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5. A trucking company determined that, onan annual basis, the distance traveled per truck is normallydistributed with a mean of 60 thousand miles and a populationst...

Question

5. A trucking company determined that, onan annual basis, the distance traveled per truck is normallydistributed with a mean of 60 thousand miles and a populationstandard deviation of 16 thousand miles. If a sample of 16 trucksis selected,(a) What is the probability that the average distance traveledis less than 58 thousandmiles? (b) What is the probability that the average distance traveledis more than 62 thousand miles?(c) What is the probability that the average distance traveledis betwe

5. A trucking company determined that, on an annual basis, the distance traveled per truck is normally distributed with a mean of 60 thousand miles and a population standard deviation of 16 thousand miles. If a sample of 16 trucks is selected, (a) What is the probability that the average distance traveled is less than 58 thousand miles? (b) What is the probability that the average distance traveled is more than 62 thousand miles? (c) What is the probability that the average distance traveled is between 58 thousand miles and 62 thousand miles? (d) Do we need the Central Limit Theorem to solve (a), (b), and (c)? Why or why not? Explain.



Answers

5. A trucking company determined that, on an annual basis, the distance traveled per truck is normally distributed with a mean of 60 thousand miles and a population standard deviation of 16 thousand miles. If a sample of 16 trucks is selected, (a) What is the probability that the average distance traveled is less than 58 thousand miles?


(b) What is the probability that the average distance traveled is more than 62 thousand miles? (c) What is the probability that the average distance traveled is between 58 thousand miles and 62 thousand miles? (d) Do we need the Central Limit Theorem to solve (a), (b), and (c)? Why or why not? Explain.

The speed of the motor vehicles anomaly distributed with the mean of 64.2 Under Standard Division of 8.44. Computer these probabilities, we can transform them into a standard normal distribution. That is we subtract the mean and then divided by the standard deviation. Hence This probability equal this one. Since they is a standard normal random railroad, we can use the table in appendix C enhance it's probability vehicles 0.46 Well, what for power BI This probability because 0.3 085 prepare to see the probability is a difference of two probabilities, which equals points 75 For nine miners. Point For 641. Make she calls point 29 08

In problem 20 the speeds of course on the road are approximately normally distributed with new equals 58 km/h and the standard deviation sigma equals four kilometer parole birthday. We want to find the probability that randomly selected car is going to be between 60 and 65 kilometre brow first we should define the probability Tennessee function four. The variable X where X denotes the speed. Of course here we have P of X equals one divided by sigma, Multiplied by the Square Root of two. Boy applied by it was a bar of minus x minus meal, oil squared divided by two are deployed by sigma squared. This defines the dynasty function for the variable X. Then to get the probability for ex to be between 60 and 65, we integrate P of X From 60 to 65. This integration is calculated numerically. Then we can use a graphing calculator such as this must to get this probability let's entered isthmus, we have here P of X or F of X and we integrate from 60 to 65. Then we bought a 60 and would be 65. Then this a. Here represents the area which represents the probability and here we have the shaded area represent this. The probability It's about four point to succeed Equals open 268 or 26.8 for barbie. We want to determine the fraction That all cars are going slower than 52 km/h, which is the same as party but we won't the fraction which is the probability for ex To be smaller than 52. We integrate from the start of the interval to the end and the start of course is A steady state call from zero 2 52 for B of X dx. We can use this move again to get the integration limit like this and to get the barbershop or the fraction We bought equal zero and we would be equals 52. It's the shaded area here, We can see that it has an area of 0.067 0.067 and this is of an answer of party and this is an answer of party.

And this question we're told the useful life of a real type of tire is normally distributed with means 50,500 mi and standard deviation in 915 mi. And whereas for the probability that her tire will have a useful life between 57,000 and 58,000 miles, so we're going to convert that to our standard normal random variable so that Z Between nine is .53 10.53, which is the same as probability the less than 0.53 minus, probably TZ less than minus 0.53 And from our table, those values are 27 019 -1981, which is .403. Now in B, we're told that Hamlet pies for such tires, assuming their lifetimes our independence. So this is looking like a binomial distribution. So advise four tires, perhaps for the probability that all four will last for this long. So the probability we have P in our binomial distribution, .403, 8. And so our probability here that all four will last is basically .403, 8 to the power of four, Which gives us .0266.


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