# Stat Asst 2

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8 / 8 points
When the probability of event B is affected by the occurrence
of event A, the events are not independent. Let P(B | A)
denote the probability of B given the condition that A has
occurred. This is called a conditional probability.
Type | by holding down Shift and type \
For independent events A and B, P(B | A) = P(B), and P(A |
B) = P(A)
For dependent events A and B
o P(B | A) ≠ P(B). The occurrence of A has changed
the probability of B.
o P(A | B) ≠ P(A). The occurrence of B has changed
the probability of A.
For dependent events, P(A and B) = P(A) x P(B | A) = P(B) x
P(A | B). This is the General Multiplication Rule.
Assume the following joint and marginal probabilities:
When we know the condition that some event has occurred,
the table reduces to a row or column matching the condition.
For example, when we know that the party is Democrat, the
table reduces to the Democrat column:
P(Yes | Democrat) is the probability of event Yes given the
condition that the event Democrat has occurred. In condition
Democrat, Yes occurs at a rate of 0.15 in 0.40. So P(Yes |
Democrat) = 0.15/0.40 = 0.375.
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Supplemental
Course Content Session 3 Lecture beginning p. 13
Video
P(Male | Republican) is a ________ probability.

Question options:
Conditional
Given the following partial relative frequency table
Republica
n
Democra
t
Independen
t
Male 0.144 0.184 0.129
Femal
e
0.108 0.185 ?
decimal places.
0.501
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P(Female | Democrat) = P(Female and Democrat) ÷
P(Democrat)
8 / 8 points
Given the following partial relative frequency table
Republica Democra Independen
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n t t
Male 0.145 0.161 0.169
Femal
e
0.121 0.100 ?
decimal places.
0.305
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P(Republican | Male) = P(Republican and Male) ÷
P(Male)
12 / 12 points
Assume breast cancer affects 0.001 of the female population
between 45 and 55 years of age.
There are two kinds of positive test results:
True positive (the test indicates you have a disease, and
you actually have it)
False positive (the test indicates you have a disease, but
you actually do not).
Assume mammograms are
0.90 accurate detecting people who actually have breast
cancer (true positive rate)
0.91 accurate for people who do not have breast cancer
(true negative rate).
Compute the probability that a female between the ages of 45
and 55 who tests positive for breast cancer has breast cancer,
0.009
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https://betterexplained.com/articles/an-intuitive-and-short-explanation-of-bayes-theorem/
Probability = desired event / all possibilities
In this case the desired event is a true positive test.
All possibilities = true positive test + false positive test
Probability of a true cancer positive test = probability of having cancer * the probability that the
test found the cancer.
The false positive = (1 – probability of having cancer) * the probability that the test found cancer
(1 – true negative rate.)
P(Cancer)=0.001
P(No cancer)= 1-P(cancer)
=1-0.001
=0.999
P(Test Positive |no cancer)= 1-P(Negative | no cancer)
= 1-0.91
=0.09
[0.001*.90] / [(0.001*0.90) + (0.999*0.09)]
0.0009 / (0.0009) + (0.08991)
0.0009 / 0.9891
= 0.009
8 / 8 points
Misa earned a score of 95 on her latest exam.The professor
determined that all of the scores from this exam had a normal
distribution with a mean of 87 and a standard deviation of 3.
What is Misa’s Z score on her exam?
Round to three decimal digits.
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2.667
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Z = (Value – Mean) / standard deviation
8 / 8 points
Blue Crab lengths have a normal distribution with a mean of 5
inches and a standard deviation of 2 inches.
What is the Z value of 5 inches?
Round to three decimal digits.
0
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Z = (Value – Mean) / Standard deviation
8 / 8 points
In every normal distribution
Question options:
The mean is larger than the standard deviation
The mean and the median are equal
The interquartile range covers 68% of the values
The mean = 0 and the standard deviation = 1
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In a normal distribution the probability of a value larger than
one standard deviation above the mean is:
Question options:
There is not enough information to answer the question
In a normal distribution the
probability of a value between 2 and
3 standard deviations below the
mean is:
Question options:
.15%
-2.35%
-.15%
2.35%
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8 / 8 points
The Standard Normal Distribution is a Normal Distribution with
the following characteristics:
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Question options:
The mean is 1 and the standard deviation is 0
The mean is 0 and the standard deviation is 1
The mean and the standard deviation are 0
The mean and the standard deviation are 1
Choose all of the correct answers for the characteristics of a
Binomial Probability Distribution.
Question options:
The variables are discrete.
An example of a binomial trial is rolling a die.
There can be three or more outcomes of each binomial trial.
Each binomial trial has only 2 possibilities.
The binomial trials are independent of each other.
An example of a binomial trial is flipping coins.
The variables are continuous.
The binomial trials are dependent on each other.
Select all correct answers for the Poisson probability
distribution.
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Question options:
The Poisson distribution’s mean=median=mode.
The Poisson distribution describes a situation where the probability of success does not remain the same from trial to trial.
The Poisson distribution describes the number of events occurring in a unit of space or time.
The Poisson distribution is always positively skewed.
An example of a Poisson trial is selecting the members of a committee from one business.
An example of a Poisson trial is counting the number of misspelled words in a magazine article.
The variables in a Poisson trial are counted.
The variables in a Poisson trial are measured.
100 / 100 – 100 %
100 / 100 – 100
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