An airline reports that it has been experiencing a 20% rate of no-shows on advanced reservations. Among 230 advanced reservations, find the probability that there will be less than 42 no-shows. Use continuity correction and the standard normal distribution table to approximate the binomial probability.Options:A. $ P(X < 41.5) = 0.3014 $B. $ P(X < 41.5) = 0.2296 $C. $ P(X < 42.5) = 0.3046 $D. $ P(X < 42) = 0.2546 $E. $ P(X < 42.5) = 0.2810 $

Answered: Image /qna-images/answer/f6def6bf-b8a9-4f94-bddd-772f6f7da63c.jpg

Answered: 4. An airline reports that it has been…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The average student loan debt for college graduates is $25,750. Suppose that that distribution is normal and that the standard deviation is $10,350. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. Find the probability that the college graduate has between $22,700 and $30,000 in student loan debt. (------) The middle 20% of college graduates' loan debt lies between what two numbers? Low: $(______) High: $(_____)
A dishwasher has a mean lifetime of 10 years with an estimated deviation of 2.05 years. The lifetime of a dishwasher is normally distributed. Supposed you are manufacturer and you take a sample of 10 dishwashers that you made. Find the probability that the sample would have a mean lifetime less than 9.73 years? Round to four decimal places.
Let XX represent the full height of a certain species of tree. Assume that XX has a normal probability distribution with a mean of 97.9 ft and a standard deviation of 5.4 ft.A tree of this type grows in my backyard, and it stands 86 feet tall. Find the probability that the height of a randomly selected tree is as tall as mine or shorter.P(X<86)P(X<86) = My neighbor also has a tree of this type growing in her backyard, but hers stands 112.5 feet tall. Find the probability that the full height of a randomly selected tree is at least as tall as hers.P(X>112.5)P(X>112.5) = Enter your answers as decimals accurate to 4 decimal places.
The average age of all employees is 25 years, and the standard deviation of the ages of all employees is 3 years. Suppose 100 employees are randomly selected.
A random number generator picks a number from 6 to 65 in a uniform manner. Round answers to 4 decimal places when possible. The mean of this distribution is . The standard deviation is . The probability that the number will be exactly 48 is P(x=48)=P(x=48)= . The probability that the number will be between 17 and 22 is P(17<x<22)=P(17<x<22)= . The probability that the number will be larger than 50 is P(x>50)=P(x>50)= . P(x>24∣x<57)=P(x>24∣x<57)= . Find the 21st percentile. Find the minimum for the upper quarter.
The percent of fat calories that a person consumes each day is normally distributed with a mean of about 35 and a standard deviation of about ten. Suppose that 9 individuals are randomly chosen. Let X = average percent of fat calories. For the group of 9, find the probability that the average percent of fat calories consumed is more than six. (Round your answer to four decimal places.)
Let XX represent the full height of a certain species of tree. Assume that XX has a normal probability distribution with a mean of 320 ft and a standard deviation of 45 ft.Find the probability that the height of a randomly selected tree is between 334 ft and 388 ft. P(334 < X < 388) = Enter your answer as a number accurate to 3 decimal places.
Suppose that the duration of a particular type of criminal trial is known to have a mean of 18 days and a standard deviation of 6 days. We randomly sample 9 trials.Give the distribution of ΣX.Find the probability that the total length of the 9 trials is more than 188 days. 70 percent of the total of 9 of these types of trials will last more than how long? (Round your answer to two decimal places.)days
Suppose that the weight of an newborn fawn is Uniformly distributed between 2.2 and 3.8 kg. Suppose that a newborn fawn is randomly selected. Round answers to 4 decimal places when possible. The mean of this distribution is The standard deviation is The probability that fawn will weigh exactly 2.4 kg is P(x = 2.4) = The probability that a newborn fawn will be weigh between 2.9 and 3.7 is P(2.9 < x < 3.7) = The probability that a newborn fawn will be weigh more than 2.72 is P(x > 2.72) = P(x > 2.8 | x < 3.5) = Find the 25th percentile.

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