- AnimalLover8

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?

- katieb

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- AnimalLover8

82%
2%
34%
16%

- jim_thompson5910

Do you have a TI calculator?

- AnimalLover8

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- jim_thompson5910

ok we'll use a table then

- jim_thompson5910

first convert the raw score x = 95 to a z-score
use the formula
z = (x-mu)/sigma

- jim_thompson5910

in this case, mu = 80 and sigma = 15

- AnimalLover8

What is the "raw score"?

- jim_thompson5910

x = 95 is the raw score

- jim_thompson5910

it's the score on the test
the z-score is the transformed score to the standard normal distribution

- jim_thompson5910

z = (x-mu)/sigma
z = (95-80)/15
z = ???

- AnimalLover8

So z=...1?

- jim_thompson5910

yes z = 1

- jim_thompson5910

ie, z = 1.00

- jim_thompson5910

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)

- AnimalLover8

.84134

- AnimalLover8

I don't understand what this number actually is.

- jim_thompson5910

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

- jim_thompson5910

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

- AnimalLover8

0.269... ?

- jim_thompson5910

compute 1 - 0.84134

- AnimalLover8

Wait so how do you get the probability from there?

- jim_thompson5910

1 - 0.84134 = 0.15866

- jim_thompson5910

0.15866 = 15.866% which rounds to 16%

- jim_thompson5910

we have some normal curve
|dw:1440808310147:dw|

- AnimalLover8

OOOOOOH!

- jim_thompson5910

|dw:1440808329487:dw|

- jim_thompson5910

|dw:1440808367344:dw|

- AnimalLover8

Yeah. :) Thank you so much!

- jim_thompson5910

I'm glad that everything is making sense now

- AnimalLover8

Me too, I usually pick things up very quickly, but I just couldn't hack this question. :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.