Asked by Ed

The Random Variable X is normally distributed with mean 560 and standard deviation 20.

Find P(X<530) which I calculated as 0.0668.

However... Next question asks

It is known that p(|X-560|<a) = 0.6
Find the value of a
Answered by Steve
using the web site I gave you, I get 16.829

check "Value from area"
enter mean and std and area=0.6
click "between"

see graph and go "Aha!"
Answered by Ed
So I got mean as 0 and SD as 20 and area being 0.6 and got 16.8.

How do I formulate working out step by step for this question without using online calculator
Answered by Steve
you have to look up values in the Z table and see what area they define. Better review how to use the table.
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