Question

The measure of test anxiety has a mean of 16 and a standard deviation of 4. Using a normal curve table, what percentage of students have scores:

a. Above 15

b. Above 17

c. Above 13

d.Below 18

e. Below 14

f. Below 16.25

Answer 1

We know that, in Excel

P(X<x) = NORMDIST(x,mean,standard_dev,cumulative)

a. Above 15 = P(X>x) =1-P(X<x)=1-NORMDIST(15,16,4,TRUE) = 1-0.4012 = 0.5987 = 59.87%

b. Above 15 = P(X>x) =1-P(X<x)=1-NORMDIST(17,16,4,TRUE) = 1-0.5987 = 0.4012 = 40.12 %

c. Above 13 = P(X>x) =1-P(X<x)=1-NORMDIST(13,16,4,TRUE) = 1-0.2266 = 0.773 = 77.3%

d.Below 18 =P(X<x) =NORMDIST(18,16,4,TRUE) = 0.6914 = 69.14 %

e. Below 14 = P(X<x) =NORMDIST(14,16,4,TRUE) =  0.3085 = 30.85%

f. Below 16.25 =P(X<x) = =NORMDIST(16.25,16,4,TRUE) = 0.5249 = 52.49%