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Key
1) Matching - Vocabulary
7___a. Normal distribution
8___b. Uniform distribution
4___c. Density curve
3___d. Standard normal dist.
2___e. Sampling distribution
of sample means
1___f. Central limit theorem
6___g. Standard error of the
mean
5___h. Continuity correction
1. As sample size  , dist. of sample means
approaches a normal distribution
2. Obtained when repeatedly draw samples
from same population
3.  = 0,  = 1
4. Total area under curve = 1, vertical height  0
5. Represents x by an interval from x -.5 to
x+.5
6.  X
7. Distribution is symmetric, bell-shaped
8. Every random variable is equally likely
For #’s 2, 3, 4: Assume that the readings on a thermometer are normally distributed with a mean
of 0°C and a standard deviation of 1.00°. A thermometer is randomly selected and tested. Draw a
sketch and find the probability of each reading in degrees.
USE TABLE A-2 since you are given that the mean is 0 and the standard deviation is 1 (scores
are already standardized!).
2) Between 0 and 2.00 (= Area between 0 and 2.00)
0.4772
3) Between -2.0 and 0 (=Area between -2.0 and 0)
0.4772
4) Between -1.73 and .48 (Area between -1.73 and .48)
0.6426
For #’s 5,6: Find the indicated probability where z is the reading in degrees:
5) P (z > 1.8)
1-Area below 1.8 = .0359
6) P (-1.5 < z < 2.8)
Subtract the area found by -1.5 from the area found by 2.8 = .9306
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For #’s 7, 8, 9: Find the indicated percentile. = Go backwards! The percentile gives you the area
below the curve (for these, you are trying to find z). Locate the area in the body of the table, then
the corresponding z-score.
7) Find P 40 (the 40th percentile)
Z = -0.25
8) Find D 20 (the 20th percentile)
Z = -0.84
9) Find Q 2 (the 50th percentile)
Z=0
10) A bank’s loan officer rates applicants for credit. The ratings are normally distributed with a
mean of 200 and a standard deviation of 50. Find the z-score using the area corresponding
to the information given, then use the formula z 
x

to solve for x.
(a)
Find D6 (the sixth decile)
Look up the z-score corresponding with area = .60; this gives you z = .25. Plug this into the
formula along with the mean (200) and standard deviation (50). Answer: 212.5
(b)
If 40 different applicants are randomly selected, find the probability
that their mean is above 215.
This is a problem that uses the central limit theorem. N=40, x=215,  X    200 and
X 

n

50
 7.9 . Want: P( X >215) = P(Z>(215-200)/7.9 = 1.9 = .0287 (don’t forget to
40
find the area for 1.9 then subtract this area from 1 since the table always gives you the area to the
left).
11) In one region, the September energy consumption levels for single-family homes are found
to be normally distributed with a mean of 1050 kwh and a standard deviation of 218 kwh.
(a) For a randomly selected home, find the probability that the September energy
consumption level is between 1050 kwh and 1250 kwh.
a. .3212
Find P(1050<x<1250); convert both scores into z-scores using the formula z 
x

:
P(0<z<.92), then subtract the areas found when you look up .92 and 0. Answer: .3212
(b) For a randomly selected home, find the probability that the September energy
consumption level is below 1200 kwh.
b. .7549
Find P(x<1200); convert 1200 into a z-score using the formula z 
x

; this gives
you P(z<.69). Look up .69; this gives you the answer of .7549.
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(c) For a randomly selected home, find the probability that the September energy
consumption level is above 1175 kwh.
c. .2843
Find P(x>1175); convert 1175 into a z-score using the formula z 
x

; this gives
you P(z>.57). Look up .57; this gives you the answer of .2843.
(d) Find P45 (the 45th percentile).
d. 1021.7
Look for the closest number to .4500 in the body of table A; the corresponding z-score is
-.13. Plug this number and the mean of 1050 and standard deviation of 218 into the
formula z 
x

to find the score (x). When you solve, you should get 1021.7.
(e) If 50 different homes are randomly selected, find the probability that their mean
energy consumption level for September is greater than 1075.
e. .2090
This is another problem that uses the central limit theorem. N=50, x=1075,  X    1050 and
X 

n

218
 30.8299 . Want: P( X >1075) = P(Z>(1075-1050)/30.8299 = .81. Look up
50
.81, then subtract this area from 1 which will give you .2090 (don’t forget the table always gives
you the area to the left).
For #12-13 only : Estimate the indicated probability by using the normal distribution as an
approximation to the binomial distribution.
12) In a study of the causes of death, it was found that 52% of all Americans die from heart
disease. Find the probability that in a group of 500 randomly selected Americans, at least
275 die from heart disease.
First check 2NIP: 2 = die of heart disease/don’t die of heart disease; N = 500 (set number of
trials); I – Independent (each person), P – set probability = .52. Qualifies as binomial
Second, check np  5 and nq  5: (500)(.52)  5 YES; (500)(.48)  5 YES, qualifies – we can use
the normal approximation to calculate the probability of this binomial problem.
Find the mean:   np  260 and find the standard deviation:   npq  11.17
Since we want to find P(X  275), we want to capture 275, so we will use the continuity
correction of 274.5.
P(X  274.5) = P(Z  (274.5-260)/11.17) = P(Z  1.30) = .0968.
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13) The Worthington Pottery Company manufactures beer mugs in batches of 120 and the
overall rate of defects is 5%. Find the probability of having more than 6 defects in a batch.
First check 2NIP: 2 = defect or not defect; N = 120 (set number of trials); I – Independent (each
mug is independent), P – set probability = .05. Qualifies as binomial
Second, check np  5 and nq  5: (120)(.05)  5 YES; (120)(.95)  5 YES, qualifies – we can use
the normal approximation to calculate the probability of this binomial problem.
Find the mean:   np  6 and find the standard deviation:   npq  2.387
Since we want to find P(X>6), we don’t want to capture 6, so we will use the continuity
correction of 6.5.
P(X  6.5) = P(Z  (6.5-6)/2.387) = P(Z  .10) = .4168.
14) Replacement times for CD players are normally distributed with a mean of 7.1 years and a
standard deviation of 1.4 years. Find the replacement time separating the top 45% from the
bottom 55%.
_7.282________
The z-score corresponding to the area of .55 is .13. Plug this into the formula z 
x

along
with the given mean and standard deviation. This gives you an x of 7.282.
15) A genetics experiment involves a population of fruit flies consisting of 1 male named Mike
and 3 females named Anna, Barbara, and Chris. Assume that two fruit flies are randomly
selected with replacement.
For the original population, p = ¾ = 0.75.
a) List the 16 possible samples, find the proportion of females in each sample, then
use a table to describe the sampling distribution of the proportions of females
(Hint: See Table 6-3 in your book).
The 16 equally likely samples and their proportions are given below. Each sample has a
probability of 1/16. The sampling distribution of the proportions, a list of each possible sample
proportion and its total probability, is given below as well. The distribution appears to “bunch up”
toward the upper end.
Sample
MM
MA
MB
MC
AM
AA
AB
p̂
0.0
0.5
0.5
0.5
0.5
1.0
1.0
condensed
p̂
0.0
0.5
1.0
P( p̂ )
1/16
6/16
9/16
16/16
p̂ *P( p̂ )
0/16
3/16
9/16
12/16
4
AC
BM
BA
BB
BC
CM
CA
CB
CC
1.0
0.5
1.0
1.0
1.0
0.5
1.0
1.0
1.0
12.0
b) Find the mean of the sampling distribution.
 pˆ  [ pˆ * P( pˆ )]  12 /16  0.75
c) Is the mean of the sampling distribution [from part (b)] equal to the population
proportion of females? Does the mean of the sampling distribution of proportions
always equal the population proportion?
Yes. The sample proportion is always an unbiased estimator of the population
proportion.
16) This data set includes heights (in inches) of randomly selected women:
66.2, 76.6, 68.7, 70.8, 717
Construct a normal probability plot for these five values and determine whether they appear to
come from a population that is normally distributed and explain why the sample does/does not
appear to be normal (remember: the population distribution need not be exactly normal, but it
must be a distribution that is basically symmetric with only one mode).
717 is a major outlier. When you graph this data set on your calculator, the normal probability
plot does not result in points that reasonably approximate a straight-line pattern – clearly not a
normal distribution.
17) What kind of distribution emerges when you graph the outcomes for rolling a die? (Hint:
What are the outcomes for rolling a die? What are the probabilities for each roll of the die?)
Draw the graph, then determine the following probabilities:
This is a uniform distribution – it looks like a rectangle with the numbers 1, 2, 3, 4, 5, 6
(outcomes) labeled on the x-axis and a height of 1/6 labeled on the y-axis (this is the probability
of each outcome/roll of the die – each roll has an outcome of 1/6).
a) Rolling higher than a five = length of rectangle : 6-5 = 1; height of rectangle: 1/6;
Multiply these numbers together and P(X>5) = 1.
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b) The probability of rolling between a two and a four (inclusive)
Length: Distance between 2 and 4: 2
Height: of rectangle is 1/6
Multiply these two numbers together and P(2  X  4) = 1/3
c) Rolling less than or equal to a three
Length: Distance from origin to 3 is: 3
Height: of rectangle is 1/6
Multiply these two numbers together to get P(X  3): 1/2
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