Math November 17, Test 2 Solutions

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1 Math November 7, - Test Solutions Name: Read these instructions carefully: The points assigned are not a guide to the difficulty of the problems. If the question is multiple choice, there is a penalty for wrong answers, so that your expected score from guessing at random is zero. No partial credit is possible on multiple-choice and other no-work-required questions. You must show your work to obtain full credit.. ( points) A soft-drink machine can be regulated so that it discharges an average of µ ounces per cup. If the ounces of fill are normally distributed with standard deviation. ounce, give the setting for µ so that 6-ounce cups will overflow only % of the time. You must show your work to obtain full credit. Let Y be the number of ounces of soda discharged by the machine. We are given that Y is normally distributed with mean µ and σ =.. We wish to find µ so that P (Y > 6) =.. Observe that Y > 6 is equivalent to Y µ σ > 6 µ σ and that Z = (Y µ)/σ is a standard normal random variable. To have P (Z > z) =. we must have (from the table) Z.. So if we select µ so that 6 µ σ =. we will have the required condition. Solving, it follows that µ 5... ( points) Let Z be a standard normal random variable, i.e. Z N(, ). Which of the following numbers is closest to P (Z > 4)? (a).5 (b).475 (c).95 (d).5 Observe that Z > 4 is equivalent to Z >. By the 95% rule, this probability is approximately 5%. So the answer is (d).. ( points) Let Y be a normal random variable with mean and variance 9. What is the median value of Y? (No explanation required.) The normal random variable is symmetric about its mean, i.e. 5% of the probability density is below the mean and 5% above. So the mean and median are the same and the answer is. 4. (8 points) Annual household incomes in a city have approximately a gamma distribution with parameters α = and β =. a. (4 points) Find the mean and variance of these incomes. (No explanation required.)

2 Using the results obtained for the mean and variance of the gamma distribution, we have that the mean is αβ = $, and that the variance is αβ =,, (in units of dollars-squared). b. (4 points) Would you expect to find many incomes in excess of $4, in this city? Why (or why not)? Note that the standard deviation of these incomes is $5, 5. So an income standard deviations above the mean is $4,. Using the 5% rule, and noting that the gamma distribution is approximately symmetric for these parameter values and this distance from the mean, we should expect about.5% of the incomes to be above $4,. Above $4,, we should expect more, but still a small percentage, not more than 5%. (In fact it is about 4.%; you could get an approximate value by noting that $4, is more than.8 standard deviations from the mean and using the table for the normal distribution.) If you got this far, you will get full credit. You can argue that less than 5% is not many. Or you can argue that by a city typically has a large population (say,) and that one will thus be able to find thousands of incomes at the required level, and that thousands is many. 5. A machine used to fill cereal boxes dispenses, on average, µ ounces per box. The manufacturer wants the actual ounces dispensed Y to be within ounce of µ at least 75% of the time. a. (4 points) Let σ be the standard deviation of Y, and state Tchebysheff s theorem for Y. (Either formulation is acceptable.) If Y is a random variable with mean µ and standard deviation σ, then P ( Y µ < kσ ) k. Note that the sentence beginning If Y is part of the theorem: it is the hypothesis. Not every random variable has a standard deviation, or even a mean! b. (4 points) Use your answer from a. to determine the largest value of σ that can be tolerated if the manufacturer s objectives are to be met. We need to choose k so that =.75, and σ so that = It follows that k = k kσ and σ = /. 6. ( points) A random variable Y has cumulative distribution function F given by the formula: y < y +. y < F (y) = y 4 y < y Which of the following numbers is closest to P (Y = 4 )? (a).5 (b). (c). (d) Observe that P (Y = 4 ) = lim x F ( 4 ) F ( 4 x) = =.5

3 so that the answer is (b). 7. ( points) Let Y and Y be random variables whose joint distribution function is given by { 6y y y y f(y, y ) = otherwise Are Y and Y independent? (Just a yes or no answer; no reason required.) No: note the region of definition. 8. ( points) A point (Y, Y ) is chosen at random (according to the uniform distribution) from the triangle with vertices (, ), (, ), and (, ). That is, Y and Y are random variable whose joint distribution is defined by the previous sentence. a. (4 points) Write the definition of the conditional density function f(y Y = y ). where f (y ) = f(y Y = y ) = f(y, y ) f (y ), f(y, y )dy = y and f(y, y ) is if (y, y ) is inside the triangle and outside. It follows that f(y Y = y ) = y b. (6 points) Determine the conditional density function f(y Y = y ). Indicate clearly the values of y for which the conditional density function is defined. Note that P (Y ) = P (Y < ) =, so the conditional density function is not defined for y or y <. For y < we have: f(y Y = y ) = y c. (4 points) Write P (Y Y = ) in terms of the conditional density function. P (Y Y = / ) = f(y Y = /)dy = / / / 4 dy d. (6 points) Find P (Y Y = ). You can do the integral from the previous part or use geometry to get the answer: /8. 9. (8 points) Suppose a random variable Y has a probability density function given by { ky e y/ y f(y) = y < a. (6 points) Find the value of k that makes f a probability density function. You must give a numerical answer. Observe that f is a multiple of the density function for the gamma distribution with parameters α = 4, β =, and so the constant k must be the same as it is in that density function to make the integral of f (over the whole real line) exactly. Thus we have k = β α Γ(α) = 4 Γ(4) = 4 (4 )! == 96

4 b. ( points) Does Y have a χ distribution? (Just a yes or no answer; no explanation required.) Yes. The definition of the χ -distribution with ν degrees of freedom is Γ(ν/, ), so take ν = 8. In each of the following problems you are asked to identify the probability distribution giving rise to the graph preceding the question. f (x) x. (4 points) From which of the following distributions does the probability density function graphed above arise? The near-vertical lines in the graph are meant to be vertical, or better yet nonexistent I m still having trouble getting R to display graphs in the desired form. (a) the beta distribution with parameters α = 5 and β = (b) the uniform distribution on the interval (, ) (c) the gamma distribution with parameters α = and β = (d) the beta distribution with parameters α = and β = (e) the χ distribution with 4 degrees of freedom. Answer: (d) 4

5 g (x) x. (4 points) From which of the following distributions does the probability density function graphed above arise? (a) the beta distribution with parameters α = and β = (b) the beta distribution with parameters α = 5 and β = 5 (c) the normal distribution with parameters µ =. and σ =. (d) the beta distribution with parameters α = and β = 5 (e) the normal distribution with parameters µ =.4 and σ =.5 Answer: (d) f (x) x. (4 points) From which of the following distributions does the probability density function graphed above arise? (a) the gamma distribution with parameters α = and β = 5

6 (b) the beta distribution with parameters α =.5 and β = (c) the uniform distribution on [, ] (d) the gamma distribution with parameters α = and β = (e) the chi-square distribution with 8 degrees of freedom Answer: (e). ( points) Two gamblers A and B each toss a coin until the first head appears. The coin they use is biased, so that the probability of heads is p and the probability of tails is q = p. Let Y be the number of times A tosses the coin and Y be the number of times B tosses the coin. You may assume in what follows that Y and Y are independent and you may take as given that the mean and variance of the geometric random variable with probability of success p are /p and q/p respectively. a. ( points) Find E[Y Y ]. E[Y Y ] = E[Y ] E[Y ] = p p = b. (7 points) Find V [Y Y ]. Observe that knowing Y is equivalent to knowing Y, so Y and Y are independent. Therefore, the variances add and we have V [Y Y ] = V [Y + ( Y )] = V [Y ] + V [ Y ] We showed V [ay ] = a V [Y ] in a homework problem, and using this we have: V [Y ] + ( ) V [Y ] = V [Y ] + V [Y ] = q p. Of course this also follows from the theorem on the bilinearity of covariance in Section ( points) Let Z be a normal random variable with mean and variance. a. ( points) Write an expression for E[Z 4 ] as an integral. E[Z 4 ] = z 4 e z dz b. (8 points) Find E[Z 4 ]. Suggestion: find a way to use what you know about the gamma function. The answer is. There are at least ways you could have arrived at this answer. The first two are require very specific pieces of knowledge, but once you recall the relvant facts, the work is very easy. The suggestion uses only things that you are absolutely supposed to know, but requires more work. Method : By theorem 7. in your text, which was discussed in class, Z has the χ [] distribution, which is the same as the Γ(, ) distribution, so using the equation E[Z 4 ] = E[(Z ) ] = V [Z ] + E[Z ] and the known mean and variance of the gamma distribution we get. Method : Take the known moment generating function (E t / ) of Z, take 4 derivatives, and evaluate at t =. Of course this requires that you remember the moment generating function! 6

7 Method : Work directly with the integral from part(a), and change it so that we can use what we know about the gamma function to evaluate it. First observe that the function to be integrated is even, so we may write E[Z 4 ] = z 4 e z dz Change variables in the integral above, setting y = z, dy = zdz, and we get E[Z 4 ] = y e y Note that z = y and cancel the s to get: E[Z 4 ] = y 5 dy y = e y dy = y e y dy z y 5 e y dy This integral can now be computed using what we know about the gamma function and the density function for the gamma distribution. 7