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Published byAusten Dymond Modified about 1 year ago

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How many movies do you watch?

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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.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.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.

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Does the CLT apply to means?

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Back to the show…

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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.

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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.

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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.

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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.

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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.

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Example: Awe…Nuts!

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This makes sense, because if the distribution is approximately Normal. Then the mean would divide the distribution in half.

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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.

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