Assume there is a population of steel balls whose weights are normally distribut
ID: 3351902 • Letter: A
Question
Assume there is a population of steel balls whose weights are normally distributed with a mean of 100 pounds and a standard deviation of 15 pounds. Answer at least two of the following questions:
Would it be unusual if you randomly select one steel ball that weighs 115 pounds or more? Why or why not?
Would it be unusual if you randomly select a sample of size 2 whose average weight (i.e., sample mean) is 115 pounds or more? Why or why not?
Would it be unusual if you randomly select a sample of size 4 whose average weight (i.e., sample mean) is 115 pounds or more? Why or why not?
Would it be unusual if you randomly select a sample of size 8 whose average weight (i.e., sample mean) is 115 pounds or more? Why or why not?
Hint: In order to answer the above questions, you need to calculate the probability associated with each event. Show your calculations, step by step, for the benefit of your classmates.
Note that everything is the same in all four cases above except their sample size. I constructed the above problem to help you think about the effect of sample size. Ponder on the ability of sample in representing the characteristics of the population from which it comes from. Think of it this way, as you increase the sample size (i.e., as the sample size approaches the size of population), the sample becomes a better representative of the population. Therefore, the sample characteristics should become more and more like population characteristics. For example, as we increase the sample size from 1 to 8, the sample mean should become a better estimator (i.e., will have a lower margin of error) of the population mean. Below is the information summary of the four above cases.
Summary:
Population distribution = Normal
Population mean = mu = 100
Population standard deviation = sigma = 15
Sample sizes = n = 1, 2, 4, and 8
Sample mean = xbar = 115 or more?
Explanation / Answer
In case you have doubts on the below answer, please write back to me:
Since the distribution is normal even for sample sizes less than 30 we can use the Z distribution to find out the answers , this is according to the Central Limit therom
We will normalize using the normal dist params:
Population mean = mu = 100
Population standard deviation = sigma = 15
Sample sizes = n = 1, 2, 4, and 8
X can be normalized to Z by : (X-Mu) / (Sigma/sqrt(n))
i) P(X>115) = P(Z> 115-100/(15/sqrt(1)) =P(Z>1) = .1587
ii) P(X>115) = P(Z> 115-100/(15/sqrt(2)) =P(Z>1.41) = .0787
iii)P(X>115) = P(Z> 115-100/(15/sqrt(4)) =P(Z>2) = .0228
iv) P(X>115) = P(Z> 115-100/(15/sqrt(8)) =P(Z>2.82) = .0024