A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 9 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 ?
 9 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 ?

This Question is Closed

SithsAndGiggles
 9 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
 9 months ago
Best ResponseYou've already chosen the best response.0Chebyshev's rule: \[1\frac{ 1 }{ k ^{2} }\]I've learned this, yes

SithsAndGiggles
 9 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
 9 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
 9 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
 9 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
 9 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
 9 months ago
Best ResponseYou've already chosen the best response.1When \(X=4\) you have \(0.19\) probability you turn the customer away.

wio
 9 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
 9 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
 9 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
 9 months ago
Best ResponseYou've already chosen the best response.0Oh wow... Wow. Okay thanks haha. Seemed simple enough.. Thank you!
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.