practice_midterm_solutions

practice_midterm_solutions - SDS 321 Practice midterm Note...

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SDS 321: Practice midterm Note: This practice exam is longer than your actual exam will be! The actual exam will be 5 questions 1. Find a formula for the probability that among a set of n people, at least two have their birthdays in the same month of the year (assuming the months are equally likely for birthdays). Assume n 12 (for n > 13 the probability is equal to 1.0). Let A be the event “at least one matching b-day month”. It is easier to find P ( A c ) – the probability of“no matching b-day months.” First count the number of outcomes in A c . It is 12 · 11 . . . (12 - ( n - 1)) = 12! / (12 - n )!. The number of possible ordered combinations of birthday monts is 12 n . Therefore P ( A c ) = 12! / (12 - n )! 12 n and p ( A ) = 1 - p ( A ) = 1 - 12! / (12 - n )! 12 n 2. How many ways can we arrange the letters of the word “ALPHABET” so that the first and last letters are vowels? There are three vowels - A,A,E. So, 3 possible ways of having vowels at both ends: A******A, A******E, E******A. For each of these combinations, we have 6 remaining, different, letters... so in each case, ****** can be any of 6! permutations of the remaining letters. So, there are 3 · 6! = 2160 combinations. 3. If a coin is tossed a sequence of times (infinitely many times), what is the probability that the first head will occur after the 5th toss, given that it has not occurred in the first 2 tosses? We know the first 2 coin tosses are tails. The probability of this is P ( T, T ) = (1 - p ) 2 We want the probability that the first 5 are tails, given the first two are heads. The unconditional probability P ( T, T, T, T, T ) = (1 - p ) 5 . So, P (( T, T, T, T, T ) | ( T, T )) = P ( T,T,T,T,T ) P ( T,T ) = (1 - p ) 3 4. A continuous random variable has PDF f X ( x ) = ( a + bx 2 0 < x < 1 0 otherwise (a) If E [ X ] = 3 5 , find a and b . PDF must integrate to 1, so R 1 0 a + bx 2 dx = ax + b 3 x 3 1 0 = a + b/ 3 = 1, so b = 3(1 - a ) E [ X ] = 3 / 5, so R 1 0 ax + bx 3 dx = a 2 x 2 + b 4 x 4 1 0 = a 2 + b 4 = 3 5 1
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So, a 2 + 3(1 - a ) 4 = 3 5 10 a + 15(1 - a ) = 12 a = 3 5 b = 3(1 - a ) = 6 5 (b) What is var( X )? If you couldn’t find a and b , you can leave them as algabraic variables. var( X ) = E [ X 2 ] - E [ X ] 2 . E [ X 2 ] = R 1 0 x 2 f X ( x ) dx = R 1 0 ax 2 + bx 4 dx = h ax 3 3 + bx 5 5 i 1 0 = a 3 + b 5 = 1 5 + 6 25 = 11 25 , so var( X ) = 11 25 - ( 3 5 ) 2 = 11 - 9 25 = 2 25 .
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