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 11 months ago
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 ?
 11 months ago
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 ?

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SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.0Initially, 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?

Raspberry95
 11 months ago
Best ResponseYou've already chosen the best response.0Chebyshev's rule: \[1\frac{ 1 }{ k ^{2} }\]I've learned this, yes

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.0Okay, well it looks like you'll have to calculate the mean and standard deviation for the data first: \[\begin{align*}\mu&=E(X)\\&=\sum_{x=0}^5x f(x)\\&=2.72\\\\ \sigma^2&=V(X)\\ &=E(X^2)\bigg[E(X)\bigg]^2\\ &=\sum_{x=0}^5x^2 f(x)2.72^2\\ &=9.627.4\\ &=2.22\\ \sigma&=\sqrt{2.22}\approx1.49 \end{align*}\] 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 \[\frac{1}{k^2}=0.10~~\Rightarrow~~k=3.16\] \[X2.72\ge 3.16\cdot1.49\] 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.

Raspberry95
 11 months ago
Best ResponseYou've already chosen the best response.0Thanks. 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.

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.0Hmm, it might not be a Chebyshev problem ... Maybe someone else can help out, or point out a mistake? @wio

Raspberry95
 11 months ago
Best ResponseYou've already chosen the best response.0I 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!

wio
 11 months ago
Best ResponseYou've already chosen the best response.1If you let \(X=5\) then you have \(100\%\) probability that you can serve the customer.

wio
 11 months ago
Best ResponseYou've already chosen the best response.1When \(X=4\) you have \(0.19\) probability you turn the customer away.

wio
 11 months ago
Best ResponseYou've already chosen the best response.1Seems like no matter what, you have to increase by \(3\) so that you can serve \(5\) at a time.

Raspberry95
 11 months ago
Best ResponseYou've already chosen the best response.0Why is it when X = 5 would I have an 100% probability of serving the customers?

wio
 11 months ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=0.06+%2B0.16+%2B0.26+%2B+0.23+%2B+0.1+%2B+0.19+

Raspberry95
 11 months ago
Best ResponseYou've already chosen the best response.0Oh wow... Wow. Okay thanks haha. Seemed simple enough.. Thank you!
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