A component in a device fails one time per 24 hours (on average). How many spare parts should be in order to verify that the probability they will be enough for one week is 95%? Use central limit theorem and standardization.
With standard tools (a.k.a without clt), I tried answering it with Poisson and got that minimal part of replacements should be 21.
I don't know from where to start here. I can guess that if we define $X$ to be number of replacements per week, then $\mu=7$ but I can't figure out what is $\sigma^2$? How can I proceed and use standardization here ?
Let $X_1,\dots,X_n$ be the use-lifetimes of components $1, 2, \dots, n$. We want the smallest $n$ such that $\Pr(\sum_1^n X_i\ge 7)\ge 0.95$. Then the number of replacement parts is $n-1$.
Without further information about the distribution of the $X_i$, it is not clear what to use for the variance. One could use the variance of the exponential with mean $1$. This conveniently happens to be $1$.