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To access the Internet: [0 cell phone users are Fourteen percent of cell phone users use cell phones selected at random. (out = of 10) have used their phones to acc...

Question

To access the Internet: [0 cell phone users are Fourteen percent of cell phone users use cell phones selected at random. (out = of 10) have used their phones to access the Internet is: The probability that at least 0.000008 0.O0oo8 0.008 the Internet is: &. 0 (out of 10) have used their phones to access The probability that at most 0.00008 1,0 the Intemet is: & probability that at least (out of [0) have not used their phones t0 access The .0ooo8 0.96 the Internet is: The probability that

to access the Internet: [0 cell phone users are Fourteen percent of cell phone users use cell phones selected at random. (out = of 10) have used their phones to access the Internet is: The probability that at least 0.000008 0.O0oo8 0.008 the Internet is: &. 0 (out of 10) have used their phones to access The probability that at most 0.00008 1,0 the Intemet is: & probability that at least (out of [0) have not used their phones t0 access The .0ooo8 0.96 the Internet is: The probability that at most 6 (out oC 40) have not used their phones (0 access 0.Oooo8 0.04 Internet is: &. 0 b. [ c. 2 The mcan number of people (out f 10) who use cell phones t0 accCss The variance of people (out of 10) who use cell phones t0 access Internet iS; access Internet is: 10. The standard deviation of (people (out of 10) who use cell phones to c.2



Answers

According to a report released by the National Center for Health Statistics, $51 \%$ of U.S. households has only cell phones (no land line). According to the FCC, nearly $70 \%$ of the U.S. households have high-speed Internet. Suppose of the U.S. households having only cell phones, $80 \%$ have high-speed Internet. A U.S household is randomly selected. a. What is the probability that the household has only cell phones and has high-speed Internet? b. What is the probability that the household has only cell phones or has high-speed Internet? c. What is the probability that the household has only cell phones and does not have high-speed Internet? d. What is the probability that the household does not have only cell phones and does not have high-speed Internet? e. What is the probability that the household does not have only cell phones and has high-speed Internet?

After analyzing some data from the fall of 2004, It was found that 20% of US households had some type of high speed internet. So if you envision asking people do you have Internet net to answer would be yes or no. So the probability of success would be defined as .20. In A random sample of 80 people, we want to determine different probabilities. Now this information is a binomial probability because there are only two outcomes, yes or no. Either they have internet connection or they don't. There is a defined number of trials and each of those trials would be independent and in part a we want to determine the probability that exactly 15 of those households had the high speed internet and we want to use a normal approximation to calculate this. So we are going to use the bell shaped curve. And in order to use that bell shaped curve, we need to know an average and we need to know a standard deviation and to calculate our average will multiply N times p. So we talked with 80 people or 80 households And the probability of success was .20. Therefore There was an average of 16 households that had Internet connection to find our standard deviation, we would use the formula the square root of n times p Times The Quantity of one -P. So again, we looked at 80 households, the probability of success was 0.20 and if I do one minus 10.20, I get 0.80, which would be the probability of finding someone that does not have the internet connection in 2000 and four and we end up with a Square root of 12 8 as our standard deviation. Now we could use our calculator and get a decimal approximate for that. But we highly recommend that you use the exact value throughout all our calculations, rather than a decimal approximate to get the most accurate probability. So when we're using the normal curve, our average goes in the center And that would correspond to a Z score of zero and we're trying to determine That x equals 15 or 15 would be to the left of 16 on a number line and by no meal distributions are discreet probabilities and when we model discreet probabilities, we model them with hissed a grams. So I need you to envision a hissed a gram tower For that number 15. And The low end of that tower would be at 15. Try at 14.5 as a low boundary And the upper end would be at 155. So when we're trying to determine the probability of 15, we're trying to determine the probability of this whole tower. So therefore we are going to approximate it by determining the probability that X is between 14.5 And 15 5 inclusive. Now, in order to use the normal curve, we will have to turn the boundaries of the shaded area into Z scores. So we're going to need a Z score associated with 14.5 and we're going to need a Z score associated with 15.5. And to calculate Z scores, we use the formula x minus mu over sigma. So the Z score for 14.5 could be calculated by saying 14.5 -16, which was the average or the mean and divided by the square root of 12.8, which was the standard deviation. And that Z score calculates out to be about a negative .42. Now we want the Z score for 15.5, So we'll use 15.5 minus the mean of 16 divided by the standard deviation, which was the square root of 12.8. And you will get a Z score of about negative .14. So when we're talking about the area of this curve Between 14.5 and 15.5, It's comparable to the probability that Z is between negative .42 And -14. So at that point we will need to use the table, the standard normal table in the back of your textbook to find the area that's associated with each of these Z scores. So when you use the standard normal table, we Find the units place and the 10th place along the left side. So the tense places of four. So we're going to have a units place of zero. So we're gonna be looking at negative 04 And then the hundreds place was a two. So we're going to look beneath the .02. And in doing so we are going to find an area of .3372. So that area is the area from that Z score boundary line down into the left tail. Now we want to find the Z score associated with negative 0.14. So again we want the units place and the tense place. So we'll look up negative 0.1. The hundreds place was a point oh four. And where the row and the column meet up in that chart is going to give you an area of .4443. So that means from the Z score of negative 14 into the left tail Is .4443. And we're looking for the area in between. So in order for us to find that difference we're going to have to subtract. So we're going to have to subtract the probability that Z was less than negative 0.14 minus the probability that Z was less than negative 0.42. Or will subtract those two decimals? 20.4443 minus 0.3372. For a probability of .1071. So recapping, we have surveyed people and ask them do they have an internet connection? The probability in 2004 that they had some type of high speed interconnect Internet connection was 20% When we surveyed 80 people, The probability that exactly 15 of those 80 had high speed internet Would have been approximated at .1071 for part B. We want to determine the probability that at least 20 Of those 80 households had the high speed internet. So we're going to approach this in a similar fashion. I'm going to draw that bell shaped curve. We're going to put the mean in the center, which again corresponds to that Z score of zero 20 would be located to the right of that. And I want you to envision that hist a gram tower. Again, The low class boundary would be 19.5. The upper boundary of that tower would be 25. And we're trying to determine the probability that were greater than 20 Were equal to 20. So therefore are shaded region really begins at 19.5. So we're going to approximate this by determining the probability that X is greater than or equal to 19.5. Therefore, we need the Z score associated with that 195. So we'll use that Z score formula X minus mu over sigma. And you will get a Z score of approximately .98. So when we refer to X being greater than or equal to 19 5, It's comparable to the probability that Z is greater than .98. So we will have to go back to that table in the back of the book. Again, you're standard normal table. And we look up the units place in the tents, place down the left side and the hundreds place across the top. And in doing so you are going to get an area of 8365. So that means From here into the left tail is .8365. But we want to go into the right tail. So instead we're going to do one minus the probability that Z was less than .98 or one minus that decimal of 8365 For an overall area of .1635. So the probability that At least 20 households had high speed internet access Would be approximated at .1635 in part C, you were trying to determine the probability that fewer than 10 households had that high speed internet access. So here's our bell curve again with mean in the center We're going to envision 10 to the left of that. We need to envision the tower. So if we're going to be less than that, we need to envision the next tower over as well which would have been the tower for nine. So when we want less than 10, we're really saying that we want the probability that x is less than or equal to nine. So when we focus on the tower for nine, the low boundary of that tower would be at 8.5. The upper boundary would be at 95. And we are trying to Find the area less than nine Were equal to nine, Placing our boundary of the shaded region at 9.5. So we'll have to use the probability that X is less than or equal to 9.5 to approximate this probability We will need the Z score for 95. Z score for 95 is calculated using x minus mu over sigma. And you're going to get an approximate value of -1, So therefore, the probability that X is less than or equal to 9.5 can be compared to the probability that Z is less than negative 1.82. So we'll go to our standard normal table. We'll find the Units place and the 10th place along the side. We'll find the hundreds place across the top And that yields an area of .0344. Going into the left tail. So from Zeon into the left tail is an area of .0344. So therefore our probability Is going to be .0344. So the probability that fewer than 10 Households had high speed Internet access can be approximated at .0344. And to conclude this problem, let's look at part D. In part D. We want the probability that between 12 and 18 households inclusive, had that high speed internet access. So we'll take the same approach. We'll place the mean in the center, which corresponds to that Z score of zero Will place 12 to the left of that 16 And 18 to the right. We will envision the history Graham towers and their boundaries. The low boundary of the tower for 12 would be 115. The upper boundary would be 12.5 With the tower for 18. The lower boundary would be 17 5 And the upper boundary would be 185. And we are trying to determine the probability that X is between 18 and 12 or equal to. So therefore are shaded areas are defined By the 11.5 and the 18.5. So to approximate this, we're going to determine the probability that X is between 11.5 And 18.5. Thus requiring Z scores for 11 5, An 18.5. So the z score for 11 5 will be calculated To be approximately a negative 1- six. And the 18 5 when you apply, your formula Will be approximately a 70. So therefore, if we were determining the probability that X is between 11 5 and 18 5, It's comparable to the probability that Z is between negative 1-6 and Positive .70. We will have to rely on our chart again in the back of the book, we're going to find our units and tense place and then we'll find our 100th place. So we get a corresponding area To be .1038. So that's saying from AZ of negative 1.26 into the left tail has an area of 1038. For the Z score of 70 we're going to look up the units place in the 10th place so we'll look up 07 The hundreds place was zero And you will find a corresponding area of .7580. So from the Z score of 70 into the left tail Is .7580. So if I want the area in between, I'm gonna have to look for that difference. So I'm going to need the probability that Z was less than .70 minus the probability that Z was less than -1- six. So I'm gonna take those two decimals 7580 -2038. To get an approximate probability of 6542. So the probability when selecting 80 households that between 12 and 18 will have high speed Internet access back in 2004 Was approximated at about 6542. And that concludes the four parts to this problem

Problem. Seven, we have a study conducted in the United States, which stays 25%. 25% of U. S. Households had no landline service. Gloria, then line service. We are going to big five US householders and random, and we need to calculate the following probability Big five US householders at random. This means we will define around the variable X, which follows a binomial distribution. Binomial distribution within equals five because we will select five and B equals all points 75 and X represents household with land line service. It will be a complement of no landlines. Er for birthday. We want to calculate the probability that all five of them have a landline. This means you want to calculate the probability for the random variable to equal five, which equals mm five because n equals five. And we have here five again because I equals five. Not to blow it by P, which is 1.75 to the bottom end. And it's five multiplied by one minus speed, which is open to five to the power of five minus five, which is zero. These gifts Oh boy. In 71 multiplied by open 75 out of five supplied by one gives 4.237 or in percentage 23.7%. Football to be. We want to calculate the probability that at least one of them does not have an online, which means we want to calculate the probability for the random variable X to be smaller than plate because does not have a landline means to have a landline is smaller than five. This will make the condition. At least one of them does not have an online. We can use a compliment event. It's one minus the probability of X qualified. It's all the probabilities, minus the probability to have all of them have an envoy, which means it's one minus 0.237 which equals 4.763 or in percentage equals 76.23%. Finally, for board. See, you want to calculate the probability that at least one of them does have an online, then be of the random Variable X does have a landline, which means it's greater than zero. We can again use a compliment event. It equals one minus. The probability of X equals zero. It's all the other possibilities to have no landline then equals one minus five because in equals 50 because I equals zero well deployed by the probability of success over 175 there's a lot of men multiplied, boy. Sorry, it's the amount of oil. Zero is a lot of oil and here is also to the bar of High. Because we have I qualified. Multiply it, boy. The probability over 1 to 5 to the power of n minus I, which is five minus zero, gives five. Then it equals one minus over 1 to 5 to about five gifts, all point or or one which equals 4.999 or in percentage, 99.9%. In these answers, we have used the binomial distribution function, the probability of X equals I equals and I. But to blow it by p to the power of I want to blow it by one minus B to the bottle and minus guy. And these are the final answers. Our problem

So here we're just giving a whole bunch of data to work with all these probabilities, all these groups and we're gonna be finding some ranges of probabilities. That and you have all the information you need to be doing this by hand. Problem is, is that each one of these is gonna have 10 steps to deal. If you want to do it by hand, you have to use the formula you have been using in previous problems for every case and adding them up. Not on this page in your book. It gave you some notes on doing these on a computer for a reason. You're supposed to be doing these problems on a computer to save time. Now I'm gonna be doing this in Excel. You can do this in your graphing calculator. The only figure the the figure out how to do is find the function that I'm using. After that, it will be exactly the same, and the textbooks should be able to help you there. But for me it's under formulas function, library, statistical and then it's buying Nam doctors stop range. If you can't find one, that is the range version you can always just use the regular binomial distribution. But you're gonna have to do one for each case and add them up. So I'm gonna take advantage of the range tool. Now, let's move this out of the way. So part A. So we're told that there are 50 people in each age range. Sore trials is always gonna be 50 for all of these problems, our probability. So we are in the over 65 category, so that gives us a 22% chance. But we need to convert that to a decimal. So we're gonna get point to chill and for our numbers for finding the probability a 10 to 20 people say yes. So it's gonna be 10 to 20 and we get 0.686 Now for part B, find the probability that 30 to 40 people say yes, who are between the ages of 50 and 64. So we have 50 trials between the ages of 50 and 64. That gives us a 640.58 probability and fairly true for you. And we get 0.446 All right, so for C 30 to 40 people again, But now it's in the fair T 2 49 category, which has a 75% chance of being Yes, trials are still 50.75 and 30 to 40. You get 0.8 free zero now for D. Same faint, except it's in the 18 29 category, which has a 77% chance of saying yes. So 50 0.77 uh, fair D and worry 0.74 wine. All right, e, why are the answers for part A In part D similar So A and D our first and last wine. So the 0.686 and the 0.74 wine there is a connection here to take a look. So in part A, we found the probability of 10 to 20 people said yes, and in Part D, it was fairly to 40. Now our total population is 50 and 10 plus 40 is 50 and 20 plus various 50. Also, the two probabilities for this group are 22% in 77%. That adds up to 99% which is almost 100. So that's why these two are giving us similar probabilities. For the last part, what effects of the various values of P having the probabilities. So for this one, we're gonna be mostly interested in Parts B, C and D, cause we're looking for the same range of values there. The only thing that changed each time were the probabilities probability went up. If you notice, that doesn't mean our number always went up, went up at first, then it went down a little bit. Just because we have a higher probability doesn't mean our answer is gonna be higher because they're finding the probability that fairly to 40 which is somewhere in the middle, say yes. So the probability is gonna go up as this as our probabilities up here get closer two fairly over 50 40/50. So the probability impacts it by making giving us a higher answer if it is closer to the proportion that we are looking for and we are done

So they're estimating that CNN was estimating that 7% of people basically used cell phones only and in a we have to describe if they had taken a sample of 750 people, what would the sampling distribution look like? And this is supposed to be a random sample, and we know that we anticipate that the sampling distribution that the mean of the sampling distribution will be at 1.7. We expect that the standard deviation of the sampling distribution will be 0.7 times 0.93 divided by 750. And let's find out what that is. Uh, get me clear out of this point that's square root of and square root of 0.7 times 0.93 divided by 7 50. And we've got that thing. Uh, that comes out to be 0.93 If we want those as percent, we'd say that this centers at 7% and this is 70.9 basically percent, so our distribution would be approximately normal. So I kind of drew a picture of the distribution. We would anticipate that the center of the distribution is right here at 0.7 And if we move out one standard deviation that samples of size 750 the proportion this would be like 7500.79 And if we add on another 0.9 we'd be at point 088 And if we add on another and go out three standard deviations and so on, each direction we could could show what would happen. So the distribution should be approximately normal because our sample sizes is very sufficient to cause that to be our end times p and then times one minus p are both going to be greater than or equal to 10. And let's see if we will be back it up. This white 0.7 minus 0.9 this would be down here at 0.61 And if we subtract again minus 0.9 this would be a 0.52 and I do this because then we're gonna look at the next part. It said, What is the likelihood that you take a sample upside 7 50 you get it to be at least 8% 80.8 so we can find that by converting it to a Z value, and we can see 8% is about right in here, and we want to find that. So we take point await minus the mean 0.7 and then divided by this standard deviation, which is about 0.93 And when we do that fact, I'm going to go back up and take that value that I had. And I'm going to store that value as, uh, x so that I'm using this this value in its entirety. So we have 0.1 divided by that value, and we end up getting one point. RZ value here is 1.7 At 1.733 It comes out and then you can either use the I'm going to use the second distribution, and I'm going to use my normal CDF normal CDF button. But you can also look this up in the table. And so my normal CDF I have my lower number as this value. So this is my low value, and then I'm just gonna put in a big number. And the low value is that second answer and my big number, and I'm gonna leave the mean at zero on the standard deviation at one. And we get that. That probability is about 14. It's 140.1416 so it's approximately 14.2% chance of that happening. And then we want to know. And sorry that did that. Kind of. It's kind of weird, doesn't it? Uh, and then we want to find What's the likelihood that we get a sample proportion that is 40 less than or equal to 40 out of 750? Well, I don't know what that ISS 40 divided by 7 50. That comes out to be about 0.533 So we do the same thing. What's the likelihood of the Z value being less than or equal to that number? 40/7 50 minus that 0.7 and then divided by that approximately 0.93 which again I have stored in my calculator. So I'm going to take that 40 divided by 7 50 minus 0.7 and I have that in parentheses, in the numerator and divided by that value that I square rooted and I get that that Z value is negative. 1.79 Basically, it comes out 889 so on and I'm again going to use my normal CDF and second distribution and the normal CDF. But this time, because I'm finding this value down here, I'm finding this value down here. I'm going to put in my low value is going to be like, negative 1000. My upper value is going to be this value that negative value, and I can use to second answer for that. And I'm gonna leave that as zero and one and taste. And when I do, I find out that probability is Onley 0.368 So it's approximately 3.7% chance of that happening, and that is again verified from my picture. And I'm getting that weird graf again, since I'm my digitize er my pen is kind of hit in the bottom of the screen. So hopefully that makes sense for you. Draw the picture. I always help


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