1. Let P(A) = .65, P(B) =.9, P(A and B) =.75. What is P(A or B)? (A) 1.55 (B) .5
ID: 3103410 • Letter: 1
Question
1. Let P(A) = .65, P(B) =.9, P(A and B) =.75. What is P(A or B)?(A) 1.55 (B) .585 (C) .65 (D) .8 (E) not enough information to tell
2. What is the median of the scores 58, 35, 78, 90, 45?
(A) 78 (B) 61.2 (C) 58 (D) 5 (E) 45
Use this frequency distribution for problems 3, 4, and 5.
Number of
Siblings Frequency
0? 2
1? 7
2 ? 4
3 1
3.?The approximate mean number of siblings? (A) 1.5 (B) 14 (C) 1.3 (D) 1 (E) 18
Explanation / Answer
The best way to picture problem 1 is to divide a region into 4 sections: A and B, A only, B only, and neither. Upon examination of this, there are 2 sections that P(A) account for ( A only and both A and B), so the 2 groups total to 0.65 of the entire region. Similarly, groups 'B only' and 'A and B' total to 0.9. Upon further inspection, note that 'A or B' consists of the regions 'A only', 'B only', and "A and B". So P(A or B) = P(A only) + P(B only) + P(A and B). However, note that we need a 2nd P(A and B) in the equation to use both P(A) & P(B) values; to accomodate, I'm going to stick a "+ P(A and B) - P(A and B)" at the end. After that and rearranging to get P(A or B) = [P(A only) + P(A and B)] + [P(B only) + P(A and B)] - P(A and B) = P(A) + P(B) - P(A and B) = 0.65 +0.9 -0.75 =0.8. Based on this, the answer is D. However, I must question the legetimacy of things with the following observation/question: How can P(A and B) > P(A)??. If they want you to answer E, I'd say that P(A and B) > P(A) plays a part in that. For problem 2, the median is simply the middle number (or average of middle 2, if even). 35 and 45 are the 2 small numbers, 78 and 90 the 3 big ones, leaving 58 the median. For 3, mean = total/number = [2(0) +7(1) +4(2) +1(3)]/[2+7+4+1] = [7+8+3]/14 =9/7.
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