Download Lesson 8- normal distributions

Statistics and the TI-84
Lesson 8 - Continuous Probability Distributions
Normal Distributions
2nd DISTR 1 normalpdf (value x, mean , standard deviation  ) gives the
functional value (ordinate) of the normal curve at a given value of x.
The peak of the normal distribution curve occurs at the mean , that is, when x=.
Peak= normalpdf (, ,  )
1.
Exercise 1. Find the functional value (the ordinate) of the normal function with mean =30 and
standard deviation =2.5 at x=24. Find the peak value of the normal function.
 2nd DIST 1 normalpdf(24, 30, 2.5) ENTER answer: 0.0089578121
 2nd ENTRY (change 24 to 30) ENTER
answer: 0.1595769122
Note: when graphing, an appropriate window is needed. Go to WINDOW and use,
Xmin: use approximately 4 standard deviations below the mean
Xmax: use approximately 4 standard deviations above the mean
Ymin=use -.05
Ymax=use a number higher than the peak= normalpdf(mean, mean, standard
deviation)
Also, go to the gray y= and clear all previous plots.



To clear a graph: 2nd DRAW 1 ENTER ClrDraw ENTER Done
To store a graph: 2nd DRAW
(STO) 1 ENTER (number)
nd
To recall a graph: 2 DRAW
(STO) 2 ENTER
Exercise 2 . Draw the graphs of the standard normal distribution function and the normal
distribution function with =18 and =1.5.
 gray Y=
Y1=2nd DISTR 1 normalpdf (x, 0, 1) ENTER
 2nd DISTR 1
normalpdf(0, 0,1)
ENTER
answer: 0.3989 is the peak value (use
Ymax>0.3989) when selecting an appropriate window for the graph.
 WINDOW Xmin=-4 Xmax=4 Ymin=-0.05 Ymax=0.5 GRAPH
 gray Y= deselect Y1
 Y2= 2nd DISTR 1 normalpdf(x, 18, 1.5)
 2nd DISTR 1 normalpdf(18, 18,1.5) ENTER answer: 0 .26596 (use Ymax=0.4)
 WINDOW Xmin=12 Xmax=24 Ymin=-.05 Ymax=.4 GRAPH
2. 2nd DISTR 2 normalcdf(lower, upper, mean , Stdv. ) gives the area or probability
below the normal curve and between the lower and upper values.
Note: if the mean and standard deviations are not given their defaults values are
 = 0 and stdv = 1 (the standard normal or z-distribution)
Exercise 3. Find the area below the standard normal curve, between
z=-1.28 and z=2.05
 2nd DISTR 2 normalcdf(-1.28, 2.05) ENTER answer: 0 .8795452188
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Exercise 4. The monthly rental rates for apartments in a certain city area are normally
distributed with an average rent of $450 and a standard deviation of $50. What
percentage of those apartments exceed a rent of $375?
Hint: use 1E99 to indicate infinite as the upper bound.
 2nd DISTR 2 normalcdf (375, 1E99, 450, 50) ENTER
answer: 0.933192771 or approximately 93%
Exercise 5. In a binomial n=80 and p=.15, use the normal approximation to the binomial
to find the probability that the number of successes is at least 17 (use the ±0.5 continuity
correction factor)
Use =np and  = np(1  p) and x≥ 16.5

2nd DISTR
2
normalcdf(16.5,1E99, 80*.15, 80(.15)(.85) )
Answer: 0.079416946
ENTER
Exercise 6. In a binomial problem n=100, p=0.8, use the normal approximation to the
binomial to find P(x>75)
µ = np =100(.8) = 80 ,  = np(1  p) =4


2nd DISTR
ENTER
2
normalcdf(75.5, 1E99, 100*.8,
100(.8)(.2) )
answer: 0.8697054357
3. The sampling distribution of the mean:
x distribution

n
Use: 2nd DISTR 2 normalcdf(lower, upper, ,
) ENTER
Note: we are assuming that the population size N is large compared to the sample size
n. Otherwise use the population size correction factor and obtain

n
N n
N 1
Exercise 7. The average amount of time adults spend watching TV has a mean of 6
hours and standard deviation of 2 hours a day. A random sample of 81 adults is taken.
what is the probability that their mean watching time is more than 6.5 hours? Use the
Central Limit Theorem.

2nd DISTR
2 normalcdf (6.5, 1E99, 6,
2
))
64
Answer: 0.0122244334
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ENTER
Using Shade Norm
Exercise 8. In a binomial problem n=100, p=0.8, use the normal approximation to the
binomial to find P(x>75)
µ = np =100(.8) = 80 ,  = np(1  p) =4
 WINDOW
 Xmin=60 Xmax=100
 Ymin=-.05 Ymax= 2nd DISTR 1 normalpdf(80, 80, 4)

2nd DISTR
DRAW 1 ShadeNorm(75.5, 1E99, 80, 4) ENTER
Answer: 0.869705, see graph
Exercise 9. Use Shade Norm to find the probability that X<35.2 in a normal
distribution with mean equal to 28 and standard deviation equal to 3.2.
 WINDOW Xmin=15
Xmax=41
Ymin=-.05
Ymax= 2nd DIST 1 Normalpdf(28, 28, 3.2) = .12466946
 2nd DISTR
DRAW 1 ShadeNorm(-1E99, 35.2, 28, 3.2)
ENTER
Answer: 0.987776
Using the Inverse Normal:
InvNorm calculates the x value associated with the area to the left of the
given x value.
 2nd DISTR 3 invNorm(probability to the left of desired value, mean , stdv)
Exercise 10. The grades for a large statistics class are normally distributed with mean
of 68 and a standard deviation of 5. If the top 8% of the class will receive A's in the
exam, what is the lowest A grade.
Note: the area of the left I 1-.08=.92
 2nd DISTR 3 invNorm(.92, 68, 5) ENTER
Answer: 75.0253578
Exercise 11. Miss Morgan, who is a secretary, drives everyday to work. The time
required for the trip from her apartment to the office can be approximated by a normal
distribution with mean of 20 minutes and standard deviation of 5 minutes. At what time
should Miss Morgan leave her apartment to give herself a 0.98 chance that she will be at
her office by 9 a.m.?

2nd DISTR 3 invNorm(.98, 20, 5) ENTER
Answer: 30.26874455 min.
Answer: she should leave by 8:30 AM to give herself 30 minutes.
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Exercise 12. A normally distributed random variable x possesses a mean of 14
and a standard deviation of 4. Find the probability that
a) x falls between 9 and 23.
 WINDOW Xmin=-2 Xmax=30 Ymin=-0.05 Ymax=0.1
 2nd DISTR 3
DRAW 1 ShadeNorm(9, 23, 14, 4)
ENTER
answer: 0.882126
b) Find the 90th. percentile of the distribution
 2nd DISTR 3 invNorm(0.90, 14, 4) ENTER Answer: 19.12620627

c) Find the value of the distribution below which we find 5% of the
values.
nd
2 DISTR 3 invNorm (0.05, 14, 4) ENTER Answer: 7.420585496
d) If a sample of size n=4 is taken from this population, what is the
probability that the sample mean will be larger than 11?

4
) ENTER
4
2nd DISTR 2 normalcdf (11, 1E99, 14,
Answer: .9331927713
Exercise 13 . In a normal population has mean 65 and standard deviation 4. If sample of size
n=4 is taken from the population,
a) what is the probability that the mean of the sample is less than 68?.

2nd DISTR 2 normalcdf (-1E99, 68, 65,
4
)
4
ENTER
Answer: 0.933249938
b) Find the third quartile of the distribution of sample means.

2nd DISTR 3 invNorm(.75, 64, 2)
ENTER
Answer:
Q3  65.3489795
Exercise 14. Café "El Delicioso" claims 14 ounces of coffee in each package, with a
standard deviation of 0.3 ounces. A government inspector will reject the claim and will
have the company fix the settings if a sample of 25 packages shows an average below
13.85 ounces. If the average is really 14 ounces, what is the probability that the
inspector will reject the company's claim?
 2nd DISTR 2 normalcdf(-E99, 13.85, 14, .06 ) ENTER answer: 0.0062096799
Exercise 15. In problem 14, if the average is really 13.92 ounces of coffee in each
package, what is the probability that the inspector will not reject the company's claim?
 2nd DISTR 2 normalcdf(13.85,EE99, 13.92, .06 ) ENTER answer: 0.8783274392
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