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Suppose you have a normal random variable x with mu=50 and s=15. Find the probability that x will fall within the interval 30<x<60.
The mean is given as 50 and the standard deviation as 15. We convert the raw data to z scores using `z=(a-mu)/sigma` where a is the data point, `mu` is the mean, and `sigma` the standard deviation.
Thus 30 converts to z score of -4/3, and 60 converts to 2/3.
We want to know the area under the standard normal curve between these two z values, as the area is the probability we are seeking.
Using a TI-83 I got the area to be .6563. Using the z-table found in most statistics books and finding the values closest to -4/3 and 2/3 I got the area to the left of -4/3 to be .0918, and the area to the left of 2/3 to be .7486. Taking .7486-.0918 gives .6568.
So the probability that a given x will fall in the range is approximately .6563 (The calculator computes more than the standard 4 digits in a table, so I use this value.)
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