# Finding the probability that at least one battery is alive after 4 years

1. Mar 28, 2009

### soopo

1. The problem statement, all variables and given/known data
The expected lifetime of a battery is 0.5 and the standard deviation is 0.5. At
the end of battery's lifetime, the battery will be changed so that different
batteries are independent of each other.

How many batteries do you need such that at least one of them is alive after
4 years with the probability 0.95?

Use normal approximation.

3. The attempt at a solution

I would start by calculating the number of batteries needed such that at least one of them is alive at the end of one year.

I would use normal distribution, since we have the standard deviation given.

Let
X be the number of batteries such that at least one of the batteries is alive at the end of one year

X ~ N(upperbound, expected value of batteries that break in one year, standard deviation of batteries)

I think that the above values are the following in the simplified situation.

X ~N(1, 0.5n, 0.5n)

The standard deviation is got by
st(X) = 0.5 for one battery
st(X) = 0.5n for n number of batteries

I also think that the expected value of batteries that break is 0.5n.

For example, I get the probability, 0.841 (= normalcdf(-E99, 1, 0.5, 0.5)), that
one battery is alive at the end of one year.
This result may be wrong, since the expected value is 0.5 that one battery is
alive at the end of one year.

This suggests me that the use of normal distribution may be a wrong choice at least at the beginning of the problem.

Perhaps, we need to use binomial distribution first for one battery, and then
apply normal approximation to get the number of batteries for 4 years.

Please, let me know how you would solve the problem.

2. Mar 28, 2009

### xaos

eek, a little bit too much here. my statistics is rusty. are these batteries being used in Sequence or in Parallel? it looks to be implied that they're in sequence, if they were parallel its highly unlikely any of them will reach anything beyond two years (three standard deviations from the mean?)
let me try to see this...
in sequence, one battery's lifetime is 'normally'... x_i?, so the sum(x_i) is total lifetime which we want to be...at least 4?, so we want to know...n? when i=1...n? or n+1, maybe?

3. Mar 28, 2009

### the_awesome

1/16?

4. Mar 29, 2009

### HallsofIvy

Let P(x) be the probability that a single battery is alive after x years. Then the probabilty a single battery is NOT alive after x years is 1- P(x) and the probability that none of n identical batteries are alive after x years is (1- P(x))n. Finally, the probability that at least one battery is alive after x years is 1-(1-P(x))n. You want to find n such that 1- (1-P(4))n> 0.95 which is the same as n such that (1-P(4))^n< 0.05. P(4), of course, you get from a table of the normal distribution. If the mean is 0.5 years (you didn't say in your original post) and the standard deviation is 0.5 years, then you are looking at 7 standard deviations above the mean, so p(4) will be very small.

Xaos, my interpretation of this problem is that the batteries are not being used "in paralleL" or "in series" but independently (which may be what you mean by "in parallel").

5. Mar 29, 2009

### soopo

I assume that the question means with the expected value the mean.

We need to calculate the amount of batteries such that at least one is alive after 4 years.

The question seems to ask that
put one battery to the system,
take it away after it has no energy,
put immediately a new battery to the system...
and calculate this way the amount of batteries needed.

This suggests me that the batteries are not at the same time in the machine.

Last edited: Mar 29, 2009
6. Mar 29, 2009

### Horse

You are right because "[a]t the end of battery's lifetime, the battery will be changed".

We know:
X ~ N(1/2, 1/2) per year, where the mean = 1/2 and the variance = 1/2.

Let's calculate new mean and new variance per 4 years:

mean = 1/2^4
variance = p*(1-p)*n = (1/2^4)(1-1/2^4)(4) = 15/64

So X ~ N(1/16, 15/64) per 4 years

The variance may be wrong. How can you calculate variance from the one-year system to the four-year system?

So what do we know?
1. mean 1/16
2. variance, 15/64 ?
3. probability, 95%

Needed:
4. the number of batteries (satisfying the above conditions)

Last edited: Mar 29, 2009