Question 1177812: Find the percentage of the total area under a normal curve
between the given values of z.
My first problem is z = 1.31 and z = 1.73
The book gave the answer 5.3%, which I understand by subtracting.
The second problem is z = -3.02 and z = 2.03
The book gives the answer 97.8%, which I got by adding.
However, I am confused because we are supposed to do the problems the same way. If someone could please walk me through the steps of how they got these answers that would be greatly appreciated.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you take the smaller area to the left of the z-score and subtract it from the larger area to the left of the z-score.
the larger area will be the area to the left of the more positive z-score.
the smaller area will be the area to the left of the less positive z-score.
your first z-scores are 1.31 and 1.73
you take the area to the left of 1.31 and subtract it from the area to the left of 1.73.
the result is the area between those z-scores.
i get the following, using my ti-84 plus
area to the left of 1.31 = .904902018
area to the left of 1.73 = .958184901
area in between = .958 - .905 = .053 = 5.3%
your second zs-scores are -3.02 and 2.03
area to the left of -3.02 = .0012639426
area to the left of 2.03 = .9788217992
area in between = .979 - .001 = .978 = 97.8%.
you should not have been able to get the second answer by adding.
let me know what you did and what table or calculator you used and i might be able to figure out why you got the right answer by adding.
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