Example: Awe…Nuts! - Village Christian School

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Transcript Example: Awe…Nuts! - Village Christian School

How many movies do you watch?
Does the CLT apply to means?
 The Central Limit Theorem states that the more
samples you take, the more Normal your graph will
appear for either the sample mean or sample
proportion.
 What do we need to verify in order to say the
distribution should be approximately normal?
 Independence Assumption:
• In order to be able to consider each sample an
unbiased estimator, we must insure that random
selection was used.
• If we want to insure we aren’t affecting the probability
of selecting each sample, we must insure that the
sample we take is less then 10% of our population.
Does the CLT apply to means?
Back to the show…
Back to the show…
 Suppose that the number of movies viewed in the last
year by all high school students in CA has an average of
19.3 and a standard deviation of 15.8 . If we were to
randomly sample 100 high school students many times,
what would we expect the average of the sampling
distribution to be? What about the standard deviation?
 Independence Assumption:
We are given that the selection would be random. We
can also safely assume that there are more then 1,000 high
school students in CA.
 Large Enough Condition:
Since n = 100 > 30 we can state that the distribution
would be approximately Normal.
Example: Awe…Nuts!
At the P. Nutty Peanut Co., dry-roasted, shelled peanuts are
placed in jars by a machine. The distribution of the weights in
the jars is approximately Normal with a mean of 16.1 oz and a
standard deviation of .15 oz.
 Without doing any calculations, explain which
outcome is more likely: randomly selecting a
single jar and finding that it weighs less than 16
oz. or randomly selecting 10 jars and finding that
the average is less than 16 oz.?
 Find the probability of each of these events.
Example: Awe…Nuts!
At the P. Nutty Peanut Co., dry-roasted, shelled peanuts are
placed in jars by a machine. The distribution of the weights in
the jars is approximately Normal with a mean of 16.1 oz and a
standard deviation of .15 oz.
 Without doing any calculations, explain which
outcome is more likely: randomly selecting a
single jar and finding that it weighs less than 16
oz. or randomly selecting 10 jars and finding that
the average is less than 16 oz.?
 Follow the 4 Step Process!
 1. State what you want to know:
We want to find the probability of selecting a
single jar that weighs less then 16 oz.
Example: Awe…Nuts!
At the P. Nutty Peanut Co., dry-roasted, shelled peanuts are
placed in jars by a machine. The distribution of the weights in
the jars is approximately Normal with a mean of 16.1 oz and a
standard deviation of .15 oz.
 Without doing any calculations, explain which
outcome is more likely: randomly selecting a
single jar and finding that it weighs less than 16
oz. or randomly selecting 10 jars and finding that
the average is less than 16 oz.?
 Step 2: Verify Your Assumptions
 Independence Assumption:
We are told that the jar is randomly selected. And we can
assume that on any given day the P. Nutty Company makes
more then 10 jars of peanuts.
Example: Awe…Nuts!
At the P. Nutty Peanut Co., dry-roasted, shelled peanuts are
placed in jars by a machine. The distribution of the weights in
the jars is approximately Normal with a mean of 16.1 oz and a
standard deviation of .15 oz.
 Without doing any calculations, explain which
outcome is more likely: randomly selecting a
single jar and finding that it weighs less than 16
oz. or randomly selecting 10 jars and finding that
the average is less than 16 oz.?
 Step 2: Verify Your Assumptions
 Large Enough Condition:
Since we are given that the distribution of weights is
approximately Normal, even though our sample is small, it is
safe to proceed.
Example: Awe…Nuts!
Example: Awe…Nuts!
This makes sense, because if the
distribution is approximately
Normal. Then the mean would
divide the distribution in half.
Example: Awe…Nuts!
At the P. Nutty Peanut Co., dry-roasted, shelled peanuts are
placed in jars by a machine. The distribution of the weights in
the jars is approximately Normal with a mean of 16.1 oz and a
standard deviation of .15 oz.
 Without doing any calculations, explain which
outcome is more likely: randomly selecting a
single jar and finding that it weighs less than 16
oz. or randomly selecting 10 jars and finding that
the average is less than 16 oz.?
 In your notes, follow the 4 step process to solve
this problem. I will randomly select students to
come up to the board and show their work.