MATH 1530 – Quiz # 11
(Quizpak 5)
Name _____________________________
Assume that the readings on thermometers are normally distributed with a mean of 0 C and a standard deviation of 1.00 C . One
thermometer is randomly selected and tested. In each case, label and shade the graph , then find the probability of getting the stated
readings in degrees Celsius. Show the areas to be added or subtracted as read from Table A-2 as well as the solution.
1. Between 1.22 and 2.12
0
z
P(1.22  z  2.12 ) = ___________________________ = ______________
2. Between 2.1 and 2.8
0
z
P(2.1  z  2.8 ) = ___________________________ = ______________
3. Less than 1.34
0
z
P( z  1.34 ) = ___________________________ = ______________
4. Greater than 1.56
0
z
P( z  1.56 ) = ___________________________ = ______________
MATH 1530 – Quiz # 12
(Quizpak 5)
Name _____________________________
Find the missing “z-value” given each of the following probabilities come from a Standard Normal
Distribution. In each case, label and shade the graph writing the appropriate probabilities over the
shaded region, then state the z-score that provides those probabilities as your solution.
If P(0  z  a)  .4370 , then a = _____________ .
1.
0
z
If P(b  z  0)  .3770 , then b = _____________ .
2.
0
z
If P(c  z  c)  .5222 , then c = _____________
3.
and c = _____________ .
0
z
If P( z  d )  .0287 , then d = _____________ .
4.
0
z
MATH 1530 – Quiz # 13
(Quizpak 5)
Name _____________________________
Let x be a randomly selected score from a normally distributed population with mean   100 and standard deviation   10 ,
then find each of the following probabilities. Label and shade the graphs accordingly. Show the areas to be added or subtracted as read
from Table A-2 as well as the solution.
1.
100
x
0
z
P(100  x  115) = P( ________  z  ________ ) = ___________________ = __________
2.
100
x
0
z
P( x  115) = P( z  ________ ) = ________________________ = ______________
3.
100
x
0
z
P( x  108) = P( z  ________ ) = ________________________ = ______________
4.
100
x
0
z
P(88  x  99) = P( ________  z  ________ ) = _____________________ = __________
MATH 1530 – Quiz # 14
(Quizpak 5)
Name _____________________________
Given a normally distributed population with mean   100 and standard deviation   20 , find each of the
following scores. Label the graphs accordingly. Show formulas and calculations below the graphs.
1. The score that separates the top 40% from the bottom 60%.
0
z
100
2. The score that separates the bottom 10% from the top 90%.
0
z
100
3. The score that separates the top 10% from the bottom 90%.
0
z
100
4. The score that determines the cut-off for the top 25%.
0
z
100
______________
x
______________
x
______________
x
______________
x
MATH 1530 – Quiz # 15
(Quizpak 5)
Name _____________________________
Assume that women’s heights are normally distributed with a mean of   63.6 inches and a standard
deviation   2.5 inches (based on data from the National Health Survey).
1. If 1 woman is selected at random, find the probability that her height is above 63
inches.
P(
) = P(
) = ________________________ = _______________
2. If 100 women are selected at random, find the probability that their mean height is
greater than 63 inches.
P(
) = P(
) = ________________________ = _______________
3. Consider the sampling distribution of randomly selecting 100 women at a time and
recording the mean heights, then this distribution of sample means has a mean and a
standard deviation that are equal to:
(give the numerical values here)
 x = _______________
 x = _______________
MTH 1050 – STATDISK WORKSHEET - CHAPTER 5
5-1
Name _______________________
a. Use STATDISK to generate 106 values from a normally distributed population with a mean of 98.6 and a
standard deviation of 0.62. Use Data/Descriptive Statistics to find the mean of the generated sample.
b.
Record the 10 sample means here: (Use decimals = 1 and seed = 1, 2, … , 10)
_____
c.
_____
_____
_____
_____
_____
_____
_____
_____
_____
After examining those 10 sample means, what do you conclude about the likelihood of getting a sample mean
of 98.20? Explain.
_______________________________________________________________________________
_______________________________________________________________________________
d.
Given that researchers did obtain a sample of 106 temperatures with a mean of 98.20 F, what does their
result suggest about the common belief that the population mean is 98.6 F?
_______________________________________________________________________________
5-2
a. Sample mean: __________
b.
Record the 10 sample means here. (Use decimals = 2 and seed = 1, 2, … , 10)
_____
5-3
Standard deviation: __________
_____
_____
_____
_____
_____
_____
_____
_____
_____
a. P(
) = _____________________________ = _____________________
b. P(
) = _____________________________ = _____________________
c. P(
) = _____________________________ = _____________________
d. P(
) = _____________________________ = _____________________
e. P(
) = _____________________________ = _____________________
5-5
a. One Die:
Mean:
__________
(seed = 5)
Standard Deviation:
__________
Distribution shape:
_______________________________
Sketch histogram here.
b. Two Dice:
Mean:
__________
(seed = 5)
Standard Deviation:
__________
Distribution shape:
_______________________________
Sketch histogram here.
c. 10 Dice:
Mean:
__________
(seed = 5)
Standard Deviation:
__________
Distribution shape:
_______________________________
Sketch histogram here.
d. 20 Dice:
Mean:
__________
(seed = 5)
Standard Deviation:
__________
Distribution shape:
_______________________________
Sketch histogram here.
e.
General conclusions:
What happens to the mean as the sample size increases from 1to 2 to 10 to 20?
_______________________________________________________________________________
What happens to the standard deviation as the sample size increases?
_______________________________________________________________________________
What happens to the distribution shape as the sample size increases?
_______________________________________________________________________________
How do these results illustrate the central limit theorem?
_______________________________________________________________________________
_______________________________________________________________________________
The following notes will be provided for your reference as the last page of Exam 3:
The Standard Normal Distribution is a normal probability distribution that has a mean
of   0 and a standard deviation of   1 . (Utilize Table A-2 to find probabilities for
given z-scores or to find z-scores for given probabilities.)
Formula for converting x-scores to z-scores:
z
x

Formula for converting z-scores to x-scores:
x    ( z  )
The Central Limit Thereom:
For large sample sizes (n>30) drawn from ANY distribution (or for smaller sample
sizes, if the original distribution is normally distributed), the sampling distribution of the
means has the following properties:
1. The distribution of the sample means is approximately Normal.
2. The mean of the sampling distribution is equal to the mean of the population.
x  
3. The standard deviation of the sampling distribution is equal to the standard
deviation of the population divide by the square root of n.
x 

n