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Question 61ptsThe average resident of Metro City produces 580 pounds of solid waste each year;and the standard deviation is approximately 70 pounds: Use Chebyshev&#...

Question

Question 61ptsThe average resident of Metro City produces 580 pounds of solid waste each year;and the standard deviation is approximately 70 pounds: Use Chebyshev's theorem to find the weight range that contains at least 75%0 of all residents' annual garbage weights.Between 300and 860 poundsBetween 440 and 720 poundsBetween 370and 790 poundsBetween 510and 650 pounds

Question 6 1pts The average resident of Metro City produces 580 pounds of solid waste each year;and the standard deviation is approximately 70 pounds: Use Chebyshev's theorem to find the weight range that contains at least 75%0 of all residents' annual garbage weights. Between 300and 860 pounds Between 440 and 720 pounds Between 370and 790 pounds Between 510and 650 pounds



Answers

The average college student produces 640 pounds of solid waste each year. If the standard deviation is approximately 85 pounds, within what weight limits will at least 88.89% of all students’ garbage lie?

So we know the mean number of pounds that are produced by college students in solid waste is supposed to be £640 with a standard deviation of £85. And we're told that we want to have the interval. That includes 88.89% at least, uh, that much. And so, if we use Chevy chefs, uh, the're, um we know that one minus one over K squared needs to be equal to the 10.8889 And so moving this term over to the right side are adding one over case where to both sides and then subtracting 0.8889 from both sides. We get this, which basically is 1/9. And so when we either cross, multiply or multiply, will end up getting case where it is equal to nine or K is equal to three. And so now we want to apply the idea that we want to add on three and subtract away three standard deviations away to find that interval. So if we take 640 subtract away three times 85 that will be the lower limit. £385 up to. And then add on the three times 85 to both to the 640 we get £895. So we would say that 88.89 or basically 88 8/9 percent of the, uh, poundage that the college students would have for solid waste is going to be at least that much is going to be between those numbers.

Oh, and coming to the question the beauty is given to us are the mean is 17.2. Standard deviation is 2.5, and they're asking the probability that X should be greater than 17 and actually less than or equal toe 18 for example, off 55 families. So coming to the graph again, this is our muse 17.2. And this would be mu minus sigma. So just a minute. Yeah, and this would be your new place Sigma. So the value will be mu minus ignites 17 points to minus 2.5. That would be close to it's kind of cute. 17 point toe minus 2.5, 14.7. So this point is 14.7 and this point would be 19.7. But we're being asked something like between 17 and 18. That means 17 would be somewhere over here and 18 would be somewhere old here. That this point to here and point it here from 17.2. So, with me, Kelly cleared. We warned the gathering off this point. So the percentage of this point order off this 2.5 from this 34% of the probability 340.8 off the whole 2.5 from this 34%. So we calculated this way 0.2 divided by 2.5 into 34%. The sport 0.2 divided by 2.5 s 0.8 into the 34% 2.72 2.7% is the probability off 17 and 17 point to and from then on we want to call it 17 point toto 18. This would be 180.8 point eight divided by 2.5 into 34%. So can we go to calculator point Date activated. The 2.5 into 34% is 10.8%. It would be 10.8% and adding the salt turns out to be you. Adding the boat tanto be 10.88 plus 2.72 is the total percent. That X would be less than requests to 18 or exists greater than or equal store 17. That would be on a calculator. 10 0.88 place 2.72 13.6 13.6 is the final answer

All right. So, students weights have a mean value of 62.5 kg. With a standard deviation of 2.7 kg. Alright. Is a normal distribution. So, we're gonna be using the normal distribution formula A minus mu over omega. Where a is going to be the weight in kilograms. All right. And this formula always assumes that we're talking the percentage of uh uh of weight that is less than than whatever you're looking for than a. It's always less than that's the key. Yeah. So, for part A I want to know How many of the 1000 total students. Yeah. Mhm. What% 55 kg? Okay, So, it just gets plugged into the formula 55 minus 62.5 Over 2.7. So that's seven point 5/2 7.0.5 inside the parentheses. So, this part inside the princess gives you a value of three and that's the Z value. We have a Z equal to three. So, if you look at your your Z. Table in the back of your book, A value of three Has a percentage of 99.87%. Okay, Excuse me. This is negative three negative three has. Yeah. A wait. 0013 or .13%. Okay, so .13% of the 1000 students. So if you multiply that by 1000, that is about 1.3 students. Mhm. You'd expect to weigh less than 55 kg. Okay. For part B we want to know. Mhm. Between 60 and 65 kg. Alright, Whenever you have a range always start with the largest and take away the smallest. Mhm. So Again, for Formula 65 -62.5 over 2.7 minus Normal distribution of 60 -62.5 -2.7. So this is 2.5 divided by 2.7 which is a z value of .93 and the second one is the z value of negative 10.93 Mhm. And then on your Your chart, .93 is 82.38%. And negative .93 is 17.62%. Mm. So, if you subtract those, 82.3, -17.62, That is 64. Mhm .76%. So 64.76% of students have a weight between 60 and 65 kg. Alright, part. See how many of these students would you expect to have? White equal to 63 kg? Yeah. So since we have a continuous distribution here, meaning that you have a continuous and infinite amount of possible weights these students could have, there's you can't make a statement about equal to some number. So the probability are that the number of students who would expect to equal 63 exactly, We have to say is 0%. If if the students could only way sick like integer values for example, um then we could say something about this, but since They could weigh 63.1, 63.001 and so on. We there's no way you can you can just say equal to some number. So that's 0%. And then finally we want to know How many students would expect to have weights greater than 61 kg. All right, so plug into the formula 61 -62.5 over 2.7. That is negative 1.5 Divided by 2.7. So this is a Z value of- .56. And if you look on your table, .56 is 71.23%. All right now, you got to keep in mind this is a greater than so you get Less than you have to subtract it from one. So the actual percentage of students greater than 61 kg Is 28.77%. And then to get an actual number of students multiply that by 1000 1000 times .2877. Yes, we would expect 287.7 students to have a weight greater than 61. Mm. Mhm. That this this is not the correct number. I looked at positive .56 negative .56 Should instead be 28.77%. But when you subtract it from one, you'll end up with 71.23%. So the final number, when you multiply it by a 1000 lined up, we would expect 712.3 students to have a weight greater than 61 kg.

Told that the mean weight of luggage checked by a randomly selected passenger flying between two cities is £40 and the stair deviation is £10. We're told that the mean and standard deviation for a business class passenger, on the other hand, that was a tourist class for a business past class passenger are £30 in £6. Mhm in part A. We're told that there are 12 business class passengers and 50 tourist class passengers on a flight were asked to find the expected value of the total luggage weight in the standard deviation of the total luggage weight. Well, but X one through extra will be the waits for the business class passengers and well, that why one through 50 b, the tourist class weights in the total weight T is going to be some overall the exes and then the sum over all the wise as well, which we can actually right as the sum of two variables ex representing the total wage for the business class and why representing the total weight for the tourist class we have that the expected value of X is going to be the same as 12 times the expected value of a single tourist business Class X one and we're given that this is 30 so this is going to be 12 times 30 which is 360 pounds. Likewise, we have that the variance of X is going to be 12 square times the variance of X one, which is going to be Oh, that's actually just sorry 12 times the variance of X one since they in your combination. So this is 12 times 36 which is equal to 432. The variance is found by squaring the standard deviation for X one. We also have the expected value of why this is going to be 50 times expected value of why one so 50 times we're told that the mean for the tourist class is 40 gives us 2000 likewise thieve variants of why this is by margin 80 50 times the variance of why one which is equal to well, we had that standard deviation of libel. Miss Tens, this is going to be 10 square which is 100 a few times 100 which is 500. I'm sorry. That's not five. Hundreds of 5000. Therefore, we have the expected value of tea since cylinder combination is simply the expected value of X plus. The expected value of why which as we kept the before, is 360 in 2000. And so we get 2360 and we have the variants of the total weight is going to be. This is simply the some of the variances for X and y, because these two variables X and Y are independent. We had that this was calculated to be 132 and 5000, so we obtained 5432 and therefore the standard deviation of tea is the square root of this, which is approximately 73 0.70 to £1. Next in Part B, we're told that individual luggage weights are independent, normally distributed random variables and were asked to find the probability that the total luggage weight is at most £2500. Well, in this case, the probability that the total weight T it's less than or equal to £2500 this is going to be he standard normal, cumulative distribution function of 2500 minus the mean tweeted determined to be 2360 over the standard deviation, which is 73.70 to 1 approximately, and this is the same as five of about 1.90 And after looking this up in a book or using a computer calculated, this is approximately 0.9713


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