Stat 400: Georgios Fellouris Homework 5 Due: Friday 4 th, 017 1. A exam has multiple choice questios ad each of them has 4 possible aswers, oly oe of which is correct. A studet will aswer all questios completely at radom, idepedetly of oe aother. Solutio: Let X be the umber of correct aswers i the = questios. Sice the studet aswers all questios completely at radom, the the probability of a correct aswer is p = 1/4 = 0.5 for each questio, ad the aswers to the differet questios are idepedet. This meas that X is a Biomial radom variable with parameter = ad p = 0.5, thus ( ) f X (x) = P(X = x) = p x (1 p) x, x = 0, 1...,. x (a) What is the expectatio ad the stadard deviatio of the umber of correct aswers? Solutio: Sice X Biomial(, p), the E[X] = p = 0.5 = 1.5 ad σ X = p(1 p) = 0.5 0.75 = 1.15 1.01. (b) What is the probability that the studet aswers more tha half of the questios correctly? Solutio: Usig the formula above, P(X > 3) = P(X = 4) + P(X = 5) + P(X = ) ( ) ( ) 1 4 ( ) 3 ( 1 = + 4 4 4 4 ) 5 3 4 + ( ) 1 4 = 15 9 + 3 + 1 4 = 154 409 = 77 048 0.038. We say that a r.v. X is uiform o {1,..., } if it is equally likely to take ay of the values i this set, i.e., if its pmf is f X (x) = 1/ for x = 1,...,. (a) Give a example of a uiform radom variable. Solutio: Whe rollig a fair die, the umber that appears o the side show as a result of the toss is a uiform radom variable o {1,..., }. (b) Compute the expected value ad the stadard deviatio of a uiform r.v. o {1,..., }.
Stat 400 Homework 5, Page of 5 Due: Friday 4 th, 017 Solutio: E[X] = E[X ] = x f X (x) = x 1 = 1 +... + = x f X (x) = x 1 ( + 1) = + 1 = 1 +... + ( + 1) ( + 1) = ( ) Var[X] = E[X ] E[X] ( + 1) ( + 1) + 1 = = + 1 [ + 1 + 1 ] 3 = + 1 1 = 1 1. = ( + 1) ( + 1) (c) Plot the cumulative distributio fuctio (CDF) of a uiform r.v. o {1,, 3}. Solutio: Figure 1: CDF of a uiform r.v. o {1,, 3} 3. Females comprise about 50.8% of the U.S. populatio of (roughly) 308 millios. A sample of 10 Americas is selected at radom. (a) What is the exact pmf of the umber of females i the sample? approximate pmf for the same radom variable? Ca you suggest a Solutio: Let X be the umber of females i the sample. The exact pmf of X is the hypergeometric distributio ( N1 ) ( x N ) x f X (x) = P(X = x) = ( N, x = 0, 1,..., ) where = 10 is the sample size, p = 0.508 the proportio of females, N = 308 10 is the total populatio size, N 1 = pn is the umber of females i the US, ad N = N N 1
Stat 400 Homework 5, Page 3 of 5 Due: Friday 4 th, 017 is the umber of males i the US. Sice N 1, N are much larger tha, the the above hypergeometric distributio ca be approximated by the Biomial with parameter ad p. (b) What is the expected umber of females i the sample accordig to each pmf i (a)? Solutio: For both pmf s i (a), the expected value is equal to p = 10 0.508 = 5.08. (c) What is the probability to have at most 8 females i the sample? approximate pmf for this computatio. You ca use the Solutio: For the approximate pmf, oe solutio is usig the complemet rule P(X 8) = 1 P(X 9) = 1 P(X = 9) P(X = 10) = 1 10 0.508 9 0.49 0.508 10 98.8% 4. Ex. 5, Sec. 3.4. The College Board reports that % of the (roughly millio) high school studets who take the SAT each year receive special accommodatios because of documeted disabilities (Los Ageles Times, July 1, 00). Cosider a radom sample of 5 studets who have recetly take the test. What is the probability that (a) exactly 1 received a special accommodatio? Solutio: Let X be the umber of studets i the sample of 5 studets who have recetly take the test that received special accommodatios because of documeted disabilities. Sice the populatio size of millio studets is much larger tha the sample size of 5, we ca assume that X is Biomial with parameters = 5 ad p = 0.0. Thus, P(X = 1) = 5 0.0 0.98 4 30.8% (b) at least received a special accommodatio? Solutio: From the complemet rule we have P(X ) = 1 P(X 1) = 8.9%, sice P(X 1) = P(X = 0) P(X = 1) = 0.98 5 5 0.0 0.98 4 0.03 + 0.308 = 91.1%. (c) the umber amog the 5 who received a special accommodatio is withi stadard deviatios of the umber you would expect to be accommodated? Remark: By withi we mea iclusive at the ed poits. Solutio: The expected value of X is µ X = E[X] = p = 5 0.0 = 0.5
Stat 400 Homework 5, Page 4 of 5 Due: Friday 4 th, 017 ad the stadard deviatio of X is σ X = Var[X] = p(1 p) = 5 0.0 0.98 = 0.49 = 0.7. We are iterested i the probability of the evet { X µ X σ X )} = {µ X σ X X µ X + σ X } = {0.5 0.7 X 0.5 0.7} = { 0.9 X 1.9} = {0 X 1}, where the last equality holds because X ca oly take iteger values. The, from (b) it follows that P( X µ X σ X ) = P(X = 0) + P(X = 1) 91.1% (d) Suppose that a studet who does ot receive a special accommodatio is allowed 3 hours for the exam, whereas a accommodated studet is allowed 4.5 hours. What would you expect the average time allowed for the 5 selected studets to be? Solutio: Call Y the total umber of hours that all studets i the sample were allowed. The, the average umber of hours the studets i the sample were allowed is Y/. Now, Y ca be expressed i terms of X as follows: Y = 4.5 X + 3 ( X) = 3 + 1.5 X, therefore from the liearity of the expectatio we have E[Y ] = 3 + 1.5 E[X] = 3 + 1.5 p ad E[Y/] = 3 + 1.5 p = 3 + 1.5 0.0 = 3.03. 5. Ex. 1, Sec. 3.4. A studet who is tryig to write a paper for a course has a choice of two topics, A ad B. If topic A is chose, the studet will order books through iterlibrary loa, whereas if topic B is chose, the studet will order 4 books. The studet believes that a good paper ecessitates receivig ad usig at least half the books ordered for either topic chose. If the probability that a book ordered through iterlibrary loa actually arrives i time is p, ad books arrive idepedetly of oe aother, which topic should the studet choose to maximize the probability of writig a good paper whe (a) p = 0.5, (b) p = 0.9? Solutio: Let X A (resp. X B ) the umber of books that will arrive i time amog the oes that the studet will order if he does project A (resp. B). Sice books arrive idepedetly of oe aother, ad they all have the same probability to arrive i time, the X A is Biomial(, p) ad X B is Biomial(4, p). The, the probability of a good paper is P(X A 1) (resp. P(X B ) for topic A (resp. topic B). Specifically, P(X A 1) = 1 P(X A = 0) = 1 (1 p) P(X B ) = 1 P(X B 1) = 1 P(X B = 0) P(X B = 1) = 1 (1 p) 4 4p(1 p) 3.
Stat 400 Homework 5, Page 5 of 5 Due: Friday 4 th, 017 Whe p = 0.5, P(X A 1) = 1 0.5 = 75% P(X B ) = 1 0.5 4 4 0.5 4 = 8.75%, i which case i order to maximize the probability to write a good paper the studet should choose topic A. O the other had, whe p = 0.9, P(X A 1) = 1 0.1 = 99% P(X B ) = 1 0.1 4 4 0.99 0.1 3 99.% i which i order to maximize the probability to write a good paper the studet should choose topic B.