QUESTION 3shlpment of 1000 scnt t0 a local dlstributor. 20 are slightly blemished If Ahmed purchases 200 o/ ihese tlresJi manulacturer of automobile tres reported t...

In a random sample of 200 cars of a particular model,
3 have a manufacturing defect. At this rate, how
many of $10,000$ cars of the same model will have a
manufacturing defect?
A) 150
B) 200
C) 250
D) 300

Welcome to enumerate in the current problem we are considering the lifetimes of tires which is distributed normally with mean one like kilometers and the sigma to be 10,000 kilometers. Now we are given a sample of size 5000, okay, and we are asking what what number of tires out of these will exceed 118 1000. That is what we are looking for. The probability of x exceeding 118000 This will be a probability of x minus mu by sigma better than that's the c 118 triple zero minus mu by sigma. Do you understand rather compared to the previous problems that we were solving in this one I'm directed I think music MMA you know why because new and signal way too big to write and it will take the entire screen space and also slow down the slow down our thinking so shortcut always to meet Any mathematical thing easier. Now this minus this will be very easy, 18,000 so we don't need to write that step pattern and this is So we have probability of zed greater than 1.8. So that will be one minus probability zero less than 1.8. Now, what is this probability if I bring this table a little up? Okay, now 1.8 over here. So it is 9641. So oh 1 -0.9641 which is equal to nine five 73 zero. Okay, so it is 0.0359. And the percentage if you ask multiplied by 100 and we will get 3.59%. Now, what is 3.59% of 5000? So it is 3.59 divided by 1000. So this cancels this, cancel. So we are left with 3.59 in 2 50 that is 35.9 into five, so this will be two, So it is 179.5, which can be rounded off to 180, correct? So that means on an average out of 1 5080 tires. If this distribution is really following normal, on an average, 1 80 tires will exceed 118,000 km.

Welcome to numerous in the current problem. We are considering the lifetimes of tigers, Which is distributed normally between 10 to 10, uh relax kilometer and sorry, one like kilometer and sigma to be 10,000 kilometers. Now we are asked the question that what percentage of tires will last between 95,000 km, 2, 1, these many kilometers. So for that the first thing that we will start is funded rising. So we will have 95 triple zero minus 1, 2, 3, 4, 5, less than x minus late, Less than 115 000 -100000 divided by 10,000. Now this after being simplified, it will become minus this, correct? Because 100 minus 95 is five and 33 zeros and then less than this will be zed and this will be 15,000 divided by 10,000, which will give -0.5, less than zero, Less than 1.5. That is if there is a Standard normal distribution like this centered at zero. So this is something -0.5 and this is something say one. So this will be 1.5. So we want to know what percent will be this. So what we will do, we will take the entire area left to this, that is this, and then take the entire area left to this. And then we will find that probability Some probability said less than 1.5 Minour probability of zero, less than -0.5, Which is equivalent to probabilities said less than 1.5. This stays common but this becomes zero, greater than 0.5. Due to symmetry of normal distribution, which will be probability zed less than 1.5 -1 minus probability that less than 0.5. This is because the regular table, Yes, the regular table table that we have this will generally be giving the area less than if you can get a table that gives the value more than this. And you can directly that evaluate this. Now, going back to the values probabilities hit less than 1.5. years over here. So this is the value we are considering 0.9332 -1 plus. Probably to sit less than 0.5. So which is this? So it is 0.6915 6915. So how much will this value? B let us calculate. So For that I will simply use Excel. So I have a point line 33. Cool .9 metric -1 plus .6915. Eric. So we get 01 62 oh Little .6247. So we have 0.6247. Or if we do a percentage, it will be 62.47%. Right? So 62.47 percent. People will sort of tires Okay. This percent of tires will have a longevity of this.

Welcome to memory. In the current problem. We are given the lifetimes the lifetimes of certain types of one years tires. Okay, so the sudden by himself touches lifetime most fires and we are also given that X follows a normal distribution with new equals one like kilometer, been lax kilometer and is equal to 10,000 no reliability. The question asking is what percentage? Well what is a bit of fires? What percentage of barriers between 85 1000 kilometers and then kilometers? Okay, then lack kilometers. So that is equivalent to asking this. Now, if we do a standardization, we get Divided by 10,000. Now, if we simplify this, we get probability now 100 -85 gives us 15, correct, This is it value and this is zero. So we will have -1.5, less than zero, less than zero. So in standard normal distribution like this, This will be zero And this will be -1.5, correct. Somewhere we're here. So, but what happens is we have 1.5 given over here that too this area completely. So the table gives okay, the table gives then less than area, correct. So this area. So what can we do? We want this area because this area and this area will be cool, correct. So we'll be right probability of That. Less than zero Minour probability of that is the -1 point faith, Correent. So probability of said less than zero Minour probability of zed greater than 1.5, which is equals two, probability zero, less than zero minus one minus probability zero less than 1.5. Because our table gives value less than that. So for 1.5 if I bring the diagram, little town, no 1.5 We have this value, correct? So we and for zero we have this value. It's we know already minus one plus zero point I don't want 9332, correct? So this and this will give 0.933 to minus 0.5 because 0.5 minus one is minus zero point right? So we get 0.4332 which Will be 43.32%. So 43.2% of all tires will have uh huh. Lifetime between 85,000 km two then black kilometers.

So in this problem the probability is 0.2 And the number of sample is 1000. Then we can calculate n times P is telling and one minus P is 10.98. So first problem, the probability that X is squared and 25 can be approximate. Um so this is equivalent to exit square record in 25.5. And we can approximate using uh caution distribution. So this is justified from prive -20 over squared off Journey Times .98. So this is probably secretly at 1.24 And this can be calculated as .107. This is a problem. A We have problems be calculate you probably that access between 20 and 30 again we can first Use it another equivalent form. So this is from 25.20.5 to 29.5. And this Purple taste Can be approximate by 20.5 -20 or squared off 20-plus times .98. Smalling than they which is a number standard number distribution and 29.5 -20 Over squared off 20 times .98. And this is From .112 2.15. And this privilege to is .44. So that that's the answer.