Question

The time required to finish a test in normally distributed with a mean of 80 minutes and a standard
deviation of 15 minutes. What is the probability that a student chosen at random will finish the test in more
than 95 minutes?

Answer 1

82%

2%

34%

16%

Answer 2

Do you have a TI calculator?

Answer 3

No, if I did I would be able to do many more questions.

Answer 4

ok we'll use a table then

Answer 5

first convert the raw score x = 95 to a z-score

use the formula

z = (x-mu)/sigma

Answer 6

in this case, mu = 80 and sigma = 15

Answer 7

What is the "raw score"?

Answer 8

x = 95 is the raw score

Answer 9

it's the score on the test

the z-score is the transformed score to the standard normal distribution

Answer 10

z = (x-mu)/sigma

z = (95-80)/15

z = ???

Answer 11

So z=...1?

Answer 12

yes z = 1

Answer 13

ie, z = 1.00

Answer 14

Now use a table like this one

https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

locate the row that starts with `1.0`
and find the column that has `.00` at the top
the intersection of this row and column has the number ______ (fill in the blank)

Answer 15

.84134

Answer 16

I don't understand what this number actually is.

Answer 17

this number represents the probability of getting a z-score less than 1.00

in notation form, we say `P(Z < 1.00) = 0.84134`

so there's approx a 84.134 % of getting a z-score less than 1.00

in other words, there is a 84.134 % chance of getting a test score less than 95

Answer 18

subtract 0.84134 from 1 to get the probability of getting a score larger than z = 1 or x = 95

Answer 19

0.269... ?

Answer 20

compute 1 - 0.84134

Answer 21

Wait so how do you get the probability from there?

Answer 22

1 - 0.84134 = 0.15866

Answer 23

0.15866 = 15.866% which rounds to 16%

Answer 24

we have some normal curve

Answer 25

OOOOOOH!