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Answer:
a) 0.9938 = 99.38% probability that the boxes will end up with at least 1 pound of cereal
b) 16.121 ounces.
c) 16.0664 ounces.
d) Between 16.0342 and 16.1658 ounces.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean of 16.1 ounces, standard deviation of 0.04 ounces. So
[tex]\mu = 16.1, \sigma = 0.04[/tex]
(a) What is the probability of the boxes will end up with at least 1 pound of cereal?
One pound is 16 ounces, so this is 1 subtracted by the pvalue of Z when X = 16. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 16.1}{0.04}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062
1 - 0.0062 = 0.9938
0.9938 = 99.38% probability that the boxes will end up with at least 1 pound of cereal.
(b) The seventy percent of the boxes will contain less than what number of ounces?
Less than the 70th percentile, which is X when Z has a pvalue of 0.7. So X when Z = 0.525.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.525 = \frac{X - 16.1}{0.04}[/tex]
[tex]X - 16.1 = 0.525*0.04[/tex]
[tex]X = 16.121[/tex]
16.121 ounces.
(c) Eighty percent of the boxes will contain more than what number of ounces?
More than the 100 - 80 = 20th percentile, which is X when Z has a pvalue of 0.2. So X when Z = -0.84.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X - 16.1}{0.04}[/tex]
[tex]X - 16.1 = -0.84*0.04[/tex]
[tex]X = 16.0664[/tex]
16.0664 ounces.
(d) The middle 90% of the boxes will be between what two weights?
Between the 50 - (90/2) = 5th percentile and the 50 + (90/2) = 95th percentile.
5th percentile:
X when Z has a pvalue of 0.05, so X when Z = -1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.645 = \frac{X - 16.1}{0.04}[/tex]
[tex]X - 16.1 = -1.645*0.04[/tex]
[tex]X = 16.0342[/tex]
95th percentile:
X when Z has a pvalue of 0.95, so X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 16.1}{0.04}[/tex]
[tex]X - 16.1 = 1.645*0.04[/tex]
[tex]X = 16.1658[/tex]
Between 16.0342 and 16.1658 ounces.
It can be deduced that the probability of the boxes ending up with at least 1 pound of cereal will be 99.38%.
How to calculate the probability
The Z score based on the information will be:
Z = (16 - 16.1) / 0.04
Z = -2.5
Therefore, the p value is 0.0062. The probability will be:
= 1 - 0.0062
= 0.9938
The seventy percent of the boxes will contain less than 16 ounces. The ounces that will be contained in the eighty percent will be:
= -0.84 = (X - 16.1) / 0.04
X = 16
Lastly, the middle 90% of the boxes will be between 16.03 ounces 16.17 ounces.
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