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QUESTION: 1. The continuous random variable has the cumulative distribution function defined by these three parts:  F(y)=0 for y<0, F(y)=(1/4)(y^3-4Y^2+ky) for 0<=y<=2, F(y)=1 for y>2.  P(y>1)=0.2A.  Find A.

2. A politician, Mr Muggle, claims that 45% of people would vote for him in an election. N people are selected at random for an opinion poll, and none of them said that the would vote for him.  Using a 1% level of significance, B=the smallest value of N for which the null hypothesis, p=0.45, will be rejected in favour of the alternate hypothesis, p<0.45.

3. A gps manufacturer finds that the life expectancy of its units has a standard deviation of 5000 hours of use.  On a random sample of 100 worn out units, it finds that the total hours of use for all units is 1740000 hours.  If the 95% confidence interval for the mean lifetime of a unit is written as [p,q], C=(q-p)/140.

4. At an event cache, the male cachers have heights which follow a normal distribution with mean 177cm and a standard deviation of 5.  The female cachers heights also follow a normal distribution, but with a mean of 163cm and a standard deviation of 4.  A male and a female cacher are chosen at random.  When written to 2 decimal places, the probability that the male cacher is taller than the female cacher is 0.9D.  Find D.

5. The width of 35mm film canisters is found to be normally distributed with mean Wmm and a standard deviation of 0.5mm.  The probability that a randomly selected canister is within 0.6mm of W is 0.E698.  Find E.

It is safe to say that I don't have a clue about this lot!

And my apologies for my mistake with the Matrix inversion. I simply wasn't understanding what I was looking at! With hindsight I should have searched out an on-line solver - they do exist, I just found one and it, of course, agreed with your answer.

ANSWER: It seems like this one is all about probability.

First, let's start out by explaining density functions:

When a random event takes finitely many values, each value can simply be assigned a probability. But when that event is continuous (takes a range of values), you can usually only define the probability of falling between two values (i.e. "what is the probability of being between 1.4 and 1.5m tall?").

The probability density function (PDF) is the one that looks like what you expect -- it is a curve you can draw where the total area under the curve is 100% (i.e. 1) and you can cut up the range of values to define whatever other probabilities you want. To do this, you normally perform the calculus operation of integration.

The cumulative density function (CDF) is basically the indefinite integral of the PDF. In this case, you can simply use F(b)-F(a) to determine the probability of falling into the interval a<x<b.

So for #1, you can compute P(y>1) by really finding P(y<1), which is F(1).

F(1) = (k-3)/4 = 0.2A

Of course, you need to know what k is! But you can do that by realizing that F(2)=1 (it has to top-off at 1, and it also has to bottom out - going to the left - at zero).

F(2) = (8-16+2k)/4 = 1

This gives k = 6 which means F(1) = 3/4.

However, the probability we wanted was really 1-F(1), which is 1/4.

Thus 0.2A = 1/4, so A = 5/4 = 1.25. Or maybe "A" is a digit, in which case A=5.



For #2, we are looking at a "binomial" distribution -- data is either "yes" or "no" (well, apparently all of it is "no"). The "p" value is the probability of our outcome occurring, meaning that something "normal" is going on when the p values are not too small, based to our assumptions. For example, if we flip a coin 10 times and get 6 heads, that's normal (even though you would maybe expect 5 heads) because p=0.2 (a 20% event is not especially unlikely, especially considering the most likely outcome -- 5 heads -- has probability 25%).

If the actual probability of yes votes is 45% (and so 55% for no), then every time we select a no vote, that is a 55% likely event. So, for example, the probability of selecting three no votes is 0.55^3. After eight no votes, you wind up with 0.55^8 = 0.0083... meaning this is below the 1% threshold. So B=8.



In #3 the topic is confidence intervals. We have some data, as well as the standard deviation. There is <a href=https://en.wikipedia.org/wiki/Confidence_interval#Practical_example">an example</a> on Wikipedia. There are some very standard formulas here, including some here. We are using an exponential distribution because time-to-fail statistics are exponentially distributed -- not normally distributed.

The two chi-squared values are 162.7279 and 241.0578 (try here or here) and so you get as your intervals:

2*1740000/162.7279 = 21385
2*1740000/241.0578 = 14436

(21385 - 14436)/140 = 6949/140 = 49.64

I do not know how you are supposed to interpret this, since your answers are supposed to be integers. I am also highly suspicious of this question because it provides too much information. The standard deviation is not required to solve this problem. See another reference here which very clearly explains how to derive these formulas, which require only the sample mean and number of samples. I suspect whoever wrote these problems might be treating this as a normally distributed random variable, which as I noted, is not correct. If you can't find any typos in the question, I will take another look and try to find my error. I will also do the problem as if the variable is normally distributed, anticipating that this might be the error the question-writer has made.



For #4 you have two separate (independent) variables -- male height and female height.

The sums (and differences) of normal random variables are also normal random variables, so in particular if M is the male height and F the female, you want to really examine M-F. This has mean 177-163 = 14 and sd = √(5^2 + 4^2) = √(41), approximately 6.4 (see here).

You just want to compute when is our new distribution M-F>0, which you can compute on any standard table. Here is one of many tools online you can play with to do this computation. The z-score here is 14/√(41) which is about 2.1864 (so this is two standard deviations away from average). Accordingly, the probability is 98.57. I'm not sure if D is a digit or a number, so I guess D=9 if you round it up? Or if D is supposed to be 0.9*D=98.57 then it's something else.



#5 can be done on the same tool. Here you want an interval within 0.6/0.5 = 1.2, which according to the same calculator as the previous problem will be 0.7698... so E (clearly a digit in this case) is 7.

---------- FOLLOW-UP ----------

QUESTION: This is not really a follow up question.

A, D and E are digits.

For Q4 he states "When written to 2 decimal places...". In past questions when a number of decimal places has been specified, rounding to that number of decimal places has been what was required.

Learning my lesson, I trusted your answers for A, B, D and E and deduced the possible values for C (just 6). I calculated all possible final co-ordinates, plotted them in GoogleEarth and realised that only one looked plausible, and that was the correct one.

So: A=5, B=8, C=14, D=9, E=7.

That's all I need. If you want to look at question 3 again it would only really be for your own interest....

Answer
Based on my reading of the question, I am fairly sure the person who wrote it assumed that the time-to-failure is normally distributed. It's a common mistake -- assuming all random events have some normal distribution. Technically, nothing is wrong with whatever math the person is doing, but by not even telling you to use a normal distribution, it makes the problem a bit deceptive. Time-to-fail is almost always exponentially distributed. Glad to hear you worked it out the "old fashioned" way.

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Clyde Oliver

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I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.

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