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
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!"
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
you have to look up values in the Z table and see what area they define. Better review how to use the table.
To find the value of a in the equation p(|X-560| < a) = 0.6, we can use the properties of the normal distribution.
First, let's understand the meaning of |X-560| < a. This inequality implies that the absolute value of the difference between X and the mean (560) is less than a. In other words, it represents the range within which X will fall given a certain probability.
Since |X-560| < a has a probability of 0.6, we need to find the corresponding z-scores (standardized values) for this probability. We can then use these z-scores to determine the value of a.
To find the z-score, we can use the formula z = (X - μ) / σ, where X is the value of interest, μ is the mean, and σ is the standard deviation.
Rearranging the equation, we have |X-560| = a * σ.
Using the z-score table or a calculator, we can find the z-score that corresponds to a cumulative probability of 0.6. Let's denote this value as z. The corresponding z-score for a cumulative probability of 0.6 is approximately 0.253.
Now, we can solve for a by substituting the value of z into the equation:
a * σ = z
a * 20 = 0.253
a ≈ 0.253 / 20
a ≈ 0.01265
Hence, the value of a is approximately 0.01265.