Transcript Var(B)

Adding 2 random variables
that can be described by the
normal model
AP Statistics B
1
Outline of lecture
• Review of Ch 16, pp.376-78 (Adding or
subtracting random variables that fit
the normal curve)—last of Ch 16
• Follow along in the text
• Remember that you can download this
PowerPoint and a smaller, un-narrated
one from the Garfield web site
• Write down slide number if you don’t
understand anything
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First thing: make a picture, make
a picture, make a picture!
• Well, actually, a table, but a table IS
a kind of picture, right?
Mean
Standard
Deviation
Packing (P)
9 min
1.5 min
Boxing (B)
6 min
1 min
3
Preliminaries: set-up and
assumptions
Setting up:
NOTE: NOT time to get an entire box packed
for sending; rather, we are only packing,
not boxing!
1. P1=time for packing 1st stereo system
2. P2=time for packing NEXT stereo system
3. T=P1+P2
Assumptions:
• Normal models for each RV
• Both times independent of each other
4
Step one: calculate expected
value (aka find the mean)
• Remember that the expected value
(EV) is a fancy word for finding the
mean.
• And with the mean, the EV sum of
two random variables is the sum of
their means:
5
Application to the packing and
boxing problem
• Mean (EV) of packing 1ST system is 9
min
• Mean (EV) of packing 2nd system is
also 9 min
• Therefore:
6
Calculate standard deviation
just like we have before
• Nothing new, same old formula:
7
So what?
• (You should always ask yourself “so what?”
when somebody tells you to do something)
• Well, we know that we had two RVs
(random variables, not recreational
vehicles; this is statistics, after all, not an
auto show) that have a normal distribution.
• So we now have a normal distribution and
know the mean and standard deviation.
• In statistical terms: N(18, 2.12)
• Now we can evaluate this using what we
learned in Ch. 6! A seriously big deal!
8
Q: What is the probability that
packing 2 consecutive systems
takes over 20 min?
• This is the question we need to answer.
• Remember the z-formula from Ch 6?
• Write it down, and we’ll apply it on the
next slide.
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Setting up the problem
• We already know the mean and
standard deviation from our earlier
calculations: 18 and 2.12,
respectively.
• The “y” value we are looking for is
20, so we set up the solution thusly:
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Are we done yet?
• Of course not.
• Our goal is to find the probability that it
will take MORE than 20 minutes to
package 2 consecutive stereo systems.
• The z-value of 0.94 simply means that
the area to the LEFT of that point on
the z-table (text, A79-A80) will be the
probability that packaging will take
LESS than 20 minutes.
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Draw a picture!
• The probability we
get for z=0.94 is
the dark blue area
on the left.
• However, we’re
interested in the
light blue area
that’s ABOVE
z=0.94
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But first…..
• Remember that our table only
measures the cumulative probability
to the LEFT of the value.
• That’s all we have, so let’s find it,
and then answer the question more
directly.
13
How to use the table
• It’s been a while, but get the X.x value
on the inner column, and the 0.0x
value across the top. The intersection is
where the value lies, and looks like
this:
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Finding the probability to the
left of 0.94
• Since 0.94>0, look on p. A-80
• Find 0.90 along the z-column on the
far left
• Read across the top row to 0.04
• The intersection of the 0.90 row and
the 0.04 is the percentage of the
normal curve to the LEFT of 0.94…..
• ……which should be 0.8294
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What are we looking for?
• Not the blue area
to the LEFT, but
the clear area to
the RIGHT of 0.94
• Calculate by
subtracting the
area to LEFT from
1:
• 1.00000.8294=0.1706
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Why the difference from the
textbook?
• Beats me.
• But 0.1706 isn’t all
THAT different
from 0.1736…3
parts in a
thousand.
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Back to interpretation
• We get a z-value of just over 17%.
• That means, in everyday language,
that there’s only a 17% chance
(probability) that it will take more
than 20 minutes to package 2
consecutive stereo systems.
• NOTE: the AP exam expects you to
write out things like this
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NEXT QUESTION
(BOTTOM OF PAGE 377)
What percentage of
stereo systems
take longer to back
than to box?
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Set-up on questions like this is
crucial
• The key is to realize that you don’t
set it up as an inequality, exactly.
• That is, the question is NOT P>B
• Rather, the question is whether PB>0.
• We pick a different variable (D for
“difference”) and define it as
D=P-B
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Why do we do it like this?
• We now can ask a specific question,
namely what’s the expected value of
D?
• In statistical terms, we have
E(D)=E(P-B).
• We can now calculate these values
using what we’ve learned in Ch 16
and combining it with the normal
model from Ch 6.
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First the mean,
then the standard deviation
1. E(D)=E(P-B)=E(P)-E(B)
2. E(P) we get from reading the mean
for packing right off the table
3. We get E(B) the same way.
4. E(P)-E(B)=9 min – 3 min = 6 min
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Calculating the standard
deviation
• I like to calculate the SD directly, but
you can start with the variance, and
then take the square root.
• Var(D)=Var(P-B)=Var(P)+Var(B)
• =1.52+ 1.02 (from the
table)=2.25+1=3.25
• σP-B=(3.25)½=1.8 min
(approximately)
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Now we have the normal
model
• i.e., N(3, 1.80)
• We are interested only in the values
that are GREATER than 0, i.e., to the
RIGHT of the value, like the purple:
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Now, to calculate
• Same as we did before (Slide 9):
• By table A-79, z=-1.67 has 0.0475
to its LEFT, but we want the area to
the RIGHT
• So subtract from 1 and get 0.9525.
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Interpreting the result
• We have determined the percent of
the time where P-B>0
• In other words, 95.25% of the time,
it takes longer to pack the boxes
than to box them.
• How do we know? Because P-B is
positive ONLY when P>B (otherwise,
the difference would be negative)
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Exercise
• Do Exercise 33 on p. 383-84 of the
textbook
• Take 10-15 minutes to complete all
parts.
• Review your answers with Ms. Thien,
who has them all worked out.
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