Question

An orange juice producer buys all oranges from a large orange-tree farm in Florida. The amount of juice squeezed from a kilogram of these oranges is approximately normally distributed with a mean of 600 gr. and a standard deviation of 30 gr.

Note: Make sure to specify the random variable and its distribution. Use the appropriate cumulative distribution table to compute the probability.

(a) Using the Empirical Rule, evaluate approximately the probability that a given kilogram of oranges yields more than 570 gr. of juice. (5 pts)

b) Consider two possibilities for the juice squeezed out of a kilogram of oranges: (i) the juice weight is between 600 gr. and 630gr. and (ii) the juice weight is between 540 gr. and 570 gr. Which of the two possibilities is more likely? Explain without computing any probabilities. (5 pts)

(c) For the next year’s budget projections, the orange juice producer wants to compute an interval for the weight of juice squeezed from one kilogram of oranges in the typical case: the lower bound of the interval would represent the worst-case scenario, the upper bound—the best-case scenario. Compute and use the first and third quartiles of juice weights to build that interval. (10 pts)

Answer 1

Solution:
Given in the question
Mean = 600
Standard deviation = 30
Solution(a)
We need to calculate
P(X>570) = 1-P(X<=570)
Z = (570-600)/30 = -1
From Z table we found p-value
P(X>570) = 1-P(X<=570) = 1-0.0.1587 = 0.8413
So there is 84.13% that oranges yields more than 570 gr. of Juice.
Solution(b)
P(540<X<570) = P(X<570) - P(X<540)
Z=(540-600)/30 = -2
Z=(570-600)/30 = -1
From Z table we found p-value
P(540<X<570) = P(X<570) - P(X<540) = 0.1587 - 0.0228 = 0.1359
P(600<X<630) = P(X<630) - P(X<600)
Z = (600-600)/30 = 0
Z = (630-600)/30 = 1
From Z table we found p-value
P(600<X<630) = P(X<630) - P(X<600) = 0.8413 - 0.5 = 0.3413
From the above calculations we can see that the juice weight is b/w 600 gr. and 630 gr. is more likely.
Solution(c)
For First quartile p-value = 0.25 so Z-score from the z table is -0.6745
So Lower bound of interval is = 600 - 0.6745*30 = 579.8
For Third quartile p-value = 0.75 so Z-score from the z table is 0.6745
So Lower bound of interval is = 600 + 0.6745*30 = 620.2