# Adding two independent normal random variables

• March 27th 2008, 07:11 PM
indian
Adding two independent normal random variables
Hi guys,
I couldn't figure this one out:
Bulbs fail with normal distribution: mean = 400 days, st. dev = 100 days.

Bulbs are replaced every 365 days. What is the probability that a bulb and its replacement will fail in that period?

There's a hint though: The sum of two iid normal random variables is ______

The book also says this: The sum of independent random variables whose distributions are normal is a random variable having the normal distrbution.

I cant figure out neither the excerpt, the hint, nor the answer. Can someone please explain what it means?
• March 27th 2008, 07:53 PM
mr fantastic
Quote:

Originally Posted by indian
Hi guys,
I couldn't figure this one out:
Bulbs fail with normal distribution: mean = 400 days, st. dev = 100 days.

Bulbs are replaced every 365 days. What is the probability that a bulb and its replacement will fail in that period?

There's a hint though: The sum of two iid normal random variables is ______

The book also says this: The sum of independent random variables whose distributions are normal is a random variable having the normal distrbution.

I cant figure out neither the excerpt, the hint, nor the answer. Can someone please explain what it means?

Let $X_1$ and $X_2$ be two independent normal random variables:

$X_1$ ~ Normal $(\mu_1, \, \sigma_1)$

$X_2$ ~ Normal $(\mu_2, \, \sigma_2)$

Then $Y = X_1 + X_2$ is a normal random variable:

$Y$ ~ Normal $\left( \mu = \mu_1 + \mu_2, \, \sigma = \sqrt{\sigma_1^2 + \sigma_2^2} \right)$.

So let Y = X + X be the random variable lifetime of bulb plus replacement bulb.

Y ~ Normal $\left( \mu = 400 + 400 = 800, \, \sigma = \sqrt{100^2 + 100^2} = 100 \sqrt{2}\right)$.

You need to calculate Pr(Y < 365).

I get 0.001049 (correct to six decimal places)
• March 27th 2008, 08:29 PM
indian
Thanks a LOT! That is AWESOME! Could you please help me out with a further sub-part? Basically, now they're saying that a college has 120 such bulbs, and they want to stock up spare ones. They generally change bulbs once a year.

How many spare bulbs should they stock up to have P = 0.1 that they run out on replacements?
• March 28th 2008, 04:10 AM
mr fantastic
Quote:

Originally Posted by indian
Thanks a LOT! That is AWESOME! Could you please help me out with a further sub-part? Basically, now they're saying that a college has 120 such bulbs, and they want to stock up spare ones. They generally change bulbs once a year.

How many spare bulbs should they stock up to have P = 0.1 that they run out on replacements?

I'm not sure I quite understand the question but here's something to chew on:

The probability that a single bulb fails in less than 365 days is Pr(X < 365) = 0.36317 (correct to five decimal places).

Let B be the random variable number of bulbs that fail in less than 365 days.

Then B ~ Binomial(n = 120, p = 0.36317).

You need to find the smallest of a such that Pr( B > a) < 0.1.

This value of a will be the number of spare bulbs that should be kept in stock. The probability that a replacement bulb will fail before the year is finished is so small as to not be worth worrying about.

Finding the value of a is straightforward using standard technology: I get

Pr(B > 50) = 0.1310
Pr(B > 51) = 0.0954

I'd be inclined to give 51 spare bulbs as the answer.
• March 28th 2008, 03:00 PM
indian
ohhh! I got it now! Thanks a LOT!