STAT 230 Week 8 Quiz 2 Solution Correct Answers
- Statistics
- University of Waterloo
- 5 pages
- Shared May 2025
Week 8, quiz 2 answers for a statistics course.
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STAT 230 Week 8 Quiz 2 Solution Correct Answers 1. A pollster selected 4 of 7 available people. How many different groups of 4 are possible? 2. Your firm has a contract to make 2000 staff uniforms for a fast – food retailer. The heights of the staff are normally distributed with a mean of 70 inches and a standard deviation of 3 inches. What percentage of uniforms will have to fit staff shorter than 67inches? What percentage will have to be suitable for staff taller than 76 inches.? 3. The industry standards suggest that 20% of new vehicles require warranty service within the first year. A dealer sold 20 Nissans yesterday. Use equation for Binomial Probability for part a) and Table II for part b) & c). Show work! a) What is the probability that none of these vehicles requires warranty service? Use the Binomial equation for P(X=0). b) What is the probability that exactly one of these vehicles requires warranty service? c) Determine the probability 3 or more of these vehicles require warranty service. d) Compute the mean and std. dev. of this probability distribution. 4. Allen & Associates write weekend trip insurance at a very nominal charge. Records show that the probability a motorist will have an accident during the weekend and will file a claim is quite small (.0005). Suppose Alden wrote 400 policies for the forthcoming weekend. Compute the probability that exactly two claims will be filed using the equation for Poisson Probability. Note: The symbol λ is the mean (expected value) which we used as μ = np. So λ is nothing more than the mean number of occurrences (successes = np) in a particular interval. Get the probability that the number of claims is at least 3 from Poisson Tables. 5. Given a standard normal distribution, determine the following. Show Table Values used in each part. a) P(Z<1.4) = b) P(Z>1.4) = 1 - P(Z<1.4) = c) P(Z< - 1.4) = d) P( - 0.50<Z<1.0) = e) P(0.50<Z<1.5) =
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