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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?

Respuesta :

Answer:

a)[tex]P(X>35)=P(\frac{X-\mu}{\sigma}>\frac{35-\mu}{\sigma})=P(Z>\frac{35-25}{4})=P(Z>2.5)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z>2.5)=1-P(Z<2.5) =1-0.994=0.006 [/tex]

TI84 procedure

2nd > DISTR> Vars> normalcdf

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

b) [tex]P(12<X<37)=P(\frac{12-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{37-\mu}{\sigma})=P(\frac{12-25}{4}<Z<\frac{37-25}{4})=P(-3.25<z<3)[/tex]And we can find this probability with this difference:

[tex]P(-3.25<z<3)=P(z<3)-P(z<-3.25)[/tex]

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

[tex]P(-3.25<z<3)=P(z<3)-P(-3.25)=0.999-0.0006=0.998[/tex]

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(12;37;25;4)

c) [tex]P(X<37)=P(\frac{X-\mu}{\sigma}<\frac{37-\mu}{\sigma})=P(Z<\frac{37-25}{4})=P(Z<3)[/tex]

[tex]P(Z<3) =0.99865[/tex]

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:

[tex]X \sim N(25,4)[/tex]  

Where [tex]\mu=25[/tex] and [tex]\sigma=4[/tex]

We are interested on this probability

[tex]P(X>35)[/tex]

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

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>35)=P(\frac{X-\mu}{\sigma}>\frac{35-\mu}{\sigma})=P(Z>\frac{35-25}{4})=P(Z>2.5)[/tex]

And we can find this probability using the complement rule:

[tex]P(Z>2.5)=1-P(Z<2.5) =1-0.994=0.006 [/tex]

TI84 procedure

2nd > DISTR> Vars> normalcdf

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

Part b[tex]P(12<X<37)=P(\frac{12-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{37-\mu}{\sigma})=P(\frac{12-25}{4}<Z<\frac{37-25}{4})=P(-3.25<z<3)[/tex]And we can find this probability with this difference:

[tex]P(-3.25<z<3)=P(z<3)-P(z<-3.25)[/tex]

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

[tex]P(-3.25<z<3)=P(z<3)-P(-3.25)=0.999-0.0006=0.998[/tex]

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(12;37;25;4)

Part c

[tex]P(X<37)=P(\frac{X-\mu}{\sigma}<\frac{37-\mu}{\sigma})=P(Z<\frac{37-25}{4})=P(Z<3)[/tex]

[tex]P(Z<3) =0.99865[/tex]

TI84 procedure

2nd > DISTR> Vars> normalcdf

normalcdf(-10000;37;25;4)