Normal and binomial distribution: using Z-scores to find answer

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The discussion centers on calculating the maximum price a fruit shop owner is willing to pay for bananas, given a mean price of $1.35/kg and a standard deviation of $0.18. The owner maintains stock 8% of the time, indicating that there is an 8% probability of having stock left at the end of the day. A Z-score of approximately 0.201 corresponds to this probability, which should be interpreted as a positive value since it represents the right-hand tail of the distribution. The formula Z=(X-μ)/σ is applied to find the maximum price, leading to a calculated price of $1.31. The discussion emphasizes the correct interpretation of Z-scores in relation to probability and stock management.
jackscholar
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The prices for bananes that a fruit shop would have to pay to keep them in stock have a mean of $1.35/kg and a standard deviation of 18 cents. The owner will not pay more than a certain price, but manages to keep stock 8% of the time. What is the maximum price the ownwer will pay?

I found 0.08 on a Z-score table and it was approximately Z=0.201. This is a negative number because it is less than the mean (or so i figured). I used the formula Z=(X-μ)/σ and re-arranged to get (-0.201*0.18)+1.35= price he is willing to pay. This then gave $1.31.
 
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"...but manages to keep stock 8% of the time"

means the probability the owner will have stock left at the end of the day is 8%, or 0.08. Stated another way, there is only an 8% chance that the demand for his product will be greater than what he has on hand. This means the Z-score you seek is the one that cuts off the right-hand 8% of the distribution, so should not be a negative value. Your basic idea for attacking the problem seems correct, except for this item.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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