{5}{5}\text{ = 1 z value of 1 from normal distribution = 0.3413} \end{gathered}$

Pr( 10 <= x <=25 ) = 0.4772 + 0.3413 = 0.8185

= 0.8185 x 100%

= 81.9%

B) Pr( x >= 35 )

$\begin{gathered} z\text{ = }\frac{35\text{ - 20}}{5} \\ z\text{ = }\frac{15}{5} \\ z\text{ = 3} \\ z\text{ value from normal distribution table = 0.4987} \\ =\text{ 0.4987 x 100\%} \\ =\text{ 49.87\%} \end{gathered}$

Step 4:

A) 81.9%

B) 49.87%