1) a test has a mean of 68 and a variance of 25 in a normally distributed population. find the probability that a single score chosen at random:
a) is higher than 71
b) is between 70 and 74
c) is between 64 and 72
We are given that `mu=68` and `sigma=25` :
(1) Find P(x>71)
First convert the score to a z-score: `z=(x-mu)/sigma`
The question is now find P(z>0.12). Most standard normal tables give the area under the standard normal curve to the left of a given z score -- here we want the area to the right. So we could find the area to the left in the table and subtract from 1:
Using a table or technology we find the area to the left of z=0.12 to be .5478, so the area to the right is 1-.5478=.4522 and the probability is 45.22%.
(In a TI-83/84 calculator you can use normalcdf(0.12,E99) to get the answer directly. Or you can compute the answer without converting to a z-score by normalcdf(71,E99,68,25)
(2) Find P(70<x<74) Following the same procedure we convert 70 and 74 to z-scores to get z=.08 and z=.24
Then we want P(.08<z<.24) We find the area to the left of z=.24 and subtract the area to the left of z=.08.
.5948-.5319=.0629 so the probability is 6.29%
** You can directly input normalcdf(70,74,68,25) to get .062953 which is slightly more accurate since the calculator rounds after many more digits.
(3) P(64<x<72) Using the above procedure we get .1271 so the probability is 12.71%
** Your question was edited to be one question**