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Sagot :
Answer:
0.0143 = 1.43% probability that the amount dispensed per box will have to be increased
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
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.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean for 1 hour is 1 pound and the standard deviation is 0.2 pound
This means that [tex]\mu = 1, \sigma = 0.2[/tex]
Sample of 49:
This means that [tex]n = 49, s = \frac{0.2}{\sqrt{49}} = 0.0286[/tex]
What is the probability that the amount dispensed per box will have to be increased?
This is the probability of the sample mean being less than 15 pounds = 15/16 = 0.9375 ounces, which is the pvalue of Z when X = 0.9375.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.9375 - 1}{0.0286}[/tex]
[tex]Z = -2.19[/tex]
[tex]Z = -2.19[/tex] has a pvalue of 0.0143
0.0143 = 1.43% probability that the amount dispensed per box will have to be increased
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