Terri Vogel, an amateur motorcycle racer, averages 129.71 seconds per 2.5 mile lap (in a seven-lap race) with a standard deviation of 2.28 seconds. The distribution of herrace times is normally distributed. We are interested in one of her randomly selected laps.O Part (a)Part (b)Give the distribution of X.N)129.72.28O Part (c)Find the percent of her laps that are completed in less than 130 seconds. (Round your answer to two decimal places.)54.62X %Part (d)The fastest 5% of her laps are under how many seconds? (Round your answer to two decimal places.)125.60sec

Answered: Image /qna-images/answer/6a230d18-06d0-4344-8a11-c166fa33cb3e.jpg

Answered: Terri Vogel, an amateur motorcycle…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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uppose that the distance of fly balls hit to the outfield (in baseball) is normally distributed with a mean of 269 feet and a standard deviation of 43 feet. Let X be the distance in feet for a fly ball.b. Find the probability that a randomly hit fly ball travels less than 307 feet. Round to 4 decimal places. c. Find the 90th percentile for the distribution of distance of fly balls. Round to 2 decimal places. feet
ent will 3. The average price of a personal computer is $949. If the computer prices are approximately normally distributed with a standard deviation of $100... a. the top 20% of personal computers cost more than what amount? b. the least expensive 10% of personal computers cost less than what amount?
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A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1458 and the standard deviation was 312. The test scores of four students selected at random are 1860, 1220, 2150, and 1340. Find the z-scores that correspond to each value and determine whether any of the values are unusual. The z-score for 1860 is (Round to two decimal plaes as needed.)
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