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Based on recent records, the manager of a car painting center has determined the following probability distribution for the number of customers per day. x 0 1 2 3 4 5 f(x) 0.06 0.16 0.26 0.23 0.1 0.19 If the center has the capacity to serve two customers per day. By how much must the capacity be increased so the probability of turning a customer away is no more than 0.10 ?

Probability
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Initially, we would turn a customer away if \(X>2\), so the probability that this occurs is \(P(X>2)\): \[P(X>2)=P(X=3)+P(X=4)+P(X=5)=0.52\] Now we want to find \(k\) such that \(P(X>k)<0.10\). Hmm... have you learned about Chebyshev's (Tchebysheff's, etc) inequality?
Chebyshev's rule: \[1-\frac{ 1 }{ k ^{2} }\]I've learned this, yes
Okay, well it looks like you'll have to calculate the mean and standard deviation for the data first: \[\begin{align*}\mu&=E(X)\\&=\sum_{x=0}^5x f(x)\\&=2.72\\\\ \sigma^2&=V(X)\\ &=E(X^2)-\bigg[E(X)\bigg]^2\\ &=\sum_{x=0}^5x^2 f(x)-2.72^2\\ &=9.62-7.4\\ &=2.22\\ \sigma&=\sqrt{2.22}\approx1.49 \end{align*}\] Ok, so now you want to find \(X=x\) such that, according to the inequality, \[P(|X-\mu|\ge k\sigma)\le\color{red}{\frac{1}{k^2}}\] The red value is given to be \(0.10\), so \[\frac{1}{k^2}=0.10~~\Rightarrow~~k=3.16\] \[|X-2.72|\ge 3.16\cdot1.49\] Solving for \(X\), it looks the manager should upgrade to \(x\approx7.4284\approx8\) slots for customers, i.e. increase the current capacity by 6.

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Thanks. But apparently that isn't the right answer. I tried six then I tried 5.433... because that would be how much you would increase it by from 2, but both didn't work.
Hmm, it might not be a Chebyshev problem ... Maybe someone else can help out, or point out a mistake? @wio
I understand what you're getting at though. It does make sense to me even thought we haven't learned that inequality. The mean and the standard deviation was correct too (it was another part of the question added onto this). I just have no idea what she wants. Our prof never went through anything like this question with us. Ehh. Thanks for the help though!
If you let \(X=5\) then you have \(100\%\) probability that you can serve the customer.
When \(X=4\) you have \(0.19\) probability you turn the customer away.
Seems like no matter what, you have to increase by \(3\) so that you can serve \(5\) at a time.
Why is it when X = 5 would I have an 100% probability of serving the customers?
http://www.wolframalpha.com/input/?i=0.06+%2B0.16+%2B0.26+%2B+0.23+%2B+0.1+%2B+0.19+
Oh wow... Wow. Okay thanks haha. Seemed simple enough.. Thank you!

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