Three complicated statistics (Normal Distribution) questions

Question

1) The lengths in minutes of feature films for the current year follow a normal distribution with a mean of 109 minutes and a standard deviation of 12 minutes. What percentage of feature films in the current year last longer than two hours (120 minutes)?

2) The scores on a particular test are normally distributed with mean of 73 and standard deviation 5. What score would a student need to be among the top 10% of all test scores?

3) Explain when the Central Limit Theorem needs to be used to answer a normal probability question. Is the case of n = 1 included in the Central Limit Theorem?

Answer 1

1) Calculate a z score for the value 120

 

z = (x-μ)/σ = (120-109)/12 = 11/12 = .9167

 

The p-value for P(Z<.9167) = .8204, but we need to find P(Z>.9167) so 1 - .8204 = .1796; roughly 18 percent.

 

2) The phrase "top 10 percent" means the area under the normal curve to the RIGHT of some z score, so the area to the LEFT of this z score is 90 percent. We look up .90 and find a z score of 1.282.

 

Convert the z score to a raw score:

x = z*σ + μ = 1.282*5 + 73 = 79.41

 

A score higher than 79.41 will be among the top 10 percent.

 

3) The Central Limit Theorem deals with the distribution of the mean of a SAMPLE taken from a POPULATION. As the sample size increases the distribution of the sample mean approaches a normal distribution REGARDLESS of the distribution of the population! But in the case where the population has a normal distribution then the sample size makes no difference -- the mean of the sample will be normally distributed even if the sample size is 1.

 

Consider this example. If I have a tank full of fish whose weights are KNOWN to be normally distributed (perhaps because I've caught and weighed every one), what happens if I take many random samples of size n=1 and plot the results? It's really easy to calculate the mean of a sample of size n=1; it's the same as the weight of the fish caught, after all, the sum of a weight X is just X, and divided by one (the sample size) it's still X! The probability of the weight of my sample fish being near the mean is high and decreases as the weigh differs from the mean (either higher or lower). The distribution of the random variable X (the weight of my sample fish) MUST have the same distribution as the population.