Part lI. Multiple Choice. Circle the letter of the best answer. [30 points For P

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Part lI. Multiple Choice. Circle the letter of the best answer. [30 points For Problems 1-3: The data below are weights (in lbs) of boxes measured by certain instrument 93 96 97 99 10 102 102 103 3031 105 105 109 11 114 115 122 Summary statistics: n-16, = 167.6, D-176458 1. What is the sample standard deviation of the weilghts? A. 0.598 B. 0.773 C. 2.9 D. 10.475 2. A Normal probability plot of the data is given on the right. What are the coordinates of the point corresponding to 9.9 lbs? A. (-0.77,9.9) B. (0.22,9.9) C. (4,9.9) D. (0.77,9.9) A histogram of the data is given on the right. Suppose that the true average weight of all the boxes produced by the process is 11.75 lbs which is indicated 3. . weighing instrument? A. It is precise B. It is both indirect and inaccurate. C. It is imprecise but accurate. D. It is inaccurate. For Problems 4 and S: Suppose that P(A)-0.28, PAU B)-O.56 and P(ana)-0.13 4. What is P(B)? A. 0.14 B. 0.15 C. 0.28 D. 0.43 Are A and B mutually exclusive? A. No, because P(AnB) o B. No, because P()xP(B)- PAnB) C Yes, because P(An B)o D. Yes, because P(A)xP(B) PAn 8). 5.

Explanation / Answer

Answers below:

4.

P(A) = .28
P(A U B) = .56
P(A and B') = .13
P(A and B) = P(A) - P(A and B') = .28-.13 = .15
P(A U B) = P(A)+P(B)-P(A and B)
.56 = .28+P(B)-.15
P(B) = .56-.28+.15 = .43
P(B) = .43

Answer is D.

5. If A and B are mutually exclusive then P(A and B) = 0
So, if A and B aren't ME then they will have probability common with them
So, No, because P(A and B) !=0 is the right option

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