Reference no: EM132009194

Can you please make up your own normal distribution problem and find the area (probability) between two independent variables. 2). Then obtain samples of size n from that population and find out the probability of a sample mean being less than some value. My examples follow.

1. A normally distributed population has a mean of 42 and a std dev of 5. Find P (40 < X < 45)

P (40<X<45) = normdist(45,42,5,true) - normdist(40,42,5,true) = .7257 - .3446 = .3811

2. Now assume that samples of size 15 are obtained from that population and that a sample distribution is created. What is the probability that a sample mean would be less than 39? Remember that the mean is the best estimate of the sample mean. The std dev, however, of the new distribution is S/n^1/2 = 5/15^1/2 = 5/3.873 = 1.291.

P(x bar<39) = normdist(39,42,1.291,true) = .010