Answer:
a) 0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
b) 0.0129 = 1.29% probability that the mean weight is more than 9.0 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 normal distributions can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
Mean of 8.0 ounces and a standard deviation of 1.1 ounces.
This means that ![\mu = 8, \sigma = 1.1](https://tex.z-dn.net/?f=%5Cmu%20%3D%208%2C%20%5Csigma%20%3D%201.1)
(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?
means that ![s = \frac{1.1}{\sqrt{5}} = 0.4919](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B1.1%7D%7B%5Csqrt%7B5%7D%7D%20%3D%200.4919)
This probability is the pvalue of Z when X = 9.3. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{9.3 - 8}{0.4919}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9.3%20-%208%7D%7B0.4919%7D)
![Z = 2.64](https://tex.z-dn.net/?f=Z%20%3D%202.64)
has a pvalue of 0.9959
0.9959 = 99.59% probability that the mean weight is less than 9.3 ounces
(b) If 6 potatoes are randomly selected, find the probability that the mean weight is more than 9.0 ounces?
means that ![s = \frac{1.1}{\sqrt{6}} = 0.4491](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B1.1%7D%7B%5Csqrt%7B6%7D%7D%20%3D%200.4491)
This probability is 1 subtracted by the pvalue of Z when X = 9. So
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{9 - 8}{0.4491}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9%20-%208%7D%7B0.4491%7D)
![Z = 2.23](https://tex.z-dn.net/?f=Z%20%3D%202.23)
has a pvalue of 0.9871
1 - 0.9871 = 0.0129
0.0129 = 1.29% probability that the mean weight is more than 9.0 ounces