# Thread: normal distribution and probability

1. ## normal distribution and probability

Hi, reagrding this on a normal curve.....

Having a bit of trouble, i have an example for which i have worked out that that proabilty for x bieng bigger than 0.84 ( with a mean of 1.8 and a SD of 0.44) to be 0.9854
I take it this is right, but now it wants to know p(1.0<x<2.0), and im not sure how to approach this.

Any ideas????????

( also i posted something about confidence intervals in the advanced section and meant to post that in here, any help on that would be great aswell ).

2. Originally Posted by bobchiba
Hi, reagrding this on a normal curve.....

Having a bit of trouble, i have an example for which i have worked out that that proabilty for x bieng bigger than 0.84 ( with a mean of 1.8 and a SD of 0.44) to be 0.9854
I take it this is right, but now it wants to know p(1.0<x<2.0), and im not sure how to approach this.
You need to find the z-score.

1.8-.44z=.84
Thus,
z=2.181

Here

The probability is .4975

But you need to double that because that is only half of the normal curve to get.
.995

3. are you sure..... the table shows 2.181 is .4854 have you made a mistake or do i still not get it???

4. Originally Posted by bobchiba
are you sure..... the table shows 2.181 is .4854 have you made a mistake or do i still not get it???
Sorry, I have that table memorized and mentally went down the wrong colomun.

5. hi again, I am trying to work through these problem slooking at you r advice and some notes i have and i just cant seem to get my head around it.
this is the exact question word for word:
The Environment Protection Agency has developed a testing programme to monitor vehicle emission levels of several pollutants. Data collected under a variety of conditions suggests that the Normal curve provides an adequate approximation for the distribution of the key variable of interest, the amount of oxides of nitrogen (g/mile) emitted by a vehicle. The mean emission has been found to be 1.8 g/mile, with a standard deviation of 0.44 g/mile.

Calculate the following:

(a)The probability that a randomly selected vehicle emits more than 0.84 g/mile.

I can get up to -2.181 through z = (0.84 - 1.8)/0.44 and then i get confused