Homework SourseA candy maker produces bags of mints that have an advertised
A candy maker produces bags of mints that have an advertised weight of 20.4 ounces. Assume the distribution of mint bag weights are normally distributed with a mean of 21.87 ounces and a standard deviation of 0.40 ounces.
d) Of bags that weigh at least 21.50 ounces, what is the probability that a bag weighs at most 22.50 ounces?
e) Given a bag weighs between 21.32 and 22.82 ounces, what is the probability that a bag weighs exactly 21.99 ounces?
Solution
d)
For P(x>21.50):
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 21.5
u = mean = 21.87
s = standard deviation = 0.4
Thus,
z = (x - u) / s = -0.925
Thus, using a table/technology, the right tailed area of this is
P(z > -0.925 ) = 0.822517046
For P(21.50<x<22.50):
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 21.5
x2 = upper bound = 22.5
u = mean = 21.87
s = standard deviation = 0.4
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.925
z2 = upper z score = (x2 - u) / s = 1.575
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.177482954
P(z < z2) = 0.942371778
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.764888824
Thus,
P(x<22.50 | x > 21.50) = P(21.50<x<22.50)/P(x>21.50)
= 0.764888824/0.822517046
= 0.929936744 [ANSWER]
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e)
The probability that a bag weighs EXACTLY 21.99 OZ is 0. [ANSWER]
This is because we need an interval to get a positive probability. If we have just a point, we have 0 probability.
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Hi! If you use a special method in part e), please resubmit this question together with the method itself. That way we can continue helping you! Thanks!
