# College Statistics

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Mar 3rd 2014, 03:11 AM
AwesomeHedgehog
College Statistics
ACME manufacturing company has developed a fuel-efficient machine that
combines pressure washing with steam cleaning. It is advertised as
delivering 5 gallons of cleaner per minute. HOWEVER, a machine
delivers an amount at random anywhere between 2 and 6 gpm.

Assume that Y, the amount of cleaner dispensed per minute,
is a uniform random variable with probability density function
[p.d.f.], f(y), given below.

{ 1/4 2 <= y <= 6
f(y) = {
{ 0 otherwise

a.) Find the probability that Y, the amount of cleaner dispensed per minute, is greater than 5 gallons per minute.

b.) Find the expected value of Y, E(Y).
Find the variance value of Y, VAR(Y).

c.) Suppose a random sample of 9 pressure washing machines are to be tested and the sample mean, YBAR, calculated. Find the probability that the sample mean, YBAR, is greater than 5 gallons per minute.

What did you assume in order to calculate this probability?

d.) What happens to the probability in part c. (that the sample mean, YBAR, is greater than 5 gallons per minute) as the size n of the random sample increases? Does it stay same?

Please help. I am completely lost with where to start/how to start. Any help would be greatly appreciated!
• Mar 3rd 2014, 11:32 AM
AwesomeHedgehog
Re: College Statistics
I'm not sure if this is correct, but this is the work that I have done so far:

a.) P(Y > 5) = (5 + 6)/(6+6) = 11/12 = 0.917

b.) E(Y) = (6+2)/(1/4) = 32

Var(Y) = 1/(6 - 2)^2 = 1/16 = 0.0625

That's what I have so far, but I'm not sure if it is correct or how to do parts c and d.
• Mar 3rd 2014, 02:01 PM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
I'm not sure if this is correct, but this is the work that I have done so far:

a.) P(Y > 5) = (5 + 6)/(6+6) = 11/12 = 0.917

why use $\frac{1}{6}$ as the pdf? The problem pretty clearly states $p_Y(y)=\frac{1}{4}$

$E[y]=\int_2^6 y p_Y(y) dy$

$Var[y]=\int_2^6 \left(y-E[y]\right)^2 p_Y(y) dy$
• Mar 3rd 2014, 02:29 PM
AwesomeHedgehog
Re: College Statistics
I didn't use 1/6 as the pdf, I'm unsure what you mean by that? Are you saying that my P(Y > 5) is wrong?
• Mar 3rd 2014, 02:40 PM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
I didn't use 1/6 as the pdf, I'm unsure what you mean by that? Are you saying that my P(Y > 5) is wrong?

actually, looking more carefully at what you wrote, you have it more wrong than just that.
Why 6+6 in the denominator??

$Pr[X>5] = \int_5^{\infty} f_X(x)dx=\int_5^6 \dfrac{1}{4} dx =$

$\dfrac{x}{4}\left| \begin{array}{l} \small{6 \\ 5} \end{array}\right.=\dfrac{6-5}{4}=\dfrac{1}{4}$
• Mar 3rd 2014, 03:20 PM
AwesomeHedgehog
Re: College Statistics
Oh, alright, now I understand what you are saying.

For part c), I realized the problem does not give you the size (n) of how many pressure washing machines are developed. I know that the sample mean = (sample sizes choosen)/n. So, for this case, I would just assume how many pressure washing machines are developed, which I am going to make it be 50.

Therefore, my sample mean = 9/50 = 0.18

Then I have to find the probability that the sample mean is greater than 5. So, P[ sample mean > 5] = and this is where I am drawing a blank.
• Mar 3rd 2014, 03:29 PM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
Oh, alright, now I understand what you are saying.

For part c), I realized the problem does not give you the size (n) of how many pressure washing machines are developed. I know that the sample mean = (sample sizes choosen)/n. So, for this case, I would just assume how many pressure washing machines are developed, which I am going to make it be 50.

Therefore, my sample mean = 9/50 = 0.18

Then I have to find the probability that the sample mean is greater than 5. So, P[ sample mean > 5] = and this is where I am drawing a blank.

I'm detecting a deep misunderstanding of things. I don't know what you mean by "number of machine are developed". You have what you need to know from the individual pdf and one additional assumption, which they ask you to specify. What do you usually assume about problems like these regarding the statistics of individual machines in a sample group.
Spoiler:
That they are identically and independently distributed.
• Mar 3rd 2014, 03:38 PM
AwesomeHedgehog
Re: College Statistics
Ohhh I'm now understanding what this problem is asking, only problem is, is that I haven't had much experience with learning it. Would this be a problem involving order statistics? If so, I am lost on how to find the probability for it, which is starting to frustrate me. Ugh, I'm sorry for being so difficult. Any help for pushing me in the right direction is really appreciated.
• Mar 3rd 2014, 03:57 PM
romsek
Re: College Statistics
you're making me work!

Go back to your text and review Chebyshev's inequality. I believe it's what you are looking for.
• Mar 3rd 2014, 04:09 PM
AwesomeHedgehog
Re: College Statistics
P[ |X - μ| ≥ kσ] ≤ (1/k^2) ===> P[ |X - μ| < kσ] = 1 - (1/k^2)

So, for this problem, μ = 9, k = 5?

P(sample mean > 5) = the integral from 5 to ∞ (1/4) dx ??
• Mar 4th 2014, 01:18 AM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
P[ |X - μ| ≥ kσ] ≤ (1/k^2) ===> P[ |X - μ| < kσ] = 1 - (1/k^2)

So, for this problem, μ = 9, k = 5?

P(sample mean > 5) = the integral from 5 to ∞ (1/4) dx ??

Ok. I looked a little deeper into this.

In general when you add two independent random variables. The pdf of their sum is the convolution of their individual pdf's.

There is a formula for the convolution of n identical uniform variables. This is called an Irwin-Hall distribution. But I don't think the problem is looking for you to use this, nor do I think the problem wants you to manually calculate the convolution of 9 uniform pdfs.

I think what they want you to do is use the sample mean and variance to approximate the pdf of the sample mean as a Normal random variable.

From earlier in the problem you found $E[Y]=\overline{Y},$ and $Var[Y]$.

The sum of n identical independent random variables has sample mean $\mu=\overline{Y}$ and sample variance $\sigma^2=\dfrac{Var[Y]}{n}$

so the probability that the sample mean is greater than 5 is just $\displaystyle \int_5^\infty N(\mu,\sigma)(x) dx$

for (d) you'll notice that the variance of the sample mean shrinks as the sample size increases, i.e. the probability mass is more concentrated about the actual mean. You should be able to see what that does to the probability calculated above.

You might want to check with your professor to make sure this is how they want you to do this. But as far as I know the only other way would be to compute that convolution of 9 uniform pdfs, or to use the Irwin-Hall pdf formula (they are the same thing) and I'd be pretty surprised if that's what they have in mind.
• Mar 4th 2014, 05:47 AM
AwesomeHedgehog
Re: College Statistics
For part c, this is what I did:

μ = E(Y) = 4

σ^2 = (4/3)/n

For this problem, I'm assuming that n = 9, so that would make σ^2 = (4/3)/9 = 4/27

So, then σ = 0.3849

And N(μ, σ) = 1.5396

Then, the integral would be from 5 to ∞ (μ*σ*x) dx
• Mar 4th 2014, 11:28 AM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
For part c, this is what I did:

μ = E(Y) = 4

σ^2 = (4/3)/n

For this problem, I'm assuming that n = 9, so that would make σ^2 = (4/3)/9 = 4/27

So, then σ = 0.3849

you are correct up to here...

and then head off somewhere I don't follow

Quote:

And N(μ, σ) = 1.5396

Then, the integral would be from 5 to ∞ (μ*σ*x) dx
no idea really what those last two lines mean.
• Mar 4th 2014, 02:22 PM
AwesomeHedgehog
Re: College Statistics
Yeah, sorry about that. When you used the notation, N(μ, σ), I have never seen that before, so I wasn't completely understanding what it was.

So, after searching it up online, I found out it's the standard normal distribution. So, I'm assuming that I just plug μ = 4, σ = 0.3849, and n = 9 into the normal distribution formula.

N(μ, σ) = (x - μ)/σ = 12.99

So, plugging that into the integral, I'd get:

The integral going from 5 to 6 of 12.99*x dx = 71.445

I did 5 to 6 since f(y) is between 2 and 6
• Mar 4th 2014, 02:34 PM
romsek
Re: College Statistics
Quote:

Originally Posted by AwesomeHedgehog
Yeah, sorry about that. When you used the notation, N(μ, σ), I have never seen that before, so I wasn't completely understanding what it was.

So, after searching it up online, I found out it's the standard normal distribution. So, I'm assuming that I just plug μ = 4, σ = 0.3849, and n = 9 into the normal distribution formula.

N(μ, σ) = (x - μ)/σ = 12.99

So, plugging that into the integral, I'd get:

The integral going from 5 to 6 of 12.99*x dx = 71.445

I did 5 to 6 since f(y) is between 2 and 6

you already took the 9 into account when you found the variance. so just use $\sigma=0.3849$

and don't truncate the integral, just use the normal approximation as you usually would.

I get 0.00468737 as the answer. See if you can replicate that.
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last