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Â According to a 2014 Gallup poll 56% of uninsured Americans who plan to get health

Â insurance said they will do so through a government health insurance exchange.

Â What is the probability that in a random sample of 10 people

Â exactly 6 plan to get health

Â insurance through a government health insurance exchange?

Â Using the applet this is going to be pretty simple

Â to calculate.

Â So let's take a look at how we can do that.

Â 0:36

We have a binomial distribution.

Â So we're going to select our distribution to be binomial.

Â Our sample size is 10 and our probability of success is 0.56.

Â We're looking for an exact probability so we're

Â going to pick equality here of exactly six successes.

Â So the probability of six successes in ten trials, where the probability of

Â success is 0.56, comes out to be 0.243, or

Â 24.3% chance that in a random sample of ten people, exactly

Â six plan to get health insurance through a government health insurance exchange.

Â 1:20

Alternatively, using R and the dbinom function,

Â we set the number of successes, the first argument, to 6, the size to 10, and the

Â probability of success to 0.56 and obtain the same answer, 24.3%.

Â We also know how to calculate this by hand.

Â We're looking for the probability of exactly six successes in ten trials.

Â So we can write this out as 10 choose 6, 0.56 to the sixth power,

Â times 0.44, the probability of failure to the fourth power, n minus k.

Â We can expand the choose function 10 times 9 times 8 times 7

Â times 6 factorial and similarly expand out the denominator a little bit as well.

Â But we don't have to do a whole lot since the 6 factorials are going to cancel.

Â 2:10

And moreover the 4 and the 2 cancel with the 8.

Â We can

Â actually simplify 3 and 9 and what we get is 10 times 3 times 7.

Â 210 possible scenarios times the probability of one

Â scenario gets us to the same answer 0.243.

Â If we were to look for that once again in the binomial distribution,

Â that's going to be the height of the bar corresponding to exactly six successes.

Â This is

Â a pretty likely outcome if we think about it because

Â it's actually pretty close to what we would expect to see.

Â In 10 people, we would expect to get about 5.6 to say yes to this and

Â so the probability of exactly 6 successes is going to be not far off from the mark.

Â 2:59

Let's take a look at another question.

Â What is the probability that in a random sample of 1,000

Â people, exactly 600 plan to get health insurance through a government health

Â insurance exchange? The options are 0.243, that's the same as

Â the earlier probability we calculated where k was equal to 6 and

Â n was 10. Less than 0.243 or more than 0.243.

Â We want to answer this question without doing any calculations.

Â As it's aimed to

Â assess reasoning and conceptual understanding

Â as opposed to just computational ability.

Â Once again, the probability of success is 0.56.

Â In the previous exercise, the sample size was 10.

Â So the expected number of successes would be 10 times 0.56, 5.6.

Â 3:50

The difference between the expected and the desired

Â number of successes then would be 6 minus

Â 5.6, 0.4.

Â In this exercise, the sample size is 1,000.

Â So the expected number of successes is 1,000 times 0.56, 560.

Â The difference between the expected and the desired

Â number of successes is then 600 minus 560, 40.

Â Here, the desired outcome is much farther than the expected outcome and

Â based on our discussion of law of large numbers earlier, obtaining

Â 600 successes when the expected is 560 should be a much less likely

Â outcome that obtaining 6 successes when the expected is 5.6.

Â Therefore the answer is something less than 0.243.

Â We can actually check our answer quickly using R.

Â Probability of

Â 600 successes in 1000 trials with probability

Â of success of 0.56 is roughly 0.00098.

Â Much lower than the 243 we calculated earlier.

Â 5:00

Next we're asked to describe the probability

Â distribution of number of uninsured Americans who plan

Â to get health insurance through a government health

Â insurance exchange among a random sample of 100.

Â Once again, p is 0.56, this time n is 100. This seems like a larger sample size.

Â So let's see if it's actually large enough to yield a nearly normal distribution.

Â The rule was that we need at

Â least 10 expected successes and 10 expected failures.

Â Expected number of successes is 100 times 0.56 which is 56, which is

Â indeed greater than 10 and the expected number of failures is 100 times

Â 0.44 which is equal to 44 which is also greater than 10.

Â So we do know that the shape of the distribution will be nearly normal.

Â Normal distributions have two parameters, mean and standard deviation.

Â So to fully describe the distribution, we need to

Â calculate these parameters, which we know can be estimated

Â by the binomial, mean, and standard deviation.

Â Mean, in other words the expected number of successes, is 56.

Â We already calculated this.

Â And the standard deviation can be calculated as the square root of n times

Â p times 1 minus p. So that's the square root of 100 times

Â 0.56 times 0.44 roughly 4.96. So this binomial distribution's

Â shape actually follows a normal distribution

Â with mean 56 and standard deviation 4.96.

Â Lastly, let's consider the following question.

Â What is the probability that at least 60 out of a random sample of

Â 100 uninsured Americans plan to get health

Â insurance through a government health insurance exchange?

Â Once again, we'll present a variety of ways of solving

Â this, though you can really just pick one and stick with it.

Â One approach is, once again, to use the applet.

Â So let's take a look at how we could solve this question there.

Â Once again the distribution is binomial.

Â This time our number of trials is 100, so we're going to slide over our n to 100.

Â Probability of success is 0.56. The observation of interest

Â is 60 successes.

Â So we're going to slide our cutoff value to 60, and we're looking for

Â not just exactly 60 successes, but 60 or more successes.

Â So we want to find the upper tail area and the

Â bound we're interested in is greater than or equal to.

Â Here's the shaded area of interest, the probability comes out to be 24.1% chance.

Â 7:46

Another approach is to use R, and once again, we're

Â going to make use of two functions, the dbinom function that

Â gives us the probabilities, and the sum function that allows

Â us to add a bunch of probabilities that we're interested in.

Â So, in the dbinom function, the first argument is all successes from 60 to 100.

Â Our sample size is 100, and our probability of success is 0.56.

Â Then we wrap this around with a sum function that will basically add up all

Â the individual probabilities and yield a probability of

Â 24.1%, just like the one we obtained earlier.

Â 8:24

Lastly, we can make use of the fact that the

Â distribution is nearly normal, with mean 56 and standard deviation 4.96.

Â When finding normal probabilities, we calculate

Â the z score as the observation.

Â 60 minus the mean, 56, divided by the standard deviation, 4.96.

Â This comes out to be roughly 0.81.

Â The next step would be to find 0.81 on a

Â table, and if you're not sure how to do that, I

Â recommend that you review earlier lectures working with a normal probability

Â table, and we would find the probability to be roughly 0.209.

Â Once again, this probability is a little lower than

Â the probabilities we've calculated using the applet and R.

Â Remember that this discrepancy is mostly due to the fact that

Â under the normal distribution, probability of exactly 60 successes is undefined.

Â 9:21

To account for that, we apply a 0.5 correction to

Â the observation of interest, and can work through the problem again.

Â Calculated an updated z score and an associated probability.

Â We can calculate the updated z score as 59.5 minus 56, the mean,

Â divided by the same standard deviation 4.96, which comes out to be 0.71.

Â Then the updated probability is going to be

Â 0.239 which is much closer to the exact probability

Â that we have calculated using the exact binomial distribution.

Â