How much do you study? A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 4. Use the TI-84 PLUS calculator to answer the following. Round the answer to at least four decimal places. (a) What proportion of students studied more than 35 hours? (b) What is the probability that a randomly selected student spent between 12 and 37 hours studying? (c) What proportion of students studied less than 37 hours?

a)

And we can find this probability using the complement rule:

TI84 procedure

2nd > DISTR> Vars> normalcdf

1-normalcdf(35;10000;25;4)

b) And we can find this probability with this difference:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(12;37;25;4)

c)

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(-10000;37;25;4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the hours spent studying the week before final exams of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

And the best way to solve this problem is using the normal standard distribution and the z score given by:

If we apply this formula to our probability we got this:

And we can find this probability using the complement rule:

TI84 procedure

2nd > DISTR> Vars> normalcdf

1-normalcdf(35;10000;25;4)

Part bAnd we can find this probability with this difference:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(12;37;25;4)

Part c

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(-10000;37;25;4)

1. sometimes

2.   never

3   .08c = 7.50+.05c   250 pages

step-by-step explanation:

1.   sometimes

the theoretical probability of rolling a 1 is 1/6.   perhaps you got really lucky and never rolled a 1   in those 60 times.   your experimental probability is 0.   or perhaps you rolled a 1   20 times.   your experimental probability is 20/60 is 1/3.     it really just depends on how you rolled in that experiment.

2.   never

the probability of the event occurring is 1.   that is being 100 % sure it will happen.   that is the maximum amount of confidence we have.

3.   no customer card cost:   .08 times number of pages

customer card cost:   7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250  

the cost is the same at 250 pages

a. (14,375+570)(5.3) b. 158,417$

step-by-step explanation:

a. you are basically just setting up the equation. f= amount of debt hence 14,375. g= how much she is borrowing from her parents weekly hence 570 and t= the interest on her debt. creating the equation.

b.   14,375+570= 14,945

      14,945 x 5.3 = 79,208.5

      79,208.5 x 2(years) = 158,417(answer)