# Thread: Need help with probability

1. ## Need help with probability

I guess I was hoping if anybody could check my work? This is a 4 part question, I hope I am not breaking rules by posting so much?? I took most of them as far as I could, a couple I am completely clueless on....!

Suppose that for a school, distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles.

a) What percentage of students live farther than 6.78 miles from the school?

P(X>6.78 miles)
Z=X-mean/std --> 6.78-4.76/1.74= 1.16
P(0<z<1.16)= 0.3770
0.5- 0.3770= 0.123
Thus, 12.3% of students live farther than 6.78 miles from the school.
(yes? no? i think i may have got the right answer for that one)

b) A survey shows that 8% of students who live closest to the school chose to walk to the school. What is the maximum walking distance of these 8% of students? In other words, what is the distance below which these 8% of students walk to school?

(I really don't know how to start on this one, could anybody set me in the right direction?)

c) Suppose there is a new policy that allows students living beyond 4.50 miles to take a bus to school. There are 3,567 students enrolled. How many students are not eligible to take school buses?

Z=4.5-4.76/ 1.74= -0.15
P(-0.15<z<0)= 0.0596
0.50 + 0.0596 = 0.5596 or 55.96%
1-0.5596= 0.4404
0.4404 * 3567= 1570.91 = 1571
Thus, approximately 1,571 students are not eligible to take school buses.

d) Suppose all samples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

*Since this one is talking about sample means, to find the z-score is:

Z=Mean-Mean of sample means/ std of sample means, so

Z=6.78-4.76/ (1.74/[sqrt]12)= 4.02

I get lost here, because how am I supposed to find a z-score of 4.02 in my table? *bangs head* Anybody?

2. Hello !

Originally Posted by peanutbutter
I guess I was hoping if anybody could check my work? This is a 4 part question, I hope I am not breaking rules by posting so much?? I took most of them as far as I could, a couple I am completely clueless on....!
Breaking the rules ? You're kidding eh ?
Well, I personnally don't see why it would break the rules

Suppose that for a school, distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles.

a) What percentage of students live farther than 6.78 miles from the school?

P(X>6.78 miles)
Z=X-mean/std --> 6.78-4.76/1.74= 1.16
P(0<z<1.16)= 0.3770
0.5- 0.3770= 0.123
Thus, 12.3% of students live farther than 6.78 miles from the school.
(yes? no? i think i may have got the right answer for that one)
Good
But please use parenthesis, because it's (X-mean)/std

b) A survey shows that 8% of students who live closest to the school chose to walk to the school. What is the maximum walking distance of these 8% of students? In other words, what is the distance below which these 8% of students walk to school?

(I really don't know how to start on this one, could anybody set me in the right direction?)
That is to say $\displaystyle P(0<x<d_{max})=0.08$
I guess you can take it from here, can't you ?

c) Suppose there is a new policy that allows students living beyond 4.50 miles to take a bus to school. There are 3,567 students enrolled. How many students are not eligible to take school buses?

Z=4.5-4.76/ 1.74= -0.15
P(-0.15<z<0)= 0.0596
0.50 + 0.0596 = 0.5596 or 55.96%
1-0.5596= 0.4404
0.4404 * 3567= 1570.91 = 1571
Thus, approximately 1,571 students are not eligible to take school buses.
Good !

d) Suppose all samples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

*Since this one is talking about sample means, to find the z-score is:

Z=Mean-Mean of sample means/ std of sample means, so

Z=6.78-4.76/ (1.74/[sqrt]12)= 4.02

I get lost here, because how am I supposed to find a z-score of 4.02 in my table? *bangs head* Anybody?
I'm sorry for this one, I can't help because I have never studied it...

If you are sure you're correct, then the z-score of 4.02 is something very very near of 0.5, considered as 0.5.

3. ## so..

P(0<x<Xmax)=0.08
Z-score corresponding to 0.08 is 0.2
X=mean + z*std

X=4.76 + 0.2*1.74=5.108

yes? no?

4. ok, since i have z score for 0.08 you gave me -2.74, then is the Zmax that you are telling me I need to find, the Zscore for 0.42? since 0.5-0.08=0.42?

5. I'm sorry, I don't know what happened to my answer

P(0<x<Xmax)=0.08
Z-score corresponding to 0.08 is 0.2
X=mean + z*std

X=4.76 + 0.2*1.74=5.108

yes? no?
This is actually correct. I just hope I'm not mistaking once again... It seems that I'm tired >.<

6. Originally Posted by peanutbutter
I guess I was hoping if anybody could check my work? This is a 4 part question, I hope I am not breaking rules by posting so much?? I took most of them as far as I could, a couple I am completely clueless on....!

Suppose that for a school, distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles.

[snip]

d) Suppose all samples of size 12 are taken. What percentage of sample means has a value larger than 6.78 miles?

*Since this one is talking about sample means, to find the z-score is:

Z=Mean-Mean of sample means/ std of sample means, so

Z=6.78-4.76/ (1.74/[sqrt]12)= 4.02

I get lost here, because how am I supposed to find a z-score of 4.02 in my table? *bangs head* Anybody?
See here: http://www.mathhelpforum.com/math-he...need-help.html

7. Originally Posted by peanutbutter
I guess I was hoping if anybody could check my work? This is a 4 part question, I hope I am not breaking rules by posting so much?? I took most of them as far as I could, a couple I am completely clueless on....!

Suppose that for a school, distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles.

[snip]

b) A survey shows that 8% of students who live closest to the school chose to walk to the school. What is the maximum walking distance of these 8% of students? In other words, what is the distance below which these 8% of students walk to school?

(I really don't know how to start on this one, could anybody set me in the right direction?)

[snip]
Originally Posted by peanutbutter
P(0<x<Xmax)=0.08
Z-score corresponding to 0.08 is 0.2
X=mean + z*std

X=4.76 + 0.2*1.74=5.108

yes? no?
No. Pr(0 < X < 5.108) = 0.575, not 0.08.

I think you're probably expected to consider Pr(X < Xmax) = 0.08 rather than Pr(0 < X < Xmax) = 0.08. If you draw a simple picture of the distribution for X and put Xmax in a spot where Pr(X < Xmax) is 0.08, it's very clear that Xmax lies well left of the mean. So the corresponding z-value has to be less than 0.

You should get the z-value corresponding to Pr(Z < zmax) = 0.08 to be zmax = -1.41 and so Xmax = 2.29.

8. Originally Posted by mr fantastic
No. Pr(0 < X < 5.108) = 0.575, not 0.08.

I think you're probably expected to consider Pr(X < Xmax) = 0.08 rather than Pr(0 < X < Xmax) = 0.08. If you draw a simple picture of the distribution for X and put Xmax in a spot where Pr(X < Xmax) is 0.08, it's very clear that Xmax lies well left of the mean. So the corresponding z-value has to be less than 0.

You should get the z-value corresponding to Pr(Z < zmax) = 0.08 to be zmax = -1.41 and so Xmax = 2.29.
Ok, I have a question for this...

x can't be < 0. It's a nonsense.

So isn't $\displaystyle P(0<x<X_{max})=P(x<X_{max})$ ?

Because I once helped someone for that, and we had to consider this, id est not including the extreme values of definition...

9. Originally Posted by Moo
Ok, I have a question for this...

x can't be < 0. It's a nonsense.

So isn't $\displaystyle P(0<x<X_{max})=P(x<X_{max})$ ?

Because I once helped someone for that, and we had to consider this, id est not including the extreme values of definition...
I agree. But:

"distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles"

so the 'nonsense' is due to a defect in the given model.

I think the problem is more difficult than intended if Xmax such that Pr(0 < X < Xmax) = 0.08 is attempted.

In light of previous questions, I think Pr(X < Xmax) = 0.08 is the level of difficulty the question intends.

Peanutbutter should seek clarification on this matter from her professor.

10. Originally Posted by mr fantastic
I agree. But:

"distances between students' homes and school observe a normal distribution, with mean 4.76 miles and standard deviation being 1.76 miles"

so the 'nonsense' is due to a defect in the given model.

I think the problem is more difficult than intended if Xmax such that Pr(0 < X < Xmax) = 0.08 is attempted.

In light of previous questions, I think Pr(X < Xmax) = 0.08 is the level of difficulty the question intends.

Peanutbutter should seek clarification on this matter from her professor.
Ok, so answer should be :

$\displaystyle P(x<X_{max})=P(z<Z_{max})=0.08=1-0.92 \implies Z_{max}={\color{red}-1.41}$, etc... ?

This looks weird... and it's confusing :'(

11. Originally Posted by Moo
Ok, so answer should be :

$\displaystyle P(x<X_{max})=P(z<Z_{max})=0.08=1-0.92 \implies Z_{max}={\color{red}-1.41}$, etc... ?
[snip]

Originally Posted by mr fantastic in post #7
[snip]
You should get the z-value corresponding to Pr(Z < zmax) = 0.08 to be zmax = -1.41 and so Xmax = 2.29.
To expand:

$\displaystyle Z_{\text{max}} = \frac{X_{\text{max}} - 4.76}{1.76} \Rightarrow -1.41 = \frac{X_{\text{max}} - 4.76}{1.76} \Rightarrow X_{\text{max}} = 2.28$.

Note that Pr(X < 2.28) = 0.08 so all is well.

Originally Posted by Moo
[snip]
This looks weird... and it's confusing :'(

12. Originally Posted by Moo
Ok, I have a question for this...

x can't be < 0. It's a nonsense.

So isn't $\displaystyle P(0<x<X_{max})=P(x<X_{max})$ ?

Because I once helped someone for that, and we had to consider this, id est not including the extreme values of definition...
Just to cover all bases:

$\displaystyle \Pr(0 < X < X_{\text{max}}) = 0.08 \Rightarrow \Pr(X < X_{\text{max}}) - \Pr(X < 0) = 0.08$

$\displaystyle \Rightarrow \Pr(X < X_{\text{max}}) - 0.003 = 0.08 \Rightarrow \Pr(X < X_{\text{max}}) = 0.083$.

So the error in making the approximation is small. (For the record it turns out that Xmax = 2.32).

I'd say the difference is probably comparable to the error in the tables ....