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Don't have my book or audio? You can download a free sample of my book and audio lectures containing Lesson 1: Did you read my tips on how to study and learn this
course? If not, here is a link to those important suggestions: The department posted SOLUTIONS for Assignment 1
(these are not my solutions). Click here.The department posted SOLUTIONS for Assignment 2 (these are not my solutions). Click here.Here is a link to the actual assignment, in case you don't have it handy: Study Lesson 5: Introduction to Probability, Lesson 6: The Binomial Distribution, and Lesson 7: The Distribution of the Sample Mean in my Basic Stats 1 book to
prepare for this assignment. NOTE: You need only study to the end of question 7 in Lesson 7 (end of page 447). The section on control charts has been removed from the course.
Although this question appears to be a 3-circle Venn diagram problem (similar to my Lesson 5, #18), they have left key information out that renders it not worth the effort to try to make the diagram right away.
To make a
three-circle Venn diagram, for event A, B and C, among the givens you need the following: - The probability an outcome belongs to all three of A, B and C.
- The three "and" pairs, which is to say, the P(A and B), the P(A and C), and the P(B and C).
- Then you can assume that everything will go smoothly filling in the 8 sections of the Venn diagram.
If the above is not given, you should use the
probability formulas to solve what is missing above first.
In this problem, you have been given the probability an outcome belongs to all three of J, B, and G; you have been given P(J and B); and, you have been given P(J and G). The problem is that you have not been given P(B and G). Before I make a Venn, I would use a probability formula to solve that missing value.
But that is actually one of their questions below, so, you are
better off just taking each question as it comes and answering it directly.
This is tailor-made for the General Addition Rule P(A or B) = P(A) + P(B) - P(A and B). You have been given everything you need.
Again, use the General Addition Rule P(A or B) = P(A) + P(B) - P(A and B) and a little algebra. Note that you have been given P(J or G), P(J and G), and P(J). So, use those to figure out P(G).
Now, they want P(B and G).
Don't you dare use the formula P(B and G) = P(B)*P(G)!!!! THAT FORMULA IS ONLY TRUE WHEN TWO EVENTS ARE INDEPENDENT!!! You have no idea whether B and G are
independent or not.
One of the most common mistakes students make, is thinking that, anytime they want to know P(A and B), they merely have to compute P(A)*P(B). WRONG!!!! You can only do that when you are told that A and B are INDEPENDENT.
Here is the proper thing to do: The General Addition Rule, P(A or B) = P(A) + P(B) - P(A and B), is ALWAYS TRUE. Use that formula to solve either
"or" or "and" unknowns whenever possible. That formula has four terms: P(A), P(B), P(A and B), and P(A or B). If you are given any three of those four terms, you can use this formula and a little algebra to solve the missing fourth term. That is what you should have done to solve part (b) above, and that is what can also be done here to solve part (c). You can solve P(B and G) if you know P(B), P(G) and P(B or G).
Now, you finally know everything you need to make the three-circle Venn diagram. So, I would make the Venn and use it to solve this question.
Now, you should know P(G) from part (b) and P(B and G) from part (c). Now you can prove whether any two events in this problem are independent.
Two events are independent if and only if they satisfy the Independence
Multiplication Rule,
P(A and B) = P(A)*P(B).
For example, you were told P(J)=0.55, P(B)=0.40, and P(J and B)=0.27. If J and B are independent, then P(J and B) = P(J)*P(B). 0.27 should equal 0.55*0.40. But 0.55*0.40 = 0.22, not 0.27, proving that J and B are dependent, not independent.
You can do a similar test for the other pairs.
Part (a) I would use a two-way table to set up and solve this problem. Similar to what I do in Lesson 5. Note that they want the probability distribution of X, the number of heads, similar to my questions 3(f)
and 4(f) where I use the earlier two-way table and distribution I had come up with to then do a probability distribution for X, the number of boys.
Part (b) This certainly can be solved using binomial methods, as taught in Lesson 6, #1, for example. Keep in mind that X is the number of heads. What is n and p? They then want you to solve the probabilities for X=0, X=1, etc. using the binomial formula.
Obviously, those probabilities should match your answers in part (a).
They are simply testing you on the distribution of x̅, the sample mean. In each case, they are telling you the distribution of the population. From that, you can infer the distribution of x̅.
- If the given population is normal. Then it is always possible to
solve the probability of x̅ because it, too, will be normal. And the probability is always exact.
- If the given population is not normal, it is only possible to compute the probability of x̅ if n is large, and even then it will only be approximate, because Central Limit Theorem tells us x̅ is approximately normal for any non-normal population as long as n is large.
If you are told for a fact that a distribution is normal, then you can compute
exact probabilities because you know the exact values for the normal distribution. And if you know that X is normal, then it also follows that x-bar is normal, too. So, you can compute exact probabilities for X or x-bar, when the given population is normal.
If a population is not normal, then, you may be able to assume x-bar is approximately normal by Central Limit Theorem. However, if x-bar is only approximately
normal, you will only be able to compute approximate probabilities.
A lot of students don't understand what they are even saying in this question, so here are some clarifications. Do note, they do not want you to compute anything! They merely want you to check off the cases that are TRUE.
- For example, the first case says normal, 5, and exact. They are telling you the population is normal,
the sample size used to compute x̅ is 5, and the probability that you would compute for x̅, the sample mean, being between 95 and 120 would be exact. That is absolutely TRUE! If a population is normal, the the distribution of the sample mean will also be normal regardless of the sample size, so you will always be able to compute an exact probability, since we know exactly how the normal distribution behaves.
- Obviously, that
makes the second case, normal, 5, and approximate, FALSE, because that is saying that you would only be able to compute an approximate probability for a sample size of 5 if the population is normal, and, as we just said in the first case, we would be able to compute an exact probability.
- So, in each case, they are telling you whether the population is normal or right-skewed, what size of a sample was used to compute the sample mean, and whether the probability
you compute for the sample mean being between 95 and 120 would be exact or approximate.
- For example, case H, says right-skewed, 50, and approximate, telling us that the population is right-skewed (not normal); however, the sample size is 50 (large), so we can assume the distribution of the sample mean is approximately normal by Central Limit Theorem; therefore we can compute the approximate probability that the sample mean is between 95 and
120. This case is TRUE.
- Try the rest on your own.
You will be using Table A for much of the rest of this assignment. Here is a link where
you can download the table if you have not already done so:
Questions 7 to 13 are Lesson 7 stuff. From now on in this course, anytime they
hand you a population mean, μ, and a population standard deviation, σ, you have to always be asking yourself:
- Is the problem talking about one individual score X? Or, is it talking about the mean of n scores, x̅?
- If it is talking about just one score, X, is X normally distributed? Only if X is normally distributed can you compute the probabilities of some region for X,
since only then can we standardize X into a z score using the formula z = x - μ divided by σ.
- If it is talking about the mean of n scores, x̅, can we assume x̅ is normally distributed? Why or why not?
- Remember, if a population is known to be normal, then x̅ has a normal distribution for any sample size, n.
- If a population is not normal, it is still possible that the distribution of the sample
mean, x̅, is normal, but n must be large.
- Central Limit Theorem: If n is large, the distribution of x̅, the sample mean, is approximately normal even if the original population is not normal.
- If x̅ is normal (even approximately) then we can standardize x̅ into a z-score using the formula z = x̅ - μ divided by σ/sqrt n.
Look at my Lesson 7, #4 and #5 for
examples.
IMPORTANT CALCULATOR TIPS: When computing things like sigma/ square root n, make sure you round off to NO LESS THAN 4 decimal places. Better yet, store the exact value in the memory of your calculator.
Note that they give you a total amount in the question. However, stop and think! Never solve a bell curve problem without first visualizing it and understanding what area you are shading. If you don't understand what area they want, there is no way
you could ever get the correct probability. Are they telling you to shade left? shade right? What are they telling you to shade?
More x-bar bell curve stuff like the previous two questions. Look at my questions 4 to 7 in Lesson 7 for examples. Note that they give you a total amount in one of the questions. Note that question 11 requires that you
work backwards since you are given the probability and can read the table backwards to get a z score and solve the unknown algebraically.
Be careful, they are talking about the mean of a sample of some size n=4. Thus, you must compute the mean of x-bar and the standard deviation of x-bar, which are mu and sigma divide
by square root n first, then use those values for your 68-95-99.7 rule. Make sure you put the smaller answer first.
This is p^ bell curve stuff similar to what I do in Lesson 6, #10(c) and (d).
Be careful to note which is the true proportion, p, and which is the sample proportion p^.
Be careful that you don't lose
accuracy by rounding off too much. I suggest you round off to no less than 5 or 6 decimal places while computing things like the standard deviation of p^ to ensure that you get accurate z-scores. Better yet, store exact answers in memory in your calculator.
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