The new forums will be named Coin Return (based on the most recent vote)! You can check on the status and timeline of the transition to the new forums here.
Please vote in the Forum Structure Poll. Polling will close at 2PM EST on January 21, 2025.
Probability is doing my head in (basic)
jasonlester
Registered User regular
Registered User regular
This is driving me nuts, can anyone lend a hand? Not sure If I'm wrong or the book is
39% of people are shareholders (A)
7.1% of shareholders have been to college (B)
5% of adults have college education (A and
OKAY, so, what is the probability that an adult owns shares or has postgraduate education?
I'm getting
0.39 + 0.05 - 0.071
= 37%
what am I doing horribly wrong?
39% of people are shareholders (A)
7.1% of shareholders have been to college (B)
5% of adults have college education (A and
OKAY, so, what is the probability that an adult owns shares or has postgraduate education?
I'm getting
0.39 + 0.05 - 0.071
= 37%
what am I doing horribly wrong?
jasonlester on
0
Posts
0.39 x 0.05 = 0.0195.
If you don't understand why, just see how multiplying 0.39 by 0.05 will produce 5% of 0.39, as 0.05 is equal to 5%.
So 5% of 39% equals 0.0195%. 0.0195% is a portion of a portion. A group of people within another group of people.
I thought I was meant to be using the addition rule, and union probability?
Uh, embarrassingly enough I'm not that good at maths and have forgotten about union probability and what not. I've now learnt to never answer maths threads again, lest they become remotely complicated :P
The problem here is that the percentage in "7.1% of shareholders have been to college" is in terms of shareholders while the percentage in the other two is in terms of adults. So you can't just add them like that.
Edit: You need to figure out what percentage of adults are shareholders and college graduates and then use your adding and subtracting method.
Sorry, missed that.
So it should work
There is no way in shit I'm going to pass this subject. I've already spent like 4 hours on this subject today, and I'm just completely missing the point.
Got top 5% of my state on my highschool entrance score, but this is losing me.
It doesn't work like that. According to your logic the following works: 50% of Americans are male, 50% of americans have testicles therefore 100% of Americans are male or have testicles.
You need to add the two together and then subtract the intersection. The OP needs to figure out the size of the intersection from the information he has.
(.39 - (.39 x .071)) + (.05 - ( .05 x .071))
=.40876
No. It goes like this
0.39 + 0.05 - 0.071*0.39 = 0.41231
That's Shareholders + College Graduates - (Shareholders AND college graduates)
We get the intersection at the end by multiplying the percentage of shareholders by the percentage of (shareholder and graduates)
If you don't understand why, consider the following example: Half of all people are men (0.5), half of all men like pizza (0.5). Therefore a quarter of all people are men who like pizza (0.5*0.5 = 0.25)
Typo.
true, forgot about the double count
my book says:
P ( X union Y) = p (X) + p (y) MINUS P (x AND y)
Not plus.
arghh
Yeah, that's the right formula. I've edited all the typos out of my post.
Okay, so A intersection B is always A*B?
No not always. The trick in this question was that two of the percentages were given in terms of people, and one in terms of shareholders. The difference is like the difference between the following problems.
1) 50% of all people are women, 60% of all people work in an office, 25% of people are women working in offices. How many are either a woman or work in an office?
Here all the percentages are in terms of people so we can just apply the formula to get:
(Women) + (office workers) - (Women AND Office workers) = 0.5 + 0.6 - 0.25 = 0.85
85% of people are either women or work in offices or both.
2)50% of all people are women, 50% of all women work in offices, 60% of people work in offices. How many people are either women or work in offices?
Here what we need to do first is to express the bolded bit in in terms of all people rather than just in terms of women. To do this we multiply them together (do you get why this works?) so the percentage of people who are women working in offices is 0.5*0.5=0.25.
So (Women) + (office workers) - (women and office workers) = 0.5 + 0.6 - (0.5*0.5) = 0.85
but what about the 5% of the people who have college degrees? Some of them must invest, right?
Right now we have
People who Invest (39%) - (people who invest +went to college)
^^^^^^^^^
gives us solely the people who invest
and don't go to college
+
People who went to college (5%)
^^^^^^^^
but this is still mixed with some people who invest
.071 represents the people that invest and went to college. Than Y represents the people who went to college and invest, which can give us solely the people who went to college but don't invest.
(Y)
(.39 - .39 x .071) + (.05 - .05 x (Y))
(.39 - .02769) + (.05 - .05 x .02769)
(.05 - .05 x .02769)
+ (.05 - .0013845)
.36231 + .0486155
=
.4109255
We're trying to find the number of people who invest, who graduateed college or both.
So 7.1% of the entire shareholder population (100% of the total, but only 39% of the population) count. So this is how it matches up:
7.1% of 39% is 2.769% -- this is of the total population that counts for both Shareholders and Adults w/ college education. And since 5% is the absolute total of all people who have college education, you just subtract. That means that 2.231% are NOT shareholders... but have a college education. These people would be a part of the remaining 61%. You could be fancy and say that 3.66% of the non-stockholder population has college degrees, as well (.02231/.61).
Which means that the answer is:
39% of all people are shareholders, and 2.769% of all people are shareholders AND college educated, and 2.231 of all people are ONLY college educated. The units all match -- they're all set up as "all people."
Now you simply add for the OR -- 39+2.231=41.231
The trick to, hell, most problems is to get the units to match. In this case, it's in terms of populations.
A + B - (A AND
Now, we have 39% of people are shareholders. OF THOSE PEOPLE, 7.1% have gone to college. 39% * 7.1% = 2.769% of people are shareholders who went to college.
Let's call the percentage of people who are shareholders S and the adults with a college education as E. Our solution will be:
S + E - (S AND E) from our union formula.
39% + 5% - 2.769% = 41.231%
say you have...25% of people liking...cheese, and you randomly select say 15 people, how do you work out the probability that less than 4 of them like cheese?
Given that the population proportion of people who like cheese is 0.25, you want the probability that out of a sample of 15 that less than 4 of them like cheese.
P(X<4|p=0.25) = P(X=0|p=0.25) + P(X=1|p=0.25) + P(X=2|p=0.25) + P(X=3|p=0.25)
= 15C0*(0.25^0)*(0.75^15) + 15C1*(0.25^1)*(0.75^14) + 15C2*(0.25^2)*(0.75^13) + 15C3*(0.25^3)*(0.75^12)
Binomial distribution.
I remember using an Ncr function on my calculator a while ago. When I get home I'll have a look over my book. Thank you
Completion time (from start to finish) of a building remodelling project is normally distributed with a mean of 200 work-days and a standard deviation of 10 work-days. The probability that the project will be completed within 215 work-days is
So, in my head, I would think, oh, I need the z score. So I do
215 - 200 divided by 10 which gives me 1.5
Then I'm supposed to look that up in the table, to get my probability. The answer here is "0.9332" which is impossible judging from my table. That value doesn't even exist.
The book we use is of course, completely useless, the only way to study this is do the practice tests and trial and error until you figure out how they did it. Arghhh
Look for where 1.5 meets 0.00. Should say 0.9332 or something close to it.
...I have to be doing something majorly wrong
The standard normal curve is symmetric about zero... meaning the area under the curve left of 0 is 50% and the area to the right of it is also 50%. So the area to the left of 1.50 must be.... bigger than 50%.
If I had to guess... 0.4332 is the area between 0 and 1.5 under the curve, which makes sense, since 0.5 + 0.4332 = 0.9332
argh
0.4332 is the area between 0 and Z. So just add 0.5 or subtract as necessary ?
Thank you for your help. I'd probably be tearing my hair out without it :P
I might throw another one in here later on if I can't get it. I seem to be going okay now though.
Thank you
assuming People/adults are synonymous and that's just rough writing, people that meet the criteria are .4205 of total people
39 percent who own shares
5 percent of the 61 percent who don't (or approx. 3 percent of adults)
42.05 percent total.
The flaw in your initial phrasing is that group 3 doesn't meet a and b in all cases. You have to account for group 3 within group 2 and not count them twice.
I don't know if that's "correct" for purposes of your problem, though.
I host a podcast about movies.
.4332 is the area between the average and Z. Because the Z value is how many standard deviations you are from the mean.
Because the Z table is used for different purposes. For some problems, it's better to figure out how much is fully covered by an amount. So you'd take the value in the Z-table and add .5.
For others, you want to figure out how much of a half. Or how much is *not* covered. Regardless, you end up having to do some math to get the right answer. The one you have is pretty typical for pure statistics or mathematics, because that's how Z-values are actually calculated. The cumulative ones -- the kind that would give you the actual final answer right in the table -- are more useful if you're only dealing with a subset of those kind of problems (which, though, are usually the ones that people commonly encounter). The point of them is that they count the negative Z-values.
One thing I was taught that is very helpful, yet many people overlook it, is to simply draw a normal distribution, and put the numbers on there. It's a great way to answer the question:
Job A has an average completion time of 20 days, with a standard deviation of 3 (normally distributed). What is the probability of it finishing on the 20th day?
50%! Because if you draw a normal distribution, and put the mean down, you'll see that there's a 50/50 chance that it could occur before the deadline or after the deadline. When you're working with deadlines, the X-axis is time, so it's easy to say "Ah, ok, 22 days is more than the average. Therefore, the probability MUST be more than .5." So you look at the z-table and, with the one you have, add .5 because that counts for all of the area to the left of the average.