يناير فبراير 2024 WST 02 الربع الأول الثالث إحصاءات IAL 2

Welcome to my channel PI byMe Maths. This is by Jasan. In this video, we are going to be looking at January 2024 statistics 2 paper. Let's jump into question number one. The manager of supermarket is investigating the number of complaints they receive per day from the customers. random sample of 180 days is taken and the results are given. That means if you sum up all this frequency you will get and the number of complaints per day is also given. Calculate the mean and variance of this data. This is same like the stuffs we learned in S1. The formula to find the mean is sigma f_sub_x over sigma This is fx. This is frequency. Sigma is already given here 118. So you need to take and multiply by So the first one is zero. Leave it. 1 * 28 + 2 * 37 + 3 * 38 + 4 * 29 + 5 * 17 + 6 * 19 we get 531 divided by 180 divided by 180 is 2.95. That's the mean. That's how we find the mean. Now to find the variance of the formula is sigma f_sub_x² sigma That means you need to square this and multiply by minus sigma f_sub_x by sigma fx by sigma f² which is 2.95 square. So let's calculate this. So what I'm going to do now 0 * this is 0 1 2 * this 2 * this and square all the x's 1 2 * 28 is the same + 2 * 37 + 3 2 * 38 + 4² which is 16 * 29 + 5² * 17 + 6² * 19. So you get 2091 over 180 - 2.95 square. So the answer divided by minus 2.95 square and then round it to three significant figure. 2.91 is answer 2.91. That's all. It's three mark question. Let's move on to part Explain why the result in part suggests that poson distribution may be suitable model for the number of compliance per day. We got the mean actually 2.95 and the variance is 2.91. If you look at these values, these values are almost the same. Right? In poson distribution mean is also lambda variance is also lambda right. So since these two values are also same you can use poson. So you need to write the reason as mean is close to the variance. Just write mean is close to the variance almost equal to the variance. That's just one more question. Now part Now the manager uses posson distribution with mean three to model the number of compliance per day. So don't immediately take this as lambda because they saying part day. Maybe in the question they will talk about 3 days or 5 days. So you need to change it accordingly for randomly selected day. So just one day. So you can take it as lambda is three. follows poson distribution of three. Find at least three compliance is greater than or equal to 3. So you can rewrite it as 1 - is less than or equal to 2 and then use the table value or your calculator to get this value. So I'm going to go to shift mode scroll down go to sorry think mode scroll down go to number three distribution we need poson cumulative number three we don't need the list we just need the particular value so number two what's the value of xx is this actually so I'm going to put it as two and then the value of lamb lambda mu that's three. So immediately it will give you the answer 42 1 - 0.42 42 319 319 Let's Okay, maybe round it to four decimal place. 32. So when you subtract, you'll get 0 5768. But you need to round the final answer to three significant figures. So write it as 0.577. Now double now more than four compliance but less than eight complaints. This is basically is less than or equal to 7 because we don't need 8 minus we don't need four also is less than or equal to 4. Get these values and subtract them. Now the value is 7. Mu is the same 3 981 0.981 minus Now the value of is 4 lambda is 38153 0.8153 8153. And then you subtract these two values. 0.981 - 0.815 3. Round it to three significant figure 173. 0.1728. So round it to three significant figure 173 is the answer. Final answer. That's all. It's straightforward question from the posson distribution. Let's move on to next one. week consists of seven consecutive days. The using the managers model and suitable approximation show that the probability that there are less than 19 complaints in randomly selected week is 0.29. Okay, they're talking about 7 weeks per 7 days per day. There are three complaints. So 7 days it's 21. So what we are going to do we are going to apply normal approximation. For the normal approximation use different variable follows of lambda comma lambda. That means mu is 21. Always remember sigma is square root of 21 because this is sigma square. So they want us to find is less than less than 19. First you need to apply continuity correction. So less than or equal to 18.5. And then standardize this this normal distribution to standard normal distribution. is less than or equal to - mu / sigma. Let's calculate this value first. 18.5 - 21 / square of 21 get 0.55 is less than or equal to -0.55. We have learned this normal distribution in S1. When you have bell curve with the mean mu here 0 here let's say minus0.55 here since zed is less than they're talking about this area and due to symmetry this area will be same as this area on the right hand side 0.55 positive number right so how do we find that area you need to put 1 minus minus the area on the left hand side of the positive number. So you can take the table value in the normal distribution. Look look for 0.55. have opened the table here. 0.55 is in here actually 0.708 when you subtract you will get 0.2912 so round it to three significant figures 0.291 291 that's five mark question actually here period of 13 weeks is selected at random so is 13 find the probability that in this period there are exactly five so whenever they ask you to find something some figure out of some other figure ncr remember that's binomial so you need to look for probability of success what is the probability of success Here the probability of success is the weeks that have less than 19 compliance. So of is less than 19 is the probability of success which is what we calculated here. is less than 19. The final answer is 0.2912. So this is basically follows binomial distribution of 13, 0.29. 2 912 and they want you to find exactly five. So = 5 is use the formula from the binomial distribution. NC * ^ * is 1 - 1 - ^ 13 - 5 minus all. So first let me find 13 C5 which is 1287 times 2 912 ^ 5 get this value time 1 - 0.2912 to the power 13 - 5 is 8. So the answer is 0.17168 etc. Round it to three significant figure 172. That's the final answer. Let's look at question number two. Now the length of pregnancy for randomly selected pregnant sheep is days where follows normal distribution of mu12.4 4 comma sigma square given that 5 percentage of pregnant sheep have length of pregnancy days less than 180 that means of is less than 180 is 5% 0.05 in other words if you use the bell curve right for this normal distribution the mu is here 112.4 4 let's say 108 is here this is 108 this area is just 5% that's what given so they want us to find sigma so standardize this zed is less than - mu over sigma of what rule I'm using the coding zed is - mu / sigma the is this value that's how we standardize the normal distribution now if calc okay this number in standard normal distribution it's going to be negative number because the area on the left hand side is smaller than 0.5 so that means this is negative number you cannot use the table for negative number So first let me calculate this 180 minus is -4.4. It's negative number. I'm going to change this area is same as the right hand side area of the positive number. Right? So I'm going to write it as of is greater than 4.4 sigma is equal to 0.05. Now since zed is greater than and it's positive number and it's like rounded value neat value I'm going to use the smaller table. Now look for the value of as 0.05 and the corresponding value of zed is 1.649. So 4.4 sigma is 1.649. 649 make sigma as subject. So sigma will be 4.4 1.649. Round it to three significant figure you'll get 67 2.67 is answer. Let's move on to part Now, Chiang selects 25 pregnant sheep at random from large flock. Find the probability that more than three of these pregnant. Okay, you see they want you to find this out of this. So, as already mentioned in question one, when you get this type of question, that's binomial where is equal to 25. They want you to find the probability of more than three. Now, you need to look for the probability of success. What is the probability of success here? Pregnant sheep that have length of pregnancy of less than 108 days which is what given here 5 percentage right the probability is 0.05 right so the probability of success is 0.05 05. So the will be 1 minus which is 95. They want you to find of No, you need to define variable now. So you can use follows binomial distribution of 25, 0.05. And they want you to find more than 3 which is same as 1 minus less than or equal to 2. Sorry, less than or equal to 3. So 1 minus you can you can find this value in the table. So I'm going to go to binomial distribution table 25 on this value or we can use calculator actually go to the distribution binomial cd that's number one we don't need the list we just need particular value the value of is three and the value of is 25 and the value of probability of success is 0.05 05 it will generate the value for you 9659 0.9659 so you will get the final answer as 0.0341 0 3 4 1 That's the final answer. Okay, let's look at part Charlie takes 200 random sample of 25 pregnant sheep. Okay, is 200. Use person approximation to estimate the probability that at least two of the sample samples have more than three pregnant sheep. Okay, this is the probability of success in 25 more than three pregnancy with the length of pregnancy of less than 80 108 days. Don't take the probability as 0.05. That's wrong. That's the probability for 5% this one pregnant sheep. The probability that sheep has length of pregnancy of less than 108 days. But here the probability of success is more than three pregnant sheep. So you need to take the answer from the part So this is the answer. This is the probability of success 0.0341. is 0.0341. So to find lambda you need to write * 200 * 0.034 0341 200 * 0.0341 you get 6.82 so just write some letter I'm going to use sheep right so I'm going to put as follows distribution of 6.829 8 to know and they want you to find at least 2. So of is greater than or equal to 2 is same as 1 - of is less than two or less than or equal to 1. Okay. Now it's 1us of is 0 plus of is 1. Right? You can use the calculator to get the value. I'm just showing this working if you want to do it manually. The formula for probability the PDF function for proson distribution is power minus lambda lambda power factorial. Right? So power minus lambda lambda power 0 is 1 0 factorial is also 1 plus power minus lambda lambda power 1 / 1 factorial. So just get this value then 1us to the power minus lambda minus sorry put the minus here when open the bracket the second term also will become minus 6.82 * to the power minus lambda 6.82 982. So the answer is 0.9914 etc. Round it to three significant figure 991 is the answer. Question number three. Now row one believes that 35%age of type vacuum tube shatter when exposed to alternating high and low temperatures. Roan takes random sample of 15 of these type vacuum vacuum tubes and uses two-tailed test at 5% significant level to test his belief. Okay. Give two assumptions in context that Rowan needs to make for binomial distribution to be suitable model for this number of type vacuum tubes that shatter when exposed to alternative high and low temperature. You need to make two assumptions actually. First one, we can assume that the vacuum tubes shatter independently. That's one of the condition for binomial distribution. Right? And then you cannot write fixed number of trials. That's not the assumption. That's fact. Right? So you cannot write that but the probability of success we are going to assume that it's constant. So the probability of vacuum tube shattering is constant. There are actually four points for binomial distribution. Two outcomes. One is pro success and failure. Fixed number of trials. Success and failure. Fixed number of tries. They happened independently. And then probability of success is constant. fixed probability of success. But the assumptions are these two not the fixed number of trials. Now part have already put the table value here. Using binomial distribution find the critical region for the test. You should state the probability of rejection in each tail which should be as close as possible to 0.025. Okay. First of all, is take it from here 15. Probability of success is this. Significant level is five. So is 15. Probability of success is 0.35. Significant level is 0.05. And it stated clearly it's two-tail test, right? So for two-tail test, you need to divide the significant level by two. So you'll get 0.025, 025 which is mentioned here also even if it's not mentioned you don't forget to divide the significant level by two so you need to test the lower tail first so is less than or equal to is less than the significant level we need to find all the values of which satisfy this condition so have already put the table value here corresponding to is equal to 15 is 0.35 you can take it from the table in the binomial distribution table from s_2 the First value is smaller than the sign. This is significant level now. Okay. Second value is smaller than that. Third value is bigger than that. So this is how you show you're working. of is less than or equal to 1. You show where the value switches from less than to greater than 0142. is less than or equal to 2 is 0.0617. This value is less than the significant level. this value is greater than the significant level. Therefore, takes all these values. All these values satisfying the condition. Right? So, is less than or equal to 1. Now, for the right tail test, we need to test this actually is greater than or equal to So if you watch my video and learn this hypothesis testing have mentioned there whenever you get is less than or equal to the upper tail right take one minus of this 0.975 just for your reference and in the table value check where the value switches crosses crossing this 0.975 here right from here to here right so that's just your for your reference So I'm going to use 8 now. But when use 8 to use the value of 8, I'm going to start writing it as 9. So that when rewrite it 1 - is less than 9 or less than or equal to 8, can use this value 1 - 0.9578 which is 0.0422 bigger than the significant level. And then to use the value of 9, I'm going to start writing is greater than or equal to 10. So when rewrite, can put it as is less than or equal to 9. So 1 - 0.9876 which is 0.0124 less than the significant level. So the first value is 10 and if you take 11 12 all those values will satisfy the condition. Therefore is greater than or equal to 10. So the critical region is therefore critical region is is less than or equal to 1 or is bigger than or equal to 10. Next part find the actual significant level. How to find the actual significant level? You know, take the probability corresponding to critical value. What is the critical value? The first value that satisfies the condition. So, take the corresponding probability. If it's two-tail test, you need to take both the probabilities and then add them, change it to percentage. So 0.142 plus 0.0 1 to 4. So when you add you'll get 0.0 266 multiply by 100 it's 2.66 percentage. That's how you find the actual significant level. If it's one tail test, you will get only one value. Write the value and multiply by 100 and change it to percentage. Part Now, row one records that in the last batch of 15 type vacuum tubes exposed to alternative high and low temperature, four of them shattered, four is the observed value. So since we already have this critical region when you have observed value you need to check whether the value lies in the critical region or not four is not in the critical region in the critical region. So remember this. This is the only thing you need to remember. When your calculated value is less than the significant level, you reject Hnot or the or the observed value lies in the critical region, you reject Hnot. That's the only condition you need to remember. So if if the value does not lie in the critical region, do not reject Hn. So in this case, four does not lie in the critical region. That means four lies in the acceptance region. So you need to accept Hnot. So we don't write we accept Hnot because always we do the testing assuming that Hnot is true. So again writing accepting not doesn't make sense. So you need to write there is insufficient evidence. There is insufficient evidence. We did not write H1 for this question right here. So you can write to reject her belief insufficient evidence to reject row one's belief. That's how you write. Let's look at the last part of this question. five mark question. Row one changes to type vacuum tubes. He takes random sample of 40 of them and finds that eight of them shatter when exposed to this. So is 40. We know the probability is 0.35. Significant level is 5 percentage significant level 0.05. Now the question this is observed value. We are going to use this value. What are we testing? Test at the 5% significant level whether or not there is evidence that the proportion of type vacuum tubes that shatter when exposed to alternative high and low temperature is lower than this. State your hypothesis clearly. So you need to write your null hypothesis. Alternative hypothesis. Null hypothesis is 0.35. Alternative hypothesis is lower than 35. right? Less than 0.35. Since it's less than and the observed value is 8, you need to find of is less than or equal to 8. Okay, sorry before that you need to write assume that knot is true. So follows binomial distribution of 40 comma 0.35 but here we have used already so use some of the letter follows now you need to find of is less than or equal to the observed value. So from the table I'm going to use the calculator. Go to distribution binomial CD not list where number two is 8 is 14. Probability of success is this 0.03 03 03 which is less than the significant level. This is one tail test. So you don't divide the significant level by it. If it's less than we reject Hnot. So you need to write there is sufficient evidence to reject Hnot. You cannot just write this. You're going to lose more. You need to write what does this mean in this context actually. So not is this right? So you need to write there is sufficient evidence. You need to write there is sufficient evidence to support that the proportion of type vacuum tube shatter when exposing alternative high and low temperature is don't put lower than that's H1 is 35° 35%age you have to write this otherwise you will lose them all for every hypothesis test question you cannot Write just write reject not do not reject not and move on to the next question.