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
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 ).
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