1. Hi Martingale. Below question is hard for me can you please help me with this one. Thanks a million again for all the help.

Assume that the time (in hours) unitl a failure of a valve is a random variable X ~ exp(0.01). What is the probabilty that the valvue lasts longer than 5 days.

2. Originally Posted by leo
Hi Martingale. Below question is hard for me can you please help me with this one. Thanks a million again for all the help.

Assume that the time (in hours) unitl a failure of a valve is a random variable X ~ exp(0.01). What is the probabilty that the valvue lasts longer than 5 days.
5 days is 5*24=120 hours

so you want P(X>120)

=1-P(X<=120)

=1-int{x=0 to 120}.01*e^{-.01*x}dx=.3011942

3. thanks martingale. how come you know all of this and I don't. the solution seems easy for you. i wish i have your knowledge. what do i need to do??

4. Hi Martingale,

Can you please help me again. I am having trouble with the problem below. My answer is 153 and it's wrong as per my practice test. Can you please show me how to solve it. Thanks.

Let X1 and X2 be random variables with variances of 6 and 1 respectively. Then
var(2X1 - 3x2)

5. Originally Posted by leo
Hi Martingale,

Can you please help me again. I am having trouble with the problem below. My answer is 153 and it's wrong as per my practice test. Can you please show me how to solve it. Thanks.

Let X1 and X2 be random variables with variances of 6 and 1 respectively. Then
var(2X1 - 3x2)

if X and Y are independent random variables and a,b are constants then

Var(aX+bY)=a^2var(X)+b^2var(Y)

I get 33.

6. Thanks Martingale. My practice test gives diff question each try. I applied the formula you gave and my answer is right. thanks a million again.

7. Hi sorry i dont kno if u can help me but i have a test soon on statistics and i need desperate help.. in normal distribtutions X~N(1.3,0.15^2) how would i find the 80th percentile and what does that even mean?

8. Originally Posted by Borjana
Hi sorry i dont kno if u can help me but i have a test soon on statistics and i need desperate help.. in normal distribtutions X~N(1.3,0.15^2) how would i find the 80th percentile and what does that even mean?
Double post

1) you should start a new thread and not hijack someone else's.

2) I would look in your book to find the definition of percentile.

9. i didnt mean to "hijack" the thread i dont kno how to start a new thread i jst found this site and thought u mite be able to help me since u were the only one online.. i dont have the biostatistics text book because it costs to much for me 2 afford it and i only have 3 weeks left of this course so i dont think it would be a wise investment of buying it jst to look up 80th percentile.. i jst need help for this questions thats all

10. Originally Posted by Borjana
i didnt mean to "hijack" the thread i dont kno how to start a new thread i jst found this site and thought u mite be able to help me since u were the only one online.. i dont have the biostatistics text book because it costs to much for me 2 afford it and i only have 3 weeks left of this course so i dont think it would be a wise investment of buying it jst to look up 80th percentile.. i jst need help for this questions thats all

From what you wrote it looks like you want the value of x such that

P(X<=x)=.8

X~N(1.3,0.15^2)

so...

P[(X-1.3)/.15<=(x-1.3)/.15]=.8

P[Z<=z]=.8

where Z~N(0,1)

and z=(x-1.3)/.15

find the value of z then solve for x.

11. Hello Martingale,

this is a simple problem and I wonder why I can't get this right. Here's the question

For a sample of size 4, if x1 - x bar = 10
x2 - xbar = -6
x3 - xbar=-8 then sample variance is.

I gave 66.67 and tried diff answers but they are all wrong. Pls let me know what is right then.

12. Originally Posted by leo
Hello Martingale,

this is a simple problem and I wonder why I can't get this right. Here's the question

For a sample of size 4, if x1 - x bar = 10
x2 - xbar = -6
x3 - xbar=-8 then sample variance is.

I gave 66.67 and tried diff answers but they are all wrong. Pls let me know what is right then.
When you computed your variance you didn't include x4-xbar

note: sum{i=1 to 4}(xi-xbar)=0

I get 72 as the variance.

13. Hi Martingale,

Thanks again for giving me the right answer. Can you please help me with the two questions below. Just couldn't get it right.

Assume the amount of light (in lumens) produced by a certain type of light bulb is normally distributed with mean u = 350 and variance 2 = 400.How many lumens do 90% of the light bulbs exceed?

The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a resistance smaller than 9.7565 ohms. What are the mean value of the resistance distribution? (Hint: you will need to calculate the standard deviation first)

14. Originally Posted by leo
Hi Martingale,

Thanks again for giving me the right answer. Can you please help me with the two questions below. Just couldn't get it right.

Assume the amount of light (in lumens) produced by a certain type of light bulb is normally distributed with mean u = 350 and variance 2 = 400.How many lumens do 90% of the light bulbs exceed?

what is "variance 2 = 400"

Originally Posted by leo
The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a resistance smaller than 9.7565 ohms. What are the mean value of the resistance distribution? (Hint: you will need to calculate the standard deviation first)

Shouldn't the mean be between 10.634 and 9.7565

15. that is supposed to be sigma squared but it didn't copy the sysmbol so variance = 400 for question 1.

regarding question 2, I think it is asking for the exact mean value.