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The diameters of Red Delicious apples in a certain orchard are normally distribu

ID: 3158887 • Letter: T

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.

Explanation / Answer

a)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    1.95      
u = mean =    2.63      
          
s = standard deviation =    0.25      
          
Thus,          
          
z = (x - u) / s =    -2.72      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.72   ) =    0.003264096 [ANSWER]

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b)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2.9      
u = mean =    2.63      
          
s = standard deviation =    0.25      
          
Thus,          
          
z = (x - u) / s =    1.08      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   1.08   ) =    0.14007109 [ANSWER]

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c)

Hence, we expect around

200*0.14007 = 28 apples [ANSWER]

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d)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    2.54      
u = mean =    2.63      
          
s = standard deviation =    0.25      
          
Thus,          
          
z = (x - u) / s =    -0.36      
          
Thus, using a table/technology, the right tailed area of this is          
          
P(z >   -0.36   ) =    0.640576433 [ANSWER]

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e)

We first get the z score for the two values. As z = (x - u) / s, then as          
x1 = lower bound =    1.95      
x2 = upper bound =    2.9      
u = mean =    2.63      
          
s = standard deviation =    0.25      
          
Thus, the two z scores are          
          
z1 = lower z score = (x1 - u)/s =    -2.72      
z2 = upper z score = (x2 - u) / s =    1.08      
          
Using table/technology, the left tailed areas between these z scores is          
          
P(z < z1) =    0.003264096      
P(z < z2) =    0.85992891      
          
Thus, the area between them, by subtracting these areas, is          
          
P(z1 < z < z2) =    0.856664814   [ANSWER]

Hence, around 85.67% of apples in this orchard have diameters between 1.95 and 2.9 in. [ANSWER]

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