# central limit theorem problem 2, is this right?

• Nov 21st 2010, 10:53 AM
Sneaky
central limit theorem problem 2, is this right?
Suppose the time in days until a component fails has the gamma distribution with alpha=5 and theta = 1/10. When a component fails, it gets immediately replaced by a new one. Use the central limit theorem to estimate the probability that 40 components will together be sufficient to last at least 6 years. Assume that a year is 365.25 days.

My attempt
E(X)=0.5=mew=sample mew
Var(x)=1/20
sigma = sqrt(1/20)
sample sigma [40]= sqrt([1/20] / 40) = 0.035

so then

= 1 - P(sample X[40] <= 6*365.25)
= 1 - P(sample X[40] <= 2191.5)
= 1 - P([sample X[40] - 0.5 ] / 0.035 <= [2191.5-0.5]/0.035)
= 1 - P(X<=62600) where X~N(0,1)
~0
• Nov 22nd 2010, 09:34 AM
Guy
First, you should always include the parametrization of the gamma you are using when posting questions. Some parameterize it so that $\displaystyle E X = \alpha \theta$ whereas other parametrize it so that $\displaystyle EX = \alpha / \theta$. My guess is that this question is assuming the later parametrization, but the way that you worked the problem suggests that you are using the former. I'll assume the latter.

The sum of all our random variables is a Gamma(200, 1/10), which is approximately normal with mean 2000 and variance 20000, i.e. the standard deviation is approximately 141.42. Get a Z value:

(2191.5 - 2000) / 141.42 = 1.35