At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Explore a wealth of knowledge from professionals across different disciplines on our comprehensive platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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 [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.
Mean of 8.0 ounces and a standard deviation of 1.1 ounces.
This means that [tex]\mu = 8, \sigma = 1.1[/tex]
(a) If 5 potatoes are randomly selected, find the probability that the mean weight is less than 9.3 ounces?
[tex]n = 5[/tex] means that [tex]s = \frac{1.1}{\sqrt{5}} = 0.4919[/tex]
This probability is the pvalue of Z when X = 9.3. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9.3 - 8}{0.4919}[/tex]
[tex]Z = 2.64[/tex]
[tex]Z = 2.64[/tex] 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?
[tex]n = 6[/tex] means that [tex]s = \frac{1.1}{\sqrt{6}} = 0.4491[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 9. So
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9 - 8}{0.4491}[/tex]
[tex]Z = 2.23[/tex]
[tex]Z = 2.23[/tex] 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
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
