Prob and stats continuous random variable question

[Proggy99]

Homework Statement

Let X denote the lifetime of a radio, in years, manufactured by a certain company. The density function of X is given by

$$f(x)=\left\{\stackrel{\frac{1}{15}e^\frac{-x}{15}\ \ \ \ if\ 0\ \leq\ x\ <\ \infty}{0\\\\elsewhere}$$

What is the probability that, of eight such radios, at least four last more than 15 years?

Homework Equations

The Attempt at a Solution

P(x$$\geq15$$) = F(15) = $$\frac{1}{15}\int^{\infty}_{15}e^\frac{-x}{15}dx$$ = $$-\left[e^\frac{-x}{15}\right]^{\infty}_{15}$$
=$$-\left[e^{-\infty}\ -\ e^{-1}\right]$$ = .368

So each radio has a .368 chance of lasting more than 15 years

Next I used a Poisson variable equation using t = 8 and .368

$$\lambda$$ = 8 * .368 = 2.944

therefore $$\frac{e^{-2.944}2.944^{n}}{n!}$$

to find the probability of at least 4 of 8 radios
$$\sum^{8}_{n=4}\frac{e^{-2.944}2.944^{n}}{n!}\ =\ .3366$$

Now the book answer says .3327 and I have pretty much ruled out rounding differences. That is pretty close, but this is the first time the book and my answer have not been pretty much the same. Can anyone see anything I am doing wrong in my above solution? Any help would be greatly appreciated. Thanks

 

[tiny-tim]

So each radio has a .368 chance of lasting more than 15 years

Next I used a Poisson variable equation using t = 8 and .368

$$\lambda$$ = 8 * .368 = 2.944

therefore $$\frac{e^{-2.944}2.944^{n}}{n!}$$

to find the probability of at least 4 of 8 radios
$$\sum^{8}_{n=4}\frac{e^{-2.944}2.944^{n}}{n!}\ =\ .3366$$

Hi Proggy99!

not sure what a Poisson variable equation is …

but the radios are independent, so can't you just use ordinary combinatorial theory, with .368ⁿ?