Use the table of z-scores to find the percentage of data items in a normal distribution that lie in the given range.

Question

a. above z= -0.9

b. below z= -0.9

c. between z= .4 and z= 1.4

d. above z= -1.86

e. below z= 2.43

f. between z= 1.94 and z= 2.87

Answer 1

To compute percentages of z-values in a given range, you need to look up the probabilities (P) of the z-value being in that range (between 0 and 1), the multiply by 100 to get the percentage. The normal probability tables list the probabilities corresponding to values less than a specific z-value.

a) The probabilities corresponding to z-values in the table are cumulative below a z-value so to find

percentage of z-score above z=-0.9, you need to compute 1 - P(z < - 0.9) = 1 - 0.1841 = 0.8759.

then multiply by 100 to get percentage = 87.59%.

b) Here you just look up the probability corresponding to z < -0.9 from the table: 0.1841 and multiply by 100

to get 18.41%.

c) Here you need to find P(.4 < z < 1.4). Because we are dealing with cumulative probables, this would be the same as finding P(z<1.4) - P(z<0.4) = 0.9192 - 0.6554 = 0.2638 or 26.38%.

d) Here you need to find P(z > -1.86), so you need to compute 1 - P(z < -1.86) = 1 - 0.314 = 0.686 or 68.6%.

e) Here you need to find P(z < 2.43), which would be 0.9925 or 99.25%.

f) Here you need to find P(1.94 < z < 2.87), which would be the same as finding P(z<2.87) - P(z<1.94) = 0.9979 - 0.9738 = 0.0241 or 2.41%.

Answer 2

1) Find an appropriate z-score table: This one is good. http://www.z-table.com/

2) Look up the first non-decimal digit of the z-score in the left hand column

3) Look uo the two decimal digits of the z-score in the top row

4) Locate where the column and row intersect

5) This number is the percentage below the z-score

6) Subtract the number found in step 5 from 1 to find the percentage above the z-score

7) To find the percentage between two z-scores, you need to find the percentage associated with both z-scores. Then subtract the larger percentage from the lower percentage.