Question

Suppose that the weight of navel oranges is normally distributed with a mean µµ = 8 ounces, and a standard deviation σσ = 1.5 ounces. a. What percent of oranges weigh more than 11.5 ounces? b. What percent of oranges weigh less than 8.7 ounces? c. If you randomly select a navel orange, what is the probability that it weighs between

Answer 1

Answer:

Hello some parts of your question is missing below is the missing part

c. If you randomly select a navel orange, what is the probability that it weighs between6.2 and 7 ounces

Answer: A) 0.0099

              B) 0.6796

              C) 0.13956

Step-by-step explanation:

weight of Navel oranges evenly distributed

mean ( u ) = 8 ounces

std ( б )= 1.5

navel oranges = X

A ) percentage of oranges weighing more than 11.5 ounces

P( x > 11.5 ) = $P ( \frac{x - u}{ std} > \frac{11.5-8}{1.5} )$

                   = P ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) = $P ( \frac{x - u}{ std} > \frac{8.7-8}{1.5} )$

                    = P ( Z < 0.4667 ) = 0.6796

                    = 67.96%

C ) probability of orange selected weighing between 6.2 and 7 ounces?

P ( 6.2 < X < 7 ) = $P (\frac{6.2-8}{1.5} < \frac{x - u}{ std} < \frac{7-8}{1.5} )$

                          = P ( -1.2 < Z < -0.66 )

                          = Ф ( -0.66 ) - Ф(-1.2) = 0.13956