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Lunch break: In recent survey of 655 working Americans ages 25_ 34, the average weekly amount spent on lunch was 842.32 with standard deviation $2.65. The weekly am...

Question

Lunch break: In recent survey of 655 working Americans ages 25_ 34, the average weekly amount spent on lunch was 842.32 with standard deviation $2.65. The weekly amounts are approximately bell-shaped_Part 1 of 3(a) Estimate the percentage of amounts that are between $34.37 and $50.27.Approximately 95%of the amounts fall between $34.37 and S50.27_Part 2 of 3(b) Estimate the percentage of amounts that are between S39.67 and $44.97.Almost allof the amounts fall between $39.67 and S44.97.Part: 2 / 3

Lunch break: In recent survey of 655 working Americans ages 25_ 34, the average weekly amount spent on lunch was 842.32 with standard deviation $2.65. The weekly amounts are approximately bell-shaped_ Part 1 of 3 (a) Estimate the percentage of amounts that are between $34.37 and $50.27. Approximately 95% of the amounts fall between $34.37 and S50.27_ Part 2 of 3 (b) Estimate the percentage of amounts that are between S39.67 and $44.97. Almost all of the amounts fall between $39.67 and S44.97. Part: 2 / 3 Part 3 of 3 (c) Between what two values will approximately 95% of the amounts be? Approximately 95% of the amounts fall between and



Answers

The U.S. Bureau of Labor Statistics reports that the average annual expenditure on food
and drink for all families is $\$ 5700$ (Money, December $2003 ) .$ Assume that annual expenditure on food and drink is normally distributed and that the standard deviation is
$\$ 1500 .$
a. What is the range of expenditures of the 10$\%$ of families with the lowest annual spending on food and drink?
b. What percentage of families spend more than $\$ 7000$ annually on food and drink?
c. What is the range of expenditures for the 5$\%$ of families with the highest annual spending on food and drink?

Okay for this problem were given another normal distribution about the likelihood that people how much money people spend on fast food. So again, with the normal problems, it is good to make a sketch. Uh, what you're trying to find. Um, So the biggest fact here were given is that the mean amount of money spent us $44 and the standard deviation is 14 50 family spend. So is that the general idea like to sketch it out? It's mostly for a showing direction and weaken our steps for this over that right now when we want to find out, I like to sketch it first and then we'll use up an apple. It, in this case to to find the probabilities for this. So for part A, we want to find out how many spent less than $25 to the probability that the X is less than 25 Best show with the diagrams of 40 force here. Another 14. 50 over the side. I'm just gonna give a general sketchy of a relative idea of the size. So 25 is my mark there. Um, and for a part, be sketching. Think we're going to see your 30 to 50. So 30 is going to be roughly right here, and 50 is gonna be over here. So a decent size chunk. So we're gonna find out this area under the curve, which again there in the curve is a probability. Your likelihood that my is spent for that only use the color coding to cash show what we're doing here. But a way to rephrase that teachers probably want you to do that is you say that the probability is between 30 and 50 is written like this 30 50. And finally, for the last part, they want to know it's more than 75 fast food. The probability that the X is greater than $75 per week. We've set 25 to 34 year olds, So let's look at just 75. Before the four was 14. 50 there said I was going to get number over here. You can feel it in If something numbers are viewing more precise, a few teacher expect you to, but it's really more about direction and general idea on. And then I use these scores and scores give us the probability and use table three if you want. I prefer what you've done. A few of these. I like to stop with the one stop shop. Little kind of free apple. It's for normal distributions. You can see they do everything here, but for normal distributions. Ah, click that button. And so the area of normal curve. Well, just like we kind of sketch there. We wanna Karmakar has a center at 44 a standard deviation of 14 50. So you see this and I'm gonna start a gram and I'm gonna take some time. There's a nice cleaner picture that you can actually sketch out the 50 50 and 73 whatnot. So let's find the area first. It's going order under 25 so I don't between two guy, you don't want to the left of a value. Look at 25 you're looking at a table. You'd look at the work 25 is and get the Z score for 25 then see what that area is. Um, the calculation for that area is 250.95 to go back to my read, and it's 0.5 relatively small. Number less than 10% which makes sense. If you look at the size of it. Okay, that probably is going less than $25 per week. Let's do this again, and let's look at our green area. So we want to know between Mary and 50 are changes to between two values. If you're using a table, you take get the two values and then subtract them. But if you use the app, let you say, Well, look between 30 me and 50 messing with the areas you see the area for this problem is about 0.4933 Okay, the likelihood that you spend between 30 and 50 was gonna remark it up. Here 30 and 50 is almost 50% which looks like it's about relative size seems to make sense. And finally, let's look at one to the rights to the right of a value, which is our blue area. We want to know the probability of spending over 75. Now. If you're using a table on this, you get disease for and then be, uh, that probably let it goes bad and you do one minus. That's quirk of the table. When I look left to right. But the apple it's can look to the right of this boundary point, and the area here is 0.163 and there we have it. Just so we have a likelihood of how much money that 25 34 year olds spend free to those. If we're given the median of 44 on the sand aviation, that's the likelihood that you will get for each of those normally distributive points.

Were asked to look at some data and then find the 95% Commons interval for the data. So the data that's given to us is 30 24 38 35 27 35 23 28 28 22 straight a 22 26. Then they also have 34 29 25 28 34 24 26 28 32 29 40. So that is what is given to us for our data. So we need to look at normally assumes that it's satisfied on. Then we need to find 95% confidence. We're looking for the confidence interval at 95% so we can assume, since it's normally, um, it's a normal normality assumption that it's satisfied with a sample size. It just means it's large enough for it, um, to be able for us to kind of go through the data so it's valid. So what you do is go to your mini tab and an excel. You would enter your data into the column. So, for example, let's say you're putting it in the first column of a your column would be a 1 38 to 24 a 33 a 4 35 all the way down, etcetera, until you get to 40. So what you would dio is after you have your column all ready to go, you're going to go into that many top again. Click the stat, go to basic, and then you want to click the one a sample T Because we're only looking at this one sample. We're doing a T statistic for that competence. Terrible. So once you have that, you're gonna click the samples and columns and put in whatever column you have that end. So let's say it was just a one, so make sure that that is selected in the samples and columns. Then you're gonna go to options and you're gonna put in your confidence interval of 95.0. And then you also need to make sure your alternative I set to not evil so that we get the correct output. Then once you have all that, you can click OK, and it should spit out a one sample tea with your column of a one, and it would be something like that so it would have your variable at the top. Your en your mean your population mean? Oh, I missed the standard deviation there so you would have your end. You're mean. Then it should show a standard deviation. Then your population mean And then at the 95% confidence interval, okay, and then underneath it is gonna have whatever column you used your variables. It's calculating how many in the samples there's 22 calculated the mean of 29.3 to, and that are standard deviation, they calculated was 4.92 population mean and then right here is the confidence interval that we want. And underneath that 95 is 27.14 and 31.50 up. So that's the range in which 95% of our data would fall. So between 27.14 and five

All right. So this question gives us some information about spending on fast food and party wants us to compete the margin of error for a 95% confidence interval. So we know that margin of error equals since we only have sample standard deviation, we have to use a tea distribution. So it's t star times our sample standard deviation divided by the square root of our sample size. So we need to find our T star. So to do this, we can draw normal curve, and we know that 95% of the area it must be between negative t star and T star, so we can compete the area in each tale, which is 0.0 to 5. So to find our t star value, we just do inverse t of our area 0.975 andr degrees of freedom, which is just the sample size of minus one, which this turns out to be 1.990 So now we have everything we need to computer margin of error. We have our tea star value times air sample standard deviation divided by the square root of our sample size. And this gives us a margin of error of 1 22 point 369 Now it wants us to compete the confidence interval. So our interval is our estimate, plus or minus our error. And we know both of these. So our mean was given his 18 73 closer minus our standard margin of error from the last question. And this gives us a range of 17. 50 0.63 as are low and 1995 point 37 is our high. So we're 95% confident that the population mean lie somewhere in this interval. No, for part C, it wants us to estimate the total expenditure from this sample. So for our estimate, we estimate that a single person spends on average X bar equals 18 73 and that's in dollars. And we have 80 people in the sample, so we have $1873 per person times 80 people and that gives us a total expenditure of 1 49 a 40 for this whole group because we use our point estimate and multiply that by the number of people to get an expected total. Then Part D says that this is skewed, right? So do we expect the median to be greater or less than the mean So help? Let's draw a right skewed distribution. So this is right skewed. So for a right skewed distribution, our median would be here. But our mean would be over here because even though crest of this hill up here is the median of the data, we have these data points in the tail over here that are artificially dragging up the value of mu, so we'd expect the median to be less than the mean.

38. The amount of money spent weekly on cleaning, maintenance and repairs at a large restaurant was observed over a long period of time to be approximately normal. What the mean of $615? With the standard deviation $42? Yeah, am if 646 is budgeted for next week, what is the probability that the actual cost will exceed the budgeted amount? So the probability X will exceed 646? So we're gonna say that X is approximately normal. Draw the curve to get an idea of what's going on here. 6 15 is our centers are mean, our center deviation is 42 so I want to space this out 42 when he said, so we're gonna add 42 to that and we get 6 57 and then there's also a 2nd 3rd center deviation, there's no need to put one standard deviation below 6 15 puts us up 5 73 and I want to know what's the probability that I'm exceeding 6 46. So, somewhere in this range here, So, so I'm looking for the area to the right of 6 46 so we can use our tia 83 or 84 calculator and use the normal CDF command here, we have to insert what are lower bound. Is this essentially where we're starting our shading, which is at 6 46. Our upper bound is where we stop shedding. Technically that's never because it goes to infinity, so we're going to put in and and then just a series of nance to indicate a large number, I mean is 6 15 standard deviation here is 42. We can plug this in and we get zero point 23 B. How much should the how much should be budgeted for weekly repairs, cleaning the maintenance, so that the probability that the budget budgeted amount will be exceeded in a given week is only 10%. So what I'm looking for here is some amount of use it, read some amount so that if they go over there is only a 10% chance of going over. So I want to use my inverse norm command this time in my T 84 calculator. Um The 84 calculators are really geared to be left sided. Um A newer version will give you a choice between left, middle or right, but we're gonna focus on the left side of the version of this. So the area to the left of this red line again area to the left is 90%. So 900.90 with our mean of 6 15 and our center deviation of 42. That gives us a budgeted amount of $668.83. Now, if you have a more up to date version of the T i 84 calculator, um, you could put in 840.1 as your area, as long as you have the feature, uh to select the right sided shade.


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