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Simple question, but I am a bit confused:

Given a working month of $28$ days, and let $Z$ be the number of days on which the sales exceed the target with a probability of $0.625$. Identify the distribution of $Z$ and find the probability that the sales exceed the target on $17$ days of the month.

I know that $Z\sim \text{Binomial}(n,p)$.

My question here is the number of trials, $n$, which is $28$ or $29$? Should I take $0$ as a probability for this?

If I only take $n= 28$, then my probability is $0.150045$.

If $n = 29$, then probability is $0.135978$.

Quite a difference there, any clarification would be appreciated.

((This post can be closed/deleted after I receive an answer as I don't think it will be of much help to others?))

Mee Seong Im
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NAA
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    You have $28$ possible days, so $n=28$. Letting $Z_i$ (for $1\leq i\leq 28$) be the random variable indicating whether the sales on day $i$ exceed the target, you get $Z= \sum_{i=1}^{28} Z_i$. Now, $Z$ can take $29$ values: $0\leq Z \leq 28$, but you have $28$ summands (the value $0$ is taken by $Z$ if all of the $28$ $Z_i$'s are zero). – Clement C. Mar 19 '17 at 23:59
  • @ClementC. now that clears everything. Thank you so much! – NAA Mar 20 '17 at 10:02

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