# Find minimum number of batteries to last a week with 95% probability given mean and sd of lifetime

A TA needs a very important program on his computer but his power cable is missing. Luckily, he happens to have 100 batteries lying around and a contraption that can store batteries to use one at a time. From his understanding, the average lifetime of a battery will last 2 hours for his computer with a standard deviation of 30 minutes. How many batteries will he need to place in the contraption to keep his computer continuously running for the next 7 days (168 hours) with at least a 95% probability?

I'm trying to approach this question using confidence intervals, but i'm confused what the sample proportion would be, would it be 84 batteries? cause u need 84 to run the computer for 168 hours, assuming each battery lasts exactly 2 hours. Would my standard error just be the standard deviation given? Or is there another way to approach this question

• You use confidence intervals to find plausible bounds for an unknown mean. Here you know the true mean, so confidence intervals aren't relevant. Commented Mar 24, 2014 at 21:57
• (1) This appears to be a routine textbook-style question. Such questions are treated differently here. Please add the tag at that link, and read the guidelines there. $\,\,$ (2) "84" is not a proportion, which lie between 0 & 1. "84 batteries" is a count. Note that batteries don't last "exactly 2 hours" but on average two hours. That means that 84 batteries would only have (about) a 50-50 chance of lasting 168 hours; you'll need more batteries than that. Commented Mar 24, 2014 at 22:16

You need to use the properties of the sum of independent distributions and the Central Limit Theorem. It's a probabilistic problem, not a statistical one.

Each battery's time of use follows a distribution $X$ with $E[X]=2, Var[X]=0.5^2$. As you add batteries, you add variables (two batteries last $Y_2=X+X$ with $E[X]=2+2, Var[X]=0.25+0.25$). So $n$ batteries will last $Y_n=n\cdot X$ with $E[X]=2n, Var[X]=0.25n$. You need to find $n$ so that the probability of the joined time of use of $n$ batteries $Y_n$ is greater than 168 hours ($P(Y_n>168)=0.95$). As n will be quite large, you can use the CLT to aproximate $Y_n$ with a normal distribution.

To do that, you need to typify the distribution. You look for the 95% quantile and do the equation to find $n$. Thanks for the corrections and advices you have given.

• that's the right approach in my opinion. I'm wondering about a few things however: (1) The question doesn't say that lifetimes are normally distributed. Nevertheless you still need a way to say what the distribution is, if not for a battery lifetime then at least for the sum of the lifetimes. (2) Are you sure that adding standard deviations is the correct thing to do? I think you mean variances. (3) What condition needs to hold to add variances? (This condition doesn't always hold. E.g. to find the variance of who wins a tennis match I can't add the variances of each player together.) Commented Mar 24, 2014 at 22:05
• 1) Indeed, the Normal distribution is something I've asumed. I don't know any other way to compute the final probability (the sum of random distributions are not necessarely normal). 2) I ment variances, I'll edit it. 3) The variances can be added whenever you add two independent normal distributions. It's basic proprerties of the Normal distribution, and also from any uncorrelated variables, regardless of their distributions (Bienaymé formula). I guess that it doesn't work for tennis match because the winning of one player is correlated to the (non) winning of the other one.
– Rufo
Commented Mar 24, 2014 at 22:16
• cool, re 1) can you think of any way of saying something about the distribution of a sum (or the mean) of a large number of independent identically distributed random variables? re 3) agreed, it's good to state independence explicitly though, in particular as it's required for 1). Commented Mar 24, 2014 at 22:24
• 3) Yes, it's good to say that the time of life of one battery does not affect in the time of life of another one, so you can state that they are independent. 1) The expectation and variance of any independent random variables are the sum of their expectations and variances. Only the joint variance is corrected by the covariance if the variables are dependent. milefoot.com/math/stat/rv-sums.htm I don't know which other reason do you expect.
– Rufo
Commented Mar 24, 2014 at 22:35
• re 1) you want the central limit theorem as it applies to sums (it's more often applied to the mean but applies to sums too as the mean is just the sum/$n$). Commented Mar 24, 2014 at 22:37