Respuesta :
Answer:
99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces
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 normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, 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.
In this problem, we have that:
[tex]\mu = 16.15, \sigma = 0.3, n = 50, s = \frac{0.3}{\sqrt{50}} = 0.0424[/tex]
What is the probability that the mean weight of a 50-box case of crackers is above 16 ounces?
This is 1 subtracted by the pvalue of Z when X = 16. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{16 - 16.15}{0.0424}[/tex]
[tex]Z = -3.54[/tex]
[tex]Z = -3.54[/tex] has a pvalue of 0.0001
1 - 0.0001 = 0.9999
99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces
99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces and this can be determined by finding the z-score with the help of the given data.
Given :
- The weight of crackers in a box is stated to be 16 ounces.
- The amount that the packaging machine puts in the boxes is believed to have a Normal model with a mean of 16.15 ounces and a standard deviation of 0.3 ounces.
The probability of the mean weight of a 50-box case of crackers is above 16 ounces can be determined by using the formula f the z-score.
[tex]\rm z = \dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n} }}[/tex]
Now, substitute the known terms in the above formula.
[tex]\rm z = \dfrac{16-16.15}{\dfrac{0.3}{\sqrt{50} }}[/tex]
[tex]\rm z = \dfrac{-0.15}{0.0424}[/tex]
z = -3.54
So, z = -3.54 has a p-value of 0.0001.
P = 1 - 0.0001 = 0.9999
99.99% probability that the mean weight of a 50-box case of crackers is above 16 ounces.
For more information, refer to the link given below:
https://brainly.com/question/23017717
