# Data Manag. - Prob. Questions + My views + Checks

• Jan 5th 2009, 04:36 PM
Faisal2007
Data Manag. - Prob. Questions + My views + Checks
Thanks for looking.
I appreciate your time and effort.

Quote:

Suppose that 65% of the families in a town own computers. If eight families are surveyed at random, what is the prob. that at least four own computers ?
Ok so N = 8
X = 4
Success = 0.65 ?
Failure = 0.35 ?

I am using the formulate :
$P(X) = 8C4 * 0.65 ^4 * 0.35^0$

I know i am doing something wrong...

Quote:

Trump - airlines has dtermined 5% of its customers do not show up for their flights.
If a passenger is bumped off a flight because of overbooking, the airline pays the customer $200. What is the expected payout by the airline if it overbooks a 240 seat airplane by 5% The wording is really messing me up.... Quote: In a binomial dist., if 10 trials are conducted, and the prb. of failure is 0.3, what is the expected value ? 0.3 3 0.7 7 None N = 10 Q = 0.3 P = 0.7 $E(X) = n * p$ $E(X) = 10 * 0.7$ $E(X) = 7$ Is that correct ? Ok thanks for your time so far... This is the hardest... Quote: A manufacturer of electronics components produces precision resistors designd to have a tolerance of +/- 1%. from quality control testing, the manu. knows that about 1 resistor in 6 is actually within just 0.3% of its nominal value. A customer needs 5 of these more precise resistors. what is the prob. of finding exactly 5 such resistors aming the first 8 tested? I am twisted in there... There are 2 of them.... 1 in 6 and 5 in 8... • Jan 5th 2009, 05:38 PM Chris L T521 Quote: Originally Posted by Faisal2007 Thanks for looking. I appreciate your time and effort. Quote: Suppose that 65% of the families in a town own computers. If eight families are surveyed at random, what is the prob. that at least four own computers ? Ok so N = 8 X = 4 Success = 0.65 ? Failure = 0.35 ? I am using the formulate : $P(X) = 8C4 * 0.65 ^4 * 0.35^0$ I know i am doing something wrong... Let $X$ be the event of a family owning computers. We are looking for $P\left(X\geqslant4\right)$. Note that $P\left(X\geqslant4\right)=1-P\left(X\leqslant3\right)$ Since $X$ is $\mathcal{B}\left(8,.65\right)$, we now see that $P\left(X\geqslant4\right)=1-P\left(X\leqslant3\right)=1-\sum_{x=0}^3 {8\choose x} \left(.65\right)^x\left(.35\right)^{8-x}\approx\color{red}\boxed{.8939}$ Does this make sense? I know i am doing something wrong... Quote: Quote: Trump - airlines has dtermined 5% of its customers do not show up for their flights. If a passenger is bumped off a flight because of overbooking, the airline pays the customer$200. What is the expected payout by the airline if it overbooks a 240 sea airplane by 5%
The wording is really messing me up....
The first bit is messing me up too...maybe mr fantastic or someone else can help you with this one.

Quote:

Quote:

In a binomial dist., if 10 trials are conducted, and the prb. of failure is 0.3, what is the expected value ?

0.3
3
0.7
7
None
N = 10
Q = 0.3
P = 0.7

$E(X) = n * p$
$E(X) = 10 * 0.7$
$E(X) = 7$

Is that correct ?
Yes! (Clapping)

Quote:

Ok thanks for your time so far...
This is the hardest...

Quote:

A manufacturer of electronics components produces precision resistors designed to have a tolerance of +/- 1%. from quality control testing, the manu. knows that about 1 resistor in 6 is actually within just 0.3% of its nominal value. A customer needs 5 of these more precise resistors. what is the prob. of finding exactly 5 such resistors among the first 8 tested?
I am twisted in there...

There are 2 of them....

1 in 6 and 5 in 8...
Let $X$ be the event of receive one of these nominal resistors. The probability of getting one nominal resistor is one out of six, or $p=\tfrac{1}{6}$. The sample size/number of trials is $n=8$

Thus, $X$ is $\mathcal{B}\left(8,\tfrac{1}{6}\right)$. We can now see that $P\left(X=5\right)={8\choose5}\left(\tfrac{1}{6}\ri ght)^5\left(\tfrac{5}{6}\right)^3\approx\color{red }\boxed{0.0042}$

Does this make sense?
• Jan 5th 2009, 05:52 PM
Faisal2007
Quote:

Originally Posted by Chris L T521

Let $X$ be the event of a family owning computers.

We are looking for $P\left(X\geqslant4\right)$. Note that $P\left(X\geqslant4\right)=1-P\left(X\leqslant3\right)$

Since $X$ is $\mathcal{B}\left(8,.65\right)$, we now see that $P\left(X\geqslant4\right)=1-P\left(X\leqslant3\right)=1-\sum_{x=0}^3 {8\choose x} \left(.65\right)^x\left(.35\right)^{8-x}\approx\color{red}\boxed{.8939}$

Does this make sense?
Yes! (Clapping)

Thanks alot !

Although i was looking for methods and clarifications (Wink) you gave me the answers as well (Giggle)

the airline and the manufacturer questions are giving me a headache...

This is what i have for the electronic manufacturer...

every 1 of 6 resistors it makes is within 0.3 of its nominal value....so then the others are 99.7% within the nominal value...

I dont know where i am getting with this...
• Jan 5th 2009, 06:00 PM
mr fantastic
Quote:

Originally Posted by Faisal2007
Quote:

A manufacturer of electronics components produces precision resistors designd to have a tolerance of +/- 1%. from quality control testing, the manu. knows that about 1 resistor in 6 is actually within just 0.3% of its nominal value. A customer needs 5 of these more precise resistors. what is the prob. of finding exactly 5 such resistors aming the first 8 tested?
I am twisted in there...
There are 2 of them....
1 in 6 and 5 in 8...

Quote:

Originally Posted by Chris L T521
Let $X$ be the event of receive one of these nominal resistors. The probability of getting one nominal resistor is one out of six, or $p=\tfrac{1}{6}$. The sample size/number of trials is $n=8$

Thus, $X$ is $\mathcal{B}\left(8,\tfrac{1}{6}\right)$. We can now see that $P\left(X=5\right)={8\choose5}\left(\tfrac{1}{6}\ri ght)^5\left(\tfrac{5}{6}\right)^3\approx\color{red }\boxed{0.0042}$

Does this make sense?

Although the problem is actually sampling without replacement, if the population of resistors is large then the use of the binomial distribution is valid.

Quote:

Originally Posted by Faisal2007
[snip]
This is what i have for the electronic manufacturer...

every 1 of 6 resistors it makes is within 0.3 of its nominal value....so then the others are 99.7% within the nominal value...

dont know where i am getting with this...

• Jan 5th 2009, 06:07 PM
mr fantastic
Quote:

Originally Posted by Faisal2007
[snip]
Quote:

Trump - airlines has dtermined 5% of its customers do not show up for their flights.
If a passenger is bumped off a flight because of overbooking, the airline pays the customer \$200. What is the expected payout by the airline if it overbooks a 240 seat airplane by 5%
The wording is really messing me up....

5% of 240 is 12. So the plane is overbooked by 12. So there are 252 people booked on the plane.

Ler X be the random variable number of people who turn up.

X ~ Binomial(n = 252, p = 0.95)

$E(X) = 239.4 \leq 240$.

Therefore the expected payout is zero I would have thought.
• Jan 5th 2009, 06:45 PM
Faisal2007
Thanks alot Mr. Fantastic !
You always have the right answer !

Chris...So the +/- 1% and 0.3% were just there to confuse the solver ?
Thanks alot to you too =))))