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FINAL REVIEW
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 16 ounces
and a standard deviation of 0.2 ounces.
A. The company producing this
product can only sell jars that have
16 or more ounces of coffee. What
percentage of the jars, in the long
run, will not be sellable?
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
A. The company producing this product can only
sell jars that have 16 or more ounces of coffee.
What percentage of the jars, in the long run, will
not be sellable?
P(X < 16), where the random
variable X is the amount of coffee in
the jar in ounces.
P(X < 16) = P(Z <
16  17 )
.4
= P(Z < -2.5)
= 0.00621
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
B. If one were to randomly select 4 jars from the production
line, what is the probability of 2 or more jars being rejected?
Let the random variable X count the number
of jars out of 4 that will be rejected. From
the past problem we know that the
probability of finding one jar is 0.00621.
Note: if we assume independence then we
have a binomial situation.
So, I will calculate P(X  2) using a binomial function.
P(X  2) = 0.00029
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
C. If one were to randomly select 4 jars from the production
line, what is the probability that the average amount of coffee in
the four jars is 16.5 ounces or less?
Since we are considering the average amount
of coffee in the four jars, it is a sampling
distribution problem. We want P(X < 16.5).
So, I will calculate P(X < 16.5) = P(Z
16.5  17
< .4
4
) = P(Z < -2.5) =
0.00621.
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
D. If one were to randomly select 4 jars from the production
line, what is the probability that the amount of coffee in all four
jars is 17 ounces or more?
In this case I want all four jars to have more than
17 ounces. Let the random variable X be the
amount of coffee in a jar in ounces. The
P(X > 17) = 0.5 by looking at the graph. I want to
calculate
P(X> 17 and X > 17 and X > 17 and X > 17)
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
D. If one were to randomly select 4 jars from the production
line, what is the probability that the amount of coffee in all four
jars is 17 ounces or more?
P(X > 17) = 0.5 by looking at the graph. I want to
calculate
P(X> 17 and X > 17 and X > 17 and X > 17).
Assuming independence between the events, I can
rewrite the probability statement as
P(X>17)*P(X > 17)*P(X > 17)*P(X > 17) = (0.5)4 = 0.0625.
The amount of coffee (in ounces) filled in a jar by a machine
has a normal distribution with a mean amount of 17 ounces
and a standard deviation of 0.4 ounces.
E. If one were to randomly a jar from the production line, what
is the probability that the average amount of coffee in the jar is
more than 17.4 ounces or less than 16.6 ounces?
Notice we want to calculate
P(X< 16.6 ounces or X > 17.4 ounces). The
two events are disjoint. I also notice that
P(X< 16.6 ounces or X > 17.4 ounces) = 1 - P(16.6 < X < 17.4)
 1 - .68 = .32
During the last election about 110 million registered voters
voted for the presidency. Suppose that 45% of those
registered voters voted for George Bush. If we gathered a
simple random sample of 1000 voters what is the probability
that 400 or less voted for George Bush? Use continuity
correction to approximate the probability.
 = 1000(.45) = 450
=
Binomial parameters
1000(.45)(.55) 15.73
P(X < 400) where X is the binomial random variable.
P(X< 400)  P(Z < 400.5  450 ) = P(Z < -3.15)
15.73
= 0.0008
What is the purpose of creating a confidence interval?
To estimate the population mean with an interval.
A local bakery has determined a probability distribution for
the number of cheesecakes that they sell in a given day. The
distribution is as follows:
Number sold in one day
0
5
10
15
20
Probability of selling x
.05
.2
.30
.35
.1
Find the number of cheesecakes that this local bakery expects
to sell in a day on average in the long run.
 = 0(.05) + 5(.2) + 10(.30) + 15(.35) + 20(.1)
 = 11.25 cheesecakes per day on average.
Lets say that the probability of being born a girl is 0.55 and the
odds of being born a boy are 0.45. Let the random variable X
count the number of girls for a family of four children.
A. Write out the sample space for the random variable X.
X = {0, 1, 2, 3, 4}
B. Write the probability distribution for the random variable X
X
0
1
2
3
4
P(x)
.041
.200
.368
.299
.092
THE END ?