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The lifetime of a certain type of battery is normally distributed with mean value 14 hours and standard deviation 1 hour. There are nine batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

I'm not sure how to approach this, at first i thought i should use the equation z = (x - mean)/StandardDeviation and use a table that has all the standard normal curve areas (in my book its called table A-3)

but i can't seem to get the correct answer, can someone help me understand the steps that i need to take to solve this?

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There are $n = 9$ batteries in the pack, and it is reasonable to assume that each one has an independent and identically normally distributed lifetime, which we will call $X_1, X_2, \ldots, X_9$. Specifically, $$X_i \sim \operatorname{Normal}(\mu = 14, \sigma = 1), \quad i = 1, 2, \ldots, 9.$$ We are asked for some time $t$ such that the probability that the total of these random lifetimes exceeds $t$ is only $0.05$; that is to say, to find $t$ such that $$\Pr[X_1 + \cdots + X_9 > t] = 0.05.$$ To this end, we recall that the sum of $n$ independent normal random variables is itself normally distributed, with mean equal to the sum of the individual means, and variance equal to the sum of the individual variances. Since in your case all of the $X_i$ are IID, we find that $$T = X_1 + X_2 + \cdots + X_9 \sim \operatorname{Normal}(\mu = 126, \sigma = 3),$$ since $9(14) = 126$, and $\sqrt{9} = 3$ (note that the variances add, not the standard deviations, hence the variance of $T$ is $\sigma^2 = 9$).

So now we have simplified the problem to finding a quantile of a single normal random variable $T$, rather than the sum of $9$ random variables. Next, we standardize: $$\Pr[T > t] = \Pr\left[\frac{T - \mu}{\sigma} > \frac{t - 126}{3}\right] = \Pr\left[Z > \frac{t - 126}{3}\right] = 0.05,$$ where $Z \sim \operatorname{Normal}(0,1)$ is standard normal. Since the $0.95$ quantile of $Z$ is approximately $1.645$; that is to say, $$\Pr[Z > 1.645] = 1 - \Pr[Z \le 1.645] \approx 0.05,$$ we now can solve for $t$: $$\frac{t - 126}{3} \approx 1.645,$$ giving $130.935$ hours. This means that if we randomly selected a pack of $9$ batteries, the chance it would have a total lifetime exceeding $130.935$ hours is only $5\%$.

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  • $\begingroup$ can you explain more briefly on why you put 1.645 equal to (t-126)/3? this is the only part i dont understand $\endgroup$ – Soon_to_be_code_master Nov 29 '17 at 4:18

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