Z = (Sample mean-population mean)/SEm = (38.2-37)/SEm
SEm (Standard Error of the Mean) = SD/√(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.
Use similar process for B sections of other questions
A) Z = (score-mean)/SD = (121.8-120)/5.6
Use the same table and similar process for both A questions.
The difference between A and B is that the former deals with a distribution of scores and the latter deals with a distribution of means.
The average age of chemical engineers is 37 years with a standard deviation of 4 years. If an engineering firm employs 25 chemical engineers, find the probability that the average age of the group is greater than 38.2 years old. If this is the case, would it be safe to assume that the chemical engineers in this group are generally much older than average.
Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury and the standard deviation is 5.6. Assume the variable is normally distributed:
A)If an individual is selected, find the probability that the individual;s pressure will be between 120 and 121.8
B) If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120 and 121.8
At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed:
A) If a proofreader from the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years.
B) If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders will be between 36 and 37.5 years
1 answer
1 answer