STAT 324 Sample midterm problems1. 1/7 of men and 1/9 of women are color-blind. 60% of population are females.(a) A person is chosen at random and that person is color-blind. What is the probability that theperson is male?Solution. Let C be the event the person is color blind. Let M and F be the events the person ismale and female respectively. Then we need to computeP (M|C) =P (M ∩ C)P (C)=P (C|M)P (M)P (C|M)P (M) + P (C|F )P (F )=17·41017·410+19·610.(b) A person is chosen at random and that person is male. What is the probability that the personis color-blind?Solution. P (C|M) =17·410.2. The life time X of an electronic component follows the distribution f(x) = e−xfor x > 0 andf(x) = 0 otherwise. Three of these components operate independently in a piece of equipment.The equipment fails if at least two of the components fail. Find the probability that the equipmentwill operate for at least 2 hours without failure.Solution. The probability of a single component failure isp = P (X < 2) =Z20e−xdx = 1 − e−2.The equipment will fail if at least two out of 3 components fail. So the probability of the equipmentfailure isµ32¶p2(1 − p) +µ33¶p3= 3p2(1 − p) + p3.The probability of the equipment operating for at least 2 hours is then1 − 3p2(1 − p) − p3.3. The achievement scores for a college entrance examination are normally distributed with mean 70and variance 10. What fraction of the scores lies between 75 and 80? You may use the following Routput.> pnorm(c(1:4),1,2)[1] 0.50 0.69 0.84 0.93Solution. Let X be the achievement score and let Z =X−7010∼ N(0, 1). Need to findp = P (75 − 7010< Z <80 − 7010) = P (0.5 < Z < 1).However, the R output is given in terms of W ∼ N(1, 22). The relationship between W and Z isW = 2Z + 1. Sop = P (2 · 0.5 + 1 < W < 2 · 1 + 1) = P (2 < W < 3)= pnorm(3, 1, 2) − pnorm(2, 1, 2) = 0.84 − 0.69 = 0. 15.4. The density function of he random variable X is f(x) =x2for 0 < x < 2 and f(x) = 0 otherwise.(a) Find the cumulative distribution function of X and plot its graph.Solution. FX(x) =Rx0x2dx =x24for 0 < x < 2, FX(x) = 0 for x ≤ 0 while FX(x) = 1 for x ≥ 2.(b) Derive the probability density function of Y = 2X +1 using the cumulative distribution functionof X.Solution. FY(y) = P (Y ≤ y) = P (3X + 1 ≤ y) = FX(y −13) =14³y−13´2for 1 < y < 5. fY(y) =dFY(y )dy=16y − 13for 1 < y < 5.5. Let X and Y have joint probability densityf(x, y) = cx, for 0 < x < 1, 0 < y < 1 and 1 < x + y < 2and f(x, y) = 0 otherwise.(a) Find constant c.Solution. Draw the domain where the joint density is defined.1 =Z10Z11−ycx dx dy =c2Z10(1 − (1 − y)2) dy =c2(y +13(1 − y)3)¯¯¯10=c2(1 −13).So c = 3.(b) Find the marginal distributions of X and Y .Solution. fX(x) =R11−x3x dy = 3x2for 0 < x < 1. fY(y) =R11−y3x dx =32(1 − (1 − y)2) for0 < y < 1.(c) Find the conditional probability P (X < 1|X + Y > 1).Solution. This is triviallyP (X < 1, X + Y > 1)P (X + Y > 1)= 16. Let X1, · · · , Xnbe a random sample from Bernoulli distribution with parameter p. (a) What isE(S2/p2)? S2is the sample variance. Explain your results.Solution. The sample variance is an unbiased estimator of the population variance. Hence E(S2/p2) =E(S2)/p2= Var(Xi)/p2. The variance for a Bernoulli random variable can be esily computed asVar(Xi) = p(1 − p). So E(S2/p2) = (1 − p)/p.(b) Find an unbiased estimator for p2. Explain your results.Solution. There are many solutions to this problem. We know S2and¯X will be unbiased estimatorsof population variance p(1 − p) and mean p respecively. So E(¯X) − E(S2) = p − p(1 − p) = p2.Hence,¯X − S2is an unbiased estimator for p2.7. Let X1, X2be a random sample from N(0, 1/θ). Note that the sample size is 2 and the densityfunction for Xiis f(xi) =√θ√2πexp(−θx2i/2). Find the likelihood function and use it to obtain themaximum likelihood estimator of θ.Solution. The likelihood function is L(θ) = θ/(2π) exp(−θ(x21+ x22)/2). Get log-likelihood functionln L(θ ) = const + ln θ − θ(x21+ x22)/2. Differentiate with respect to θ to get the maximum:d ln L(θ)dθ=1θ−12(x21+ x22) = 0.Solving the equation, we getˆθ = 2/(x21+