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

**Faisal2007** · January 5th 2009, 04:36 PM

Thanks for looking.
I appreciate your time and effort.

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...

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....

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...

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...

**Chris L T521** · January 5th 2009, 05:38 PM

Originally Posted by Faisal2007

Thanks for looking.
I appreciate your time and effort.

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...

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.

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!

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

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?

**Faisal2007** · January 5th 2009, 05:52 PM

Originally Posted by Chris L T521

Warning: Answer is not complete

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!

Thanks alot !
I really appreciate your help.

Although i was looking for methods and clarifications

you gave me the answers as well

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...

**mr fantastic** · January 5th 2009, 06:00 PM

Originally Posted by Faisal2007

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...

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.

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...

The solution and answer have already been given by Chris ....

**mr fantastic** · January 5th 2009, 06:07 PM

Originally Posted by Faisal2007

[snip]

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.

**Faisal2007** · January 5th 2009, 06:45 PM

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 =))))

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