This course introduces you to sampling and exploring data, as well as basic probability theory and Bayes' rule. You will examine various types of sampling methods, and discuss how such methods can impact the scope of inference. A variety of exploratory data analysis techniques will be covered, including numeric summary statistics and basic data visualization. You will be guided through installing and using R and RStudio (free statistical software), and will use this software for lab exercises and a final project. The concepts and techniques in this course will serve as building blocks for the inference and modeling courses in the Specialization.
Working with the Binomial Distribution
Loading...
Из курса от партнера Duke University
Introduction to Probability and Data
2411 оценки
Из урока
Probability Distributions
<p>Great work so far! Welcome to Week 4 -- the last content week of Introduction to Probability and Data! This week we will introduce two probability distributions: the normal and the binomial distributions in particular. As usual, you can evaluate your knowledge in this week's quiz. There will be <b>no labs</b> for this week. Please don't hesitate to post any questions, discussions and related topics on <a href="https://www.coursera.org/learn/probability-intro/module/VdVNg/discussions?sort=lastActivityAtDesc&page=1" target="_blank"><b>this week's forum</b></a>.</p>
Познакомьтесь с преподавателями
Mine Çetinkaya-RundelAssociate Professor of the Practice
Department of Statistical Science
[MUSIC]
According to a 2014 Gallup poll 56% of uninsured Americans who plan to get health
insurance said they will do so through a government health insurance exchange.
What is the probability that in a random sample of 10 people
exactly 6 plan to get health
insurance through a government health insurance exchange?
Using the applet this is going to be pretty simple
to calculate.
So let's take a look at how we can do that.
We have a binomial distribution.
So we're going to select our distribution to be binomial.
Our sample size is 10 and our probability of success is 0.56.
We're looking for an exact probability so we're
going to pick equality here of exactly six successes.
So the probability of six successes in ten trials, where the probability of
success is 0.56, comes out to be 0.243, or
24.3% chance that in a random sample of ten people, exactly
six plan to get health insurance through a government health insurance exchange.
Alternatively, using R and the dbinom function,
we set the number of successes, the first argument, to 6, the size to 10, and the
probability of success to 0.56 and obtain the same answer, 24.3%.
We also know how to calculate this by hand.
We're looking for the probability of exactly six successes in ten trials.
So we can write this out as 10 choose 6, 0.56 to the sixth power,
times 0.44, the probability of failure to the fourth power, n minus k.
We can expand the choose function 10 times 9 times 8 times 7
times 6 factorial and similarly expand out the denominator a little bit as well.
But we don't have to do a whole lot since the 6 factorials are going to cancel.
And moreover the 4 and the 2 cancel with the 8.
We can
actually simplify 3 and 9 and what we get is 10 times 3 times 7.
210 possible scenarios times the probability of one
scenario gets us to the same answer 0.243.
If we were to look for that once again in the binomial distribution,
that's going to be the height of the bar corresponding to exactly six successes.
This is
a pretty likely outcome if we think about it because
it's actually pretty close to what we would expect to see.
In 10 people, we would expect to get about 5.6 to say yes to this and
so the probability of exactly 6 successes is going to be not far off from the mark.
Let's take a look at another question.
What is the probability that in a random sample of 1,000
people, exactly 600 plan to get health insurance through a government health
insurance exchange? The options are 0.243, that's the same as
the earlier probability we calculated where k was equal to 6 and
n was 10. Less than 0.243 or more than 0.243.
We want to answer this question without doing any calculations.
As it's aimed to
assess reasoning and conceptual understanding
as opposed to just computational ability.
Once again, the probability of success is 0.56.
In the previous exercise, the sample size was 10.
So the expected number of successes would be 10 times 0.56, 5.6.
The difference between the expected and the desired
number of successes then would be 6 minus
5.6, 0.4.
In this exercise, the sample size is 1,000.
So the expected number of successes is 1,000 times 0.56, 560.
The difference between the expected and the desired
number of successes is then 600 minus 560, 40.
Here, the desired outcome is much farther than the expected outcome and
based on our discussion of law of large numbers earlier, obtaining
600 successes when the expected is 560 should be a much less likely
outcome that obtaining 6 successes when the expected is 5.6.
Therefore the answer is something less than 0.243.
We can actually check our answer quickly using R.
Probability of
600 successes in 1000 trials with probability
of success of 0.56 is roughly 0.00098.
Much lower than the 243 we calculated earlier.
Next we're asked to describe the probability
distribution of number of uninsured Americans who plan
to get health insurance through a government health
insurance exchange among a random sample of 100.
Once again, p is 0.56, this time n is 100. This seems like a larger sample size.
So let's see if it's actually large enough to yield a nearly normal distribution.
The rule was that we need at
least 10 expected successes and 10 expected failures.
Expected number of successes is 100 times 0.56 which is 56, which is
indeed greater than 10 and the expected number of failures is 100 times
0.44 which is equal to 44 which is also greater than 10.
So we do know that the shape of the distribution will be nearly normal.
Normal distributions have two parameters, mean and standard deviation.
So to fully describe the distribution, we need to
calculate these parameters, which we know can be estimated
by the binomial, mean, and standard deviation.
Mean, in other words the expected number of successes, is 56.
We already calculated this.
And the standard deviation can be calculated as the square root of n times
p times 1 minus p. So that's the square root of 100 times
0.56 times 0.44 roughly 4.96. So this binomial distribution's
shape actually follows a normal distribution
with mean 56 and standard deviation 4.96.
Lastly, let's consider the following question.
What is the probability that at least 60 out of a random sample of
100 uninsured Americans plan to get health
insurance through a government health insurance exchange?
Once again, we'll present a variety of ways of solving
this, though you can really just pick one and stick with it.
One approach is, once again, to use the applet.
So let's take a look at how we could solve this question there.
Once again the distribution is binomial.
This time our number of trials is 100, so we're going to slide over our n to 100.
Probability of success is 0.56. The observation of interest
is 60 successes.
So we're going to slide our cutoff value to 60, and we're looking for
not just exactly 60 successes, but 60 or more successes.
So we want to find the upper tail area and the
bound we're interested in is greater than or equal to.
Here's the shaded area of interest, the probability comes out to be 24.1% chance.
Another approach is to use R, and once again, we're
going to make use of two functions, the dbinom function that
gives us the probabilities, and the sum function that allows
us to add a bunch of probabilities that we're interested in.
So, in the dbinom function, the first argument is all successes from 60 to 100.
Our sample size is 100, and our probability of success is 0.56.
Then we wrap this around with a sum function that will basically add up all
the individual probabilities and yield a probability of
24.1%, just like the one we obtained earlier.
Lastly, we can make use of the fact that the
distribution is nearly normal, with mean 56 and standard deviation 4.96.
When finding normal probabilities, we calculate
the z score as the observation.
60 minus the mean, 56, divided by the standard deviation, 4.96.
This comes out to be roughly 0.81.
The next step would be to find 0.81 on a
table, and if you're not sure how to do that, I
recommend that you review earlier lectures working with a normal probability
table, and we would find the probability to be roughly 0.209.
Once again, this probability is a little lower than
the probabilities we've calculated using the applet and R.
Remember that this discrepancy is mostly due to the fact that
under the normal distribution, probability of exactly 60 successes is undefined.
To account for that, we apply a 0.5 correction to
the observation of interest, and can work through the problem again.
Calculated an updated z score and an associated probability.
We can calculate the updated z score as 59.5 minus 56, the mean,
divided by the same standard deviation 4.96, which comes out to be 0.71.
Then the updated probability is going to be
0.239 which is much closer to the exact probability
that we have calculated using the exact binomial distribution.
Ознакомьтесь с нашим каталогом
Присоединяйтесь бесплатно и получайте персонализированные рекомендации, обновления и предложения.
Coursera делает лучшее в мире образование доступным каждому, предлагая онлайн-курсы от ведущих университетов и организаций.
© Coursera Inc., 2018 Все права защищены.
