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Fome MYUVUAssignments MG'4T--Hcmetor 7Inbax (1) _ wademisbaGe: Hemevcrk HeqPencohttps / Inewconnectmheducation com flowfconnecthtmlHomeworkSavejHelp Save Exit ...

Question

Fome MYUVUAssignments MG'4T--Hcmetor 7Inbax (1) _ wademisbaGe: Hemevcrk HeqPencohttps / Inewconnectmheducation com flowfconnecthtmlHomeworkSavejHelp Save Exit SubmitCheck my Wotk15 The weight of people smal town ceour known I0 be normally distributed win mean 186 pounds and standard ceviation 29 pcunds Tat thattakes people across the river; sign states; "Maximum capacity 434 pounds or 17 persons_ WhatE the probability that random sample 0f 17 persons will exceed the weignt limit of 3,4

Fome MYUVU Assignments MG'4T-- Hcmetor 7 Inbax (1) _ wademisba Ge: Hemevcrk Heq Penco https / Inewconnectmheducation com flowfconnecthtml Homework Savej Help Save Exit Submit Check my Wotk 15 The weight of people smal town ceour known I0 be normally distributed win mean 186 pounds and standard ceviation 29 pcunds Tat thattakes people across the river; sign states; "Maximum capacity 434 pounds or 17 persons_ WhatE the probability that random sample 0f 17 persons will exceed the weignt limit of 3,434 pounds? (You may find useful t0 reference the z table. Round lue t0 decimal Diaces and final answier decima Places ) ocints Probability cBco" Rafetence= Prev Next 3.39 PM 325/2019 Search the wveb and Windows Mo



Answers

The mean weight of luggage checked by a randomly selected tourist-class passenger flying
between two cities on a certain airline is $40 \mathrm{lb},$ and the standard deviation is 10 $\mathrm{lb.}$ The mean and standard deviation for a business-class passenger are 30 lb and 6 lb, respectively.
(a) If there are 12 business-class passengers and 50 tourist-class passengers on a particular
flight, what are the expected value of total luggage weight and the standard deviation of
total luggage weight?
(b) If individual luggage weights are independent, normally distributed rvs, what is the probability that total luggage weight is at most 2500 $\mathrm{lb} ?$

Right. We want to find sample mean X bar, sample, standard deviation S and construct a 75% confidence interval about the population, meaning you given the data below assuming the population is normally distributed, then we'll interpret our findings to start off with, let's find expire and s we have to use the appropriate definitions to find them. X bar is equal to some of the data divided by N. In this case 91 S. Is the square root of the sum of deviations about the mean square, all divided by n minus one. In this case, 30.7 next to construct our confidence interval, we need to first identify the critical T score. We can use a tea table which maps a specific degree of freedom and confidence level or the correct T score. We can use this either from google or textbook and we find for degree of freedom and minus one equals five and confidence of the level of 50.75. R t c value is 1.3. Next is find the margin of error defined by the formula on the left here, plugging in our T c r s and r M gives E equals 16.3. Now we can instruct our confidence interval using the formula expire minus E is less than you. Is that the next plus E plugging in our mean? And r E give 74.7 is less than you and that's a 107.3. We can interpret this to mean that we are 75% confident mu is between these values 74.7 and 107.3.

Right. We want to find sample mean X bar, sample, standard deviation S and construct a 75% confidence interval about the population, meaning you given the data below assuming the population is normally distributed, then we'll interpret our findings to start off with, let's find expire and s we have to use the appropriate definitions to find them. X bar is equal to some of the data divided by N. In this case 91 S. Is the square root of the sum of deviations about the mean square, all divided by n minus one. In this case, 30.7 next to construct our confidence interval, we need to first identify the critical T score. We can use a tea table which maps a specific degree of freedom and confidence level or the correct T score. We can use this either from google or textbook and we find for degree of freedom and minus one equals five and confidence of the level of 50.75. R t c value is 1.3. Next is find the margin of error defined by the formula on the left here, plugging in our T c r s and r M gives E equals 16.3. Now we can instruct our confidence interval using the formula expire minus E is less than you. Is that the next plus E plugging in our mean? And r E give 74.7 is less than you and that's a 107.3. We can interpret this to mean that we are 75% confident mu is between these values 74.7 and 107.3.

All right. So let's say we're trying to find the probability, um, that the amount of rainfall is between 1 28 and 150 millimeters, given that the average is 102 millimeters and standard deviation is 48. So we can write this, um, with our standard accumulative normal distribution function, um, as f of so 1 50 minus the average over the standard deviation minus f of 1 28 minus. I don't know why I wrote just, um, you when we actually know the value of meal? Um, so one of two. So my ass. 1 28 minus 102 over 48. And this can be evaluated. So to simplify a little, this is f of one minus f of minus f of 1 28 minus 42. I think it's negative. 14 or 48 is negative. Seven over 24. And why this could be plugged into a computer algebra system to solve. And then if you want the probability that X is greater than 1 20 that equals one minus. The probability the X is less than 1 20 Ah, Then I write less than or equal to because it doesn't affect the computation here. And this is the formula we know. So this is one minus, uh, half of 1 20 minus mu, which is under two over 48. And all of these can be plugged into a calculator Using F f Z equals one over route to pie. You know this this function negative infinity z of e to my X squared over two d X. Um, and these are the two values you got. Yeah, that's how you do this from.

Okay, we are going to answer question number 25 in your textbook water taxi safety. When a taxi, When a water taxi sank in the Baltimore harbor, an investigation revealed that safe passenger load for the water taxi was £3500. It was also noted that the mean weight of a passenger was assumed to be 100 and £40 assume a worst case scenario in which all the passengers are adult men. This could easily occur in a city that hosts conventions in which people the same gender often travel in groups, Assume the weights and men are normally distributed with a mean of £182.. a standard deviation of £40.8. So we have a mean of 182.9 And the standard deviation of 40.8. So you can see there's a lot of variability in this data set. I'm going to plot the distribution part A says, If one man is randomly selected, find the, find the probability that he weighs less than £174. Uh This is a new value suggested by the National Transportation and Safety Board. The probability that he weighs less than £174. I'm looking for the value to the left of 1 74 it's It's 41.3 per 7%. So a pretty small probability um that 174 might need to be re evaluated by the safety board. Part B says with a low limit of £3,500. How many male passengers are allowed. If we assume in me in a week of 140. So all we need to do is take 3500 And divide it by 140. And we could see that 25 passengers would be allowed on the boat. Uh Part C says with a lower limit of £3500. How many male passengers are allowed if we use a new mean of 182.9. So 3500 Divided by 1 82.9. Yeah. Is 19.1361. So approximately 19.1361 passengers would be allowed on the boat. Party says, why is it necessary to periodically review and revise the number of passengers that are allowed to board? Uh Well you can see that, you know, the old standard was 100 and £40. Unfortunately, people's weights have tend to be increasing over time. Uh So they assumed 100 and £40 per person would be good enough and they saw that that was too much. So they should re evaluate it since the um weight of people is constantly changing and they might want to re evaluate this periodically


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