# Large Set of Random Variables with Exponential Distribution

I'm struggling to understand how to solve the following problem. I have a random variable $$X$$ that represents the life time of a cellphone (in years) and I know that such variable follows an exponential distribution with $$\lambda = 0.22$$, that is $$X \sim Exponential(0.22)$$

Considering that I have a set of 500 cellphones, all following the same distribution, what is the approximate probability of no more than 80 cellphones die during the first year of utilization?

I've calculated what is the probability of one cellphone to die during the first year of utilization and it is approximately 20%. But I wasn't sure what to do with this result.

I've also tried to employ the Central Limit Theorem, hinted by "approximate probability", and I've obtained a Normal Distribution for $$\overline{X}$$, the median of the data set, such that $$\overline{X} ~ Normal(\mu, \frac{\sigma}{\sqrt{N}})$$ with $$\mu = \frac{1}{\lambda} = 4.55$$ and $$\frac{\sigma}{\sqrt{N}} = \frac{1}{\lambda^2 \sqrt{N}} = 0.92$$ However i'm still not sure on what to do here.

Is the problem related to the poisson probability?

Can someone shed some light on this issue?

• The answer is given by the Binomial distribution. After all, the event "die during the first year" doesn't care about how the deaths happen to be distributed; all that matters its chance of occurring for each phone. – whuber Apr 9 at 15:19

Maybe you're having trouble making sense of this problem because the answer is extreme.

"Failure" is that such a phone lasts less than a year. From R, we get $$P(F) = 0.1975,$$ pretty much as you say. So $$p = P(S) = 0.8025.$$ (A falure rate - $$\lambda = 0.22$$ implies that cell phones last on average for $$1/\lambda = 4.5454$$ years.

pexp(1, .22)
[1] 0.1974812


Now if you have $$n = 500$$ such phones working or failing independently, let $$Y$$ be the number that survive for at least a year. Then $$Y \sim \mathsf{Binom}(n=500, p),$$ which has $$\mu=E(Y) = 401.2594$$ and $$\sigma=SD(Y) = 8.9018.$$

As I read the problem, you seek $$P(Y > 80) = 1 -P(Y \le 80).$$ If you wish, you can use the normal approximation to a binomial distribution (a special case of the CLT) to approximate this probability. But you could also use R (where 'pbinom' is a binomial CDF) to see that the exact probability is essentially $$1:$$

1 - pbinom(80, 500, 0.8025)
[1] 1


This is no surprise because the mean (about 400) is so many standard deviations above 80.

A normal approximation would be as follows: $$P(Y > 80) = P\left(\frac{Y-\mu}{\sigma} > \frac{80-401.2594}{8.9018} = -39.17\right) \approx P(Z > =39.17) \approx 1,$$ where $$Z$$ is a standard normal random variable.

Note: What may be causing a surprise is that not many textbook problems have 'essentially 1' as an answer. Perhaps 380 was intended instead of 80.

1 - pbinom(380, 500, 0.8025)
[1] 0.9888621