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

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

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 ( Z > 2.33 ) = 0.0099

                   = 0.9%

B) percentage of oranges weighing less than 8.7 ounces

  P( x < 8.7 ) =

                    = 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 ( -1.2 < Z < -0.66 )

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

0.16 or 16%

step-by-step explanation:

lets take a coin for example.

if you toss it twice, it has a 1/4 chance of landing on heads both times because 1/2 * 1/2 is 1/4.

with the cards, the chance of picking dog is 0.4. having two people pick one is like picking a card twice in a row. so 0.4 * 0.4 is 0.16, or 16%

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier