Question
CRA CDs Inc. wants the mean lengths of the "cuts" on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10-minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs Inc. (a) What can we say about the shape of the distribution of the sample mean? Sample mean (b) What is the standard error of the mean? Standard error of the mean seconds. (c) What percent of the sample means will be greater than 140 seconds? (Round your answer to 2 decimal places. Omit the "%" sign in your response.) Percent % (e) What percent of the sample means will be greater than 128 but less than 140 seconds? (Round your answer to 2 decimal places. Omit the "%" sign in your response.) Percent %
Explanation / Answer
For any normal random variable X with mean µ and standard deviation s , X ~ Normal( µ , s ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( µ , s² ). Most software denotes the normal with just the standard deviation.) You can translate into standard normal units by: Z = ( X - µ ) / s Where Z ~ Normal( µ = 0, s = 1). You can then use the standard normal cdf tables to get probabilities. If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem. If a sample of size is is drawn from a population with mean µ and standard deviation s then the sample average xBar is normally distributed with mean µ and standard deviation s /v(n) An applet for finding the values http://www-stat.stanford.edu/~naras/jsm/… In this question we have Xbar ~ Normal( µ = 135 , s² = 64 / 16 ) Xbar ~ Normal( µ = 135 , s² = 4 ) Xbar ~ Normal( µ = 135 , s = 8 / sqrt( 16 ) ) Xbar ~ Normal( µ = 135 , s = 2 ) Find P( Xbar > 140 ) P( ( Xbar - µ ) / s > ( 140 - 135 ) / 2 ) = P( Z > 2.5 ) = P( Z < -2.5 ) = 0.006209665 Find P( Xbar > 128 ) P( ( Xbar - µ ) / s > ( 128 - 135 ) / 2 ) = P( Z > -3.5 ) = P( Z < 3.5 ) = 0.9997674 Find P( 128