Suppose manufacturer wants t0 sct 8 mlleage guarantoe on Its new XB 70 tlra. Tests revealed that the lkre'g mlleoge normally Distnbulad wiha mean 0f= 49,600 mi...

Tests of a new tire developed by a tire manufacturer led to an estimated mean tread life of 67,350 miles and standard deviation of 1,120 miles. The manufacturer will advertise the lifetime of the tire (for example, a " 50,000 mile tire") using the largest value for which it is expected that $98 \%$ of the tires will last at least that long. Assuming tire life is normally distributed, find that advertised value.

Hi This question. We need to find the prominence interval for the many and percent so offers. Here it was a point or two and here's the formula. So you plus minus C over to assemble standard deviation. Devine was squirrel defend. I used the school color on. I found that this fellow, uh, 35,534 to this one and what I really want to show is. And when you calculate this one trying to calculate that the marginal error sometimes hope me on developed chick that's matching the the confidence of terrible that's you have.

So for this problem we are going to use the normal distribution functions and I'm going to use my graphing calculator to do so Given the mean of 32,000 miles for the tires and a standard deviation of 2500 miles. The first thing we're asked to do is to determine if we can expect the tires to last 40,000. So I want to know what is the probability that our value our tires will last exactly 40,000 miles. So I've already done it in my calculator but I'll do it again just to show you I want to do a normal pdf So to find that second vars number one Under the x value type in 40,000 because that's the value in question. My standard deviation is 32,000 and I mean I'm sorry my muniz 32,000. My standard deviation is 2500 Enter, enter and this value is very small. This is a form of scientific notation. So this means to move my decimal .7 places to the left. So the probability of this occurring is actually 12, 3, 45 six and then nine. So that's not even, that's not even 1%. So no it's not likely because the probability is very small. If I change is 2% by multiplying it by a 100 That just moved my decimal point to places. So I'll have 0.12349. So it's not very likely. Now the next question, what fraction can we expect to Last? Less than 30,000 miles. So I am going to look at this as less than or equal to 30,000 miles. So with the calculator now, I'm going to do a normal CDF, the cumulative distribution function. So second bars number two and my lower value. This is a default for the calculator to do negative infinity. So I want to go as low as I possibly can negative one E. To the 99. So negative one that w is above the comma 99. And then my upper value will be 30 1000 my mean and my standard deviation paste and enter. So this With his .211 nine. So that's about 21%. Again, not exactly sure how what kind of answers you have for fractions, because there's a couple of possibilities here, but this is approximately 21% which could be 21 out of 100 You could also round it to 20%,, Which would be 20 out of a 100 or two tents or 1/5. So you could expect about 1/5 of the tires To work. Now, it's a little different if it's not including 30,000, because then if I'm using my calculator, I'm going to make my upper limit a little different. So second bars Number two, but instead of 30,000 29999, so almost 30,000 And you can see it's not much different .2117, so it still would be around 21% Or rounding to 20%, which would be 1/5 of the cars. And then moving on, what fraction can we expect to be between 30,000 And 35,000. So I'm still going to use the normal CDF function, the normal cumulative distribution function, So opening up the calculator again. Second bars number two. Now this time my lower bound will be the 30,000 And my upper bound will be 35, Say mean and standard deviation cursor down to paste, enter, enter and now I have .67 0.673. So that is about 67%. So 67%,, That's 67 over 100 and that's approximately 2/3 because 2/3 is .666, repeating. So 2/3 you could also round up, Maybe you want to round it up to 70, So that would be 7/10. But I think 2/3 is a better approximation for what fraction you can expect. Now the next one is a little trickier because you need to know your estimating the inter quartile range. So in a normal distribution, remember the inter quartile ranges? The middle 50%. So that means Q one is at the 25% mark, And Q three would be at the 75% mark. So when when that translates disease scores in our normal curve where mu is zero and the standard deviation is one, The Q one Z score Is negative .67448 and the Q to Z score is 0.67448. Now I need to do a little backwards work now that I know what those Z scores, I need to remember to find the Z score. Remember that was the X value minus the mean divided by the standard deviation. So I'm just going to work backwards and I'm going to solve these two problems for X. And that will give me the values that I want for my inter quartile range. So I'm going to go ahead and move to the next page And set up my problems this way saying that negative .6744 eight has to equal x minus mu all over the standard deviation. So working backwards, Multiplying these two values and then adding 32,000, I'm going to get an X. Value of about 30,314 rounded and now I'm going to do the same thing with the other Z value Which would be Q three. So this is Q1 Q three. So multiplying the two values That's going to give me a positive 16862 and then I'm going to add 32002 that And that gives me 3368 6.2. So you're inter quartile range then will be these between these two values. And then remember I Q are the inter quartile ranges. Q three minus Q one. So now I'm going to subtract those two and I end up with 3372.4. And then the last one your dealer only wants to take a risk one out of every 25 customers, One out of 25 is .04. So now I need to do an inverse Norm because I need to find the value that would go with an area of 4%. So that's not very much when you think about the normal curve, that's down here somewhere, so it's very small. All right, so with my calculator, second vars now inverse norm is number three. The area in question is .04. Type in the mean and the standard deviation And the value I get is 27623. So let's see. What was that? 27 7000 623. So I would say then That's a strange value to go to. So perhaps around it to 27,000 miles. That would be the mileage he should guarantee to or she should guarantee to

Mhm. Okay. So for this problem you have machines that have a average life of eight years with the standard deviation of two years. All right. They want to only replace 5% of the machines under their warranty program that they're they're using. Okay. Um and it tells us that this is a normal distribution. All right. So it follows the normal distribution formula. All right. A. Is going to be the lifespan of the machines. Okay. And this always assumes This is the probability of getting a lifespan less than whatever a. Is. Okay. Now, they tell us 5% of the machines. Since they start us out with the percentage we need to look at a Z. Table to find what Z value matches up with 5%. And the closest z value there is negative 1.65. All right. Is the value of negative 1.65 has 4.95%. All right. So, we want to know what what is the lifespan that they should guarantee their machines for? What is this a What does this egg need to be for them to only have to replace 5% of their machines. So 11 final piece of this formula. This inside part the part inside the parentheses. This is the Z. Value. So since we have a Z. Value, let's set it equal to that part inside. We're looking for A. And we know that mu is eight and omega is two. So, if we solve this formula for a, we should get the A. Is 4.7 years. So what this tells us is 95% of their machines will last longer than 4.7 years And only 5% will fall will fail after prior to 4.7 years.

We'll be using the normal curve and the empirical rule or the 68 95 99.7 Rule For this problem. So the first thing I've done is draw bell curve with the middle being 24.8 and then I've numbered by adding 6.2, Keep moving to the right and then subtracting 6, 2 to the left. Now, the first thing that we want to do is identify the 68, and 99.7% rules. So, within one standard deviation that mean above and below the mean, You should encompass 68% of the data. So there's your interval that would be for 68%, to 31 And then for 95%, that's three standard deviation, I'm sorry, that's two standard deviations above and below the mean. So here the interval for two standard deviations would be from 12.4 to 37.2, that should encompass 95% of our data and then three standard deviations above and below the mean, is what and Encapsulates 99.7% of the data. Now, for the next question, we want to know about what percent of cars get more than 31 MPG. So I'm going to highlight that 31 MPG. And essentially what we're doing is finding the area underneath the bell curve. That is more than 31, and the empirical rule will help us do that, since this falls at a nice Z score on our on our bell curve, So we already know from the left, all the way to the middle is 50%. And We also know that one standard deviation above and below the mean is 68% of the data. So that would mean there's got to be 34 To the left, between one standard deviation below the mean and mean, and 34% for one standard deviation above the mean. So now we know this part of our bell curve to the left of that marker is going to be 50 plus 34 which is 84% But that's to the left of our marker of 31, so that means the remaining 16% is going to be more than 31, so 16% of all cars get more than 31 mph. Mhm. Now, for the next question, we want to know the percentage of cars that have MPG of gas between 31 and 37.2. So, in this Section, well, we can use that 16% from the previous question to help us out. Now we can figure out this percentage to the right of 37.2, because we know This is the two standard deviation marker, 95%. So if it's 95% within two standard deviations, that means it's 5% more for greater than two standard deviations and less than two standard deviations. So if we split that 5% up into these two equal sections, This part has got to be 2.5%,, So if it's 16% all the way to the right, 16% -2.5%, will give us the percentage between 31 and 37.2, which would be 13.5%. And that 2.5 is helpful for our last part of the question because yes, if it's 2.5 More than two standard deviations above the mean, it's also got to be 2.5 percent less than two standard deviations below the mean. So the lowest 2.5% of cars, they have the worst gas mileage is from 12.4 or less.