MATH-1P98-WEBWORK-3-ATTEMPT-2

MATH-1P98-WEBWORK-3-ATTEMPT-2 - Aanjaneya Sood Assignment...

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Aanjaneya Sood MATH1P98D02FW2016 Assignment WeBWorK Assignment 3 due 10/30/2016 at 01:32am EDT Problem 1. 10. (1 pt) An automobile insurer has found that repair claims are Normally distributed with a mean of $870 and a standard deviation of $820. (a) Find the probability that a single claim, chosen at ran- dom, will be less than $850. ANSWER: (b) Now suppose that the next 90 claims can be regarded as a random sample from the long-run claims process. Find the probability that the average ¯ x of the 90 claims is smaller than $850. ANSWER: (c) If a sample larger than 90 claims is considered, there would be chance of getting a sample with an average smaller then $850. (NOTE: Enter ”LESS”, ”MORE” or ”AN EQUAL” without the quotes.) Answer(s) submitted: 0.492022 0.409046 LESS (correct) Correct Answers: 0.492022 0.409046 LESS Problem 2. 2. (1 pt) A survey of Amazon.com shoppers reveals the following probability distribution of the number of books per hit: X 0 1 2 3 4 5 6 7 P(X) 0.32 0.201 0.071 0.277 0.029 0.025 0.01 0.067 Find the following probabilities: A. What is the probability that an Amazon.com visitor will buy four books? Probability = B. What is the probability that an Amazon.com visitor will buy eight books? Probability = C. What is the probability that an Amazon.com visitor will not buy any books? Probability = D. What is the probability that an Amazon.com visitor will buy at least one book? Probability = Answer(s) submitted: 0.029 0 0.32 0.68 (correct) Correct Answers: 0.029 0 0.32 0.68
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