H2 Maths 6.3 - Normal Distribution: Linear Combinations

Step 1 - The key result for independent normal variables

When X and Y are independent normal random variables, any linear combination aX plus bY is also normally distributed.

Step 1 - The key result for independent normal variables

The mean of the combination is a mu X plus b mu Y, from standard expectation algebra.

Step 1 - The key result for independent normal variables

The variance is a squared sigma X squared plus b squared sigma Y squared - variances always add, scaled by the coefficient squared, because X and Y are independent.

Step 2 - Set up the worked example

Let X tilde N of forty, sixteen, and Y tilde N of sixty, nine, with X and Y independent.

Step 2 - Set up the worked example

We first find the distribution of X plus Y.

Step 2 - Set up the worked example

The mean of X plus Y is forty plus sixty, which is one hundred. The variance is sixteen plus nine, which is twenty five.

Step 3 - Find P(X + Y > 108)

Now find the probability that X plus Y exceeds one hundred and eight.

Step 3 - Find P(X + Y > 108)

Z = (108 − 100) / 5 = 1.6.

Step 3 - Find P(X + Y > 108)

From the GC, P of Z greater than one point six equals zero point zero five four eight.

Step 4 - The distribution of 2X − Y and a key pitfall

For the combination two X minus Y, the coefficient of Y is negative one.

Step 4 - The distribution of 2X − Y and a key pitfall

The mean is two times forty minus sixty, which is twenty. For the variance, b equals negative one so b squared equals one - the variance is four times sixteen plus one times nine, giving seventy three.

Step 4 - The distribution of 2X − Y and a key pitfall

Common error: do not subtract variances when the sign is negative. Variance is always added, scaled by the coefficient squared.