Question

The diameters of Red Delicious apples in a certain orchard are normally distributed with a mean of 2.63 inches and a standard deviation of 0.25 inches. a. What percentage of the apples in the orchard have diameters less than 1.95 inches? b. What percentage of the apples in the orchard have diameters larger than 2.90 inches? c. If a crate of 200 apples is selected from the orchard, approximately how many apples would you expect to have diameters greater than 2.90 inches? d. What is the probability that a randomly selected apple in this orchard will have a diameter larger than 2.54 inches? e. What is the probability that a randomly selected apple in this orchard will have a diameter between 1.95 inches and 2.9 inches? Interpret the meaning of this answer.

Answer 1

Given that the diameters of Red Delicious apples in a certain orchard are normally distributed with a mean of = 2.63 inches and a standard deviation of = 0.25 inches.

a. The percentage of the apples in the orchard have diameters less than X = 1.95 inches P(X<1.95) is calculated using the Z score which is calculated as:

The probability is calculated using the excel formula for normal distribution which is =NORM.S.DIST(-2.72, TRUE), thus the proportion is computed as 0.0034 and in percentage 0.0033*100= 0.33%

b) The percentage of the apples in the orchard have diameters larger than 2.90 inches, P(X>2.90) is also calculated by finding the Z score as:

The probability is calculated using the excel formula for normal distribution which is =1-NORM.S.DIST(1.08, TRUE), thus the proportion is computed as:

=1−0.8599=0.1401

= 0.1401*100= 14.01 %

c)  If a crate of n = 200 apples is selected from the orchard, and from previous calculation we saw that approximately 14.01% of apple would have a diameter greater than 2.90, thus the number of apples is calculated as:

=0.1401*200

= 28.02 or

=28 approx.

d) The probability that a randomly selected apple in this orchard will have a diameter larger than 2.54 inches, P(X>2.54) is calculated by finding the Z score as:

The probability is calculated using the excel formula for normal distribution which is =1-NORM.S.DIST(-0.36, TRUE), thus the proportion is computed as:

=1−0.3594=0.6406

d) The probability that a randomly selected apple in this orchard will have a diameter between X = 1.95 inches and X = 2.9 inches P(1.95<X<2.90) is calculated by finding the Z scores as:

and

So, the probability is computed as:

The probability is calculated using the excel formula for normal distribution which is =NORM.S.DIST(1.08, TRUE)-NORM.S.DIST(-2.72, TRUE), thus the proportion is computed as:

=0.8599−0.0033=0.8567

Interpretation:

From the above calculation, it can be said that there is 85.87% approximately chance that the diameter of the apple will be between 1.95 and 2.90.