The diameters of oranges in a certain orchard are normally distributed with a meanof 5.26 inches and a standard deviationof 0.50 inches.

a. What percentage of the oranges in this orchard have diameters less than 4.5 inches?

b. What percentage of the oranges in this orchard is larger than 5.12 inches?

c. A random sampleof 100 oranges is gathered and the mean diameter obtained was 5.12. If another sample of 100 is taken, what is the probabilitythat its sample mean will be greater than 5.12 inches?

d. Why is the z-score used in answering (a), (b), and (c)?

e. Why is the formula for z used in (c) different from that used in (a) and (b)?

The diameters of oranges in a certain orchard are normally distributed with a mean of 5.26 inches and a standard deviation of 0.50 inches.

Here, = 5.26 in and = 0.50 in

a. What percentage of the oranges in this orchard have diameters less than 4.5 inches?

Z=

P(X < 4.5) = P(Z < -1.52) = 0.064255 = 6.43%

b. What …

Normal Distributions and Z-Scores are investigated. The solution is detailed and well presented. The response received a rating of “5/5” from the student who originally posted the question.