Transcript Test Review PPT

AP Stats Review:
Probability Unit
•Unit #2 – Chapters 6, 7, and Section 8.1
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The Normal Distribution
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Symmetric, mound-shaped
Mean = median
Area under the curve = 100%
68-95-99.7% rule - also called the Empirical
Rule
• Crazy formula – no need to know it!
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The 68-95-99.7% Rule
• The following shows what the 68-95-99.7 Rule
tells us:
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For example:
1998 #4 (partial): A company is considering implementing one of
two quality control plans – plan P and plan Q – for monitoring
the weights of car batteries that it manufactures. If the
manufacturing process is working properly, the battery
weights are approximately Normally distributed with mean =
2.7 lbs and standard deviation = 0.1 lbs.
(1) Quality Control Plan P calls for rejecting a battery as defective
if its weight falls more than 1 standard deviation below the
mean. If a battery is selected at random, what is the
probability that it will be rejected by plan P?
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Solution to 1998 #4 (a):
This is the area
we’re interested
in!
So, what’s that area? It’s ½ of
the remaining 32%, so it’s 16%!
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Finding Normal Probabilities
• When a data value doesn’t fall exactly 1, 2, or
3 standard deviations from the mean, we can
use a Z Table to find the area to the left of the
value.
• The calculator can also be
used to find areas under a
normal curve – it’s easier this way!
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Finding Normal Probabilities (cont.)
• Table Z is the standard Normal table. We have to
convert our values to z-scores before using the table.
z
x

• The figure below shows how to find the area to the left
when we have a z-score of 1.80:
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Calculator Use
• Normalcdf – gives area under a normal curve. You can
use z-scores or raw data values:
normalcdf (lower, upper, mean, st dev)
• invNorm – gives z-score or raw data value associated
with a percentile:
invNorm(area to left as decimal, mean, st dev)
• Provide complete communication on Free Response questions –
use drawings and/or identify the values in your calculator
syntax. Don’t just write normalcdf(-1.52,1.28)=0.835 and that’s
all; draw a graph, shade in an area, and explain the numbers.
• Also, make sure that all explanation are in the context of the
problem. For example, if the problem is about shirt sizes, then
your answers should be about shirt sizes.
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For example, 2003 #3:
#2: Men’s shirt sizes are determined by their neck
sizes. Suppose that men’s neck sizes are
approximately normally distributed with mean 15.7”
and standard deviation 0.7”. A retailer sells men’s
shirts in sizes S, M, L, and XL, where the shirt sizes
are defined in the table below:
Shirt Size Neck Size
S
14”-15”
(b) Using a sketch of a normal
M
15”-16”
curve, illustrate and calculate
L
16”-17”
the proportion of men whose
XL
17”-18”
shirt size is Medium.
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Solution to 2003, #3 (b) (shirt sizes):
Normalcdf
(15, 16, 15.7, 0.7)
= 0.5072
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Question: When do we multiply?
• Answer: when we want to find the probability
of a series of consecutive events
• Example #3: We just found that 0.5072 of all
men wear size Medium shirts. Suppose we
select 3 men at random from a huge
population. What is the probability that all
three of them wear Medium shirts?
Answer:
(0.5072)(0.5072)(0.5072)  0.13
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What if we are counting how many
successes will occur in repeated
trials?
•The BINOMIAL
Distribution!
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#4: What does it take to be
Binomial?
(i) Fixed number of trials (“n”)
(ii) Fixed probability of success on each trial
(“p”)
(iii) Two outcomes per trials (success or
failure)
(iv) Trials are independent of each other
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Binomial on the Calculator
• Binomialpdf(n,p,x-value) > only the
probability of that specific x-value
• Binomialcdf (n,p,x-value) > the sum of the
probabilities of x=0 up through and including
that specific x-value
Example #5: We found that 50.72% of all shirts are mediums. If
we select 12 shirts at random, what is the probability that…
(i) Exactly 4 of them are Mediums?
(ii) No more than 4 of them are Mediums?
(iii) At least 4 of them are Mediums?
(i) 0.114
(ii) 0.180
(iii) 0.934
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Parameters for a BinomiallyDistributed Variable X
• “n” and “p” need to be known
> “n” is the number of trials
> “p” is the probability of success on any
single trial
***   E ( X )  np
  np(1  p)
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Question: When do we add?
• Answer: when the question can be construed as an
“OR” situation
• Example #6: We found that 0.5072 of all men wear
size Medium shirts. Out of a random sample of 12
men, what is the probability that either 3 or 4
(exactly) are Mediums?
• Pr(x=3 or x=4) = Pr(x=3) + Pr(x=4)
Do each of these as a binomialpdf
= 0.049 + 0.1139 = 0.1629
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When ADDING, be careful about
over-counting!
Chicks
Hawks
Yellow
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5
Brown
10
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• Example #7: Using the chart above, if I select one
bird at random, what is the probability that the bird
I select is either a chick or is yellow?
• Pr(chick or yellow) = Pr(chick) + Pr(yellow)
minus Pr(both chick & yellow)
= 18/25 + 13/25 – 8/25
= 23/25
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Example #8:
• Note: If there hadn’t been any yellow chicks,
then there wouldn’t be anything to subtract,
because we wouldn’t have overcounted.
• The characteristics of “yellow” and “chick”
would then be called MUTUALLY EXCLUSIVE.
• Are the characteristics of “yellow”
and “chick” mutually exclusive? No!
They CAN happen at the same time –
there are 8 of them!
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Finally, Combining Random Variables
Z=X+Y or Z=X-Y
• Means will be added or subtracted just like
the random variables     
Z
X
Y
• Variances will be added ONLY IF VARIABLES
ARE INDEPENDENT!
2
2
 Z   X  Y
(Remember: Variance = standard
deviation squared)
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Example #9:
• SAT scores are known to be Normally
distributed with   1100 and   210 .
Two students are to be selected at random and
their scores added together. What would the
distribution of their total score, T, look like?
Answer: Normal! Mean = 1100 + 1100 = 2200.
Standard Dev = 2102  2102  297
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