Edexcel IAL S2 Statistics June 2025 - Complete Paper Walkthrough | WST02/01

追加: 2026-06-05

[00:00:04]Today we will be solving June 2025 math A level statistics 2.

[00:00:12]In question one, we are told that the continuous random variable X has cumulative distribution function given by f(x) is equal to k * 6 x^2 - x ^ 4 from 0 to 1 where k is a positive constant. And we need to show that k is equal to 1 / 5. So since this is a cumulative distribution function, we know that this here must be equal to 1. So since that should be equal to one, it means that if we substitute instead of x to be one and equate that to equal to 1, then we shall find the value for k. So if we do that, if we replace x with 1, that remains 1 square remains 1 and 1 ^ 4 remains 1. So that's k * 5 is equal to 1. So k is equal to 1/5. Then in part B we are told to find the probability where X is bigger than 0.6.

[00:01:18]So that is just equal to 1us the probability of X smaller than 0.6.

[00:01:26]Therefore 1 - 1 5 * 6 * Now substituting 0.6^ 2 - 0.6 6 ^ 4. Putting that into the calculator, we get a value of 0.5939.

[00:01:43]Then in part C, I we show that the median, we should show that the median M of X satisfies a M ^ of 4 + B M^2 + C is equal to 0 where a, b, and c are integers to be found.

[00:01:59]So what is the median? So the median is the middle value. So the cumulative distribution function should be equal to 0.5 because that's the middle the 50%.

[00:02:12]So if we put m instead of x^2 and equate that to 0.5 we should get such a quadratic equation.

[00:02:23]So doing that we say f of m is equal to 0.5.

[00:02:29]So, and putting m instead of x, we're left with 1 / 5 * 6 m ^2 - m ^ 4 is equal to 0.5.

[00:02:39]And then rearranging that we're left with 6 m^ 2 - m ^ 4 is equal to 2.5.

[00:02:45]Now, we don't like the decimal place because we need to give them a, b and c integers. We multiply everything by two.

[00:02:53]So, it be and and putting everything to the right hand side. So it becomes 2 to 2 * m ^ 4 - 12 m ^ 2 + 5 is equal to 0 and that satisfies what we were told to show and then hence we need to find the value of m.

[00:03:14]How we do that? Well, if this is a hidden quadratic equation where we could say that m^2 is equal to a and then we just plug plug a into m^2. So 2 a^ 2 - 12 a + 5 is equal to 0. If we use a quadratic equation to find the value of a that goes as such minus b. B is the coefficient in front of the A plus minus roo<unk> of B ^ 2 - 4 A C. A is this coefficient and C is this over 2 A. And placing that all into the equation and putting that into the calculator, we are given two values. One value is 5 something, but that is outside the range because we see here it's from 0 to 1.

[00:04:02]The other value is 0.54505.

[00:04:06]And but don't forget we replaced m^2 for a and we are not told to find a we're told to find m. So a is equal to m^2.

[00:04:16]If m^2 is equal to 0.450 then m should be the root of that. So m is 0.671.

[00:04:27]Then in part d we are told to find using calculus the mode of x.

[00:04:32]Well, the mode of X is the highest point, the most reoccurring point.

[00:04:38]So, first we need to turn that into a PDF function. The CDF right here, we should turn it into a PDF. How we do that? It's by differentiating.

[00:04:50]So, this was the original one. If we differentiate that, we're left with this because we take away one from the power becoming one and multiply the old power by the coefficient in front becoming 12.

[00:05:03]And we do the same thing for here. And we're left with 1 5 * 12x - 4x cubed.

[00:05:12]But we're still not done. We just found the PDF. Now that we found the PDF, we need to differentiate it one more time to find the highest point, the maximum point, the turning point. So if we differentiate that once again, we take away one from the power. This was one.

[00:05:30]So if that was one, we take away one from the power. So it becomes zero. And then the old power, which was 1, we multiply it by 12. So it remains 12. And then the same thing for 4x cubed. 4 * 3 is 12 - 1 from 3 is 2. So we're left with 1 5 * 12 - 12 x^2. And we should equate that to zero to find the highest point. So that's just 12 - 12 x^2 is equal to 0. So 12= 12x^2 x^2 is = 1 and x = also 1.

[00:06:08]Then we told given that the expected value of x is 0.64 64 state giving a reason the skewess of the distribution.

[00:06:16]So the expected value is the mean and the mean we are told to be 0.64 and we found the mode to be the median to be 0.671 right here.

[00:06:31]So since 0.64 is smaller than 0.67 mean is smaller than median. So there is a negative skew.

[00:06:41]In question two, we are told that on Friday evenings, a shop opens from 7:00 p.m. to 11:00 p.m. During this time, customers are known to enter the shop at a mean rate of 20 per hour. We need to state a suitable distribution to model the number of customers that enter this shop in a 30 minute interval on Friday evenings. So as soon as we see a mean rate, we know that we're dealing with a passon distribution. Then we're told that this number this lambda occurs per hour. But we are told to give a distribution for 30 minutes. So we just have this and we get a distribution of x follows plus one distribution 10.

[00:07:23]Then we need to state a necessary assumption for the model in part a to be valid. Well, we need to say that the customers come in independently. We always need to say independently for pass on and the rate is constant so at a constant rate. In part C, we are told that the manager makes alter alterations to the shop's layout. Following these alterations, the manager wants to find out whether the mean rate of customers entering the shop has changed. To test this, the manager decides to monitor the number of customers entering the shop.

[00:07:58]The following Friday, the manager randomly selects a 30 minute interval between 700 p.m. and 11 p.m. And we need to write down a suitable null and alternative hypothesis that the manager should use.

[00:08:12]Well, we need to say that the ho the null should be that lambda is equal to 10. So this is just the same the standard one and the alternative hypothesis which we denote as H1 is that lambda is not equal to 10. Now we don't say lambda is less or bigger than 10 because in this case we're not told what we need to test for. Therefore it's it only says that the mean rate of customers has changed. Therefore we say that is not equal to 10.

[00:08:47]Then in part D, we're told using a 3% level of significance, find the critical region for the manager's test.

[00:08:56]So how we do that is we say that since the total uh significance level is 3%. Because we're dealing with two critical regions, one lower and one bigger, we divide the 3% significance level by two, leaving us with a 015 significance level.

[00:09:20]So the first one, the lower one, the lower critical region is probability of X is smaller or equal to C should be smaller than 0.015.

[00:09:32]Then if we go in the table we find that the probability of X being smaller or equal to 3 is equal to 0.0103 that is smaller than 0.015.

[00:09:43]So C critical region is equal to 3. Then for the upper critical region we say that the probability of X being bigger or equal to C should also be smaller than 0.015.

[00:09:58]Therefore, 1 minus the probability of X smaller or equal to C minus one should be smaller than 0.015.

[00:10:06]Therefore, probability of X smaller or equal to C minus 1 should be bigger than 0.985.

[00:10:13]Therefore, probability of X being smaller or equal to 17 is equal to 0.9857.

[00:10:20]If we see on the table, so C - 1 is equal to 7. So C is equal to 18.

[00:10:28]Therefore our critical region is X is smaller or equal to 3 or X is bigger or equal to 18.

[00:10:36]Then we told the actual significance level of this T test based on our critical region from part D. So our critical region was X is smaller or equal to three which we found the significance level to be 0.0103.

[00:10:53]And for 18 we found to be 1 minus this.

[00:10:57]So 0.0143.

[00:11:01]If we add these two probabilities together we get 0.0246 or 2.46%.

[00:11:10]And finally in part F we are told during the 30 minute interval that the manager monitored 16 customers entered the shop.

[00:11:18]And we need to comment on this finding based on the critical regions we found in part D.

[00:11:26]So since 16 does not [clears throat] fall within the critical region, it is in between 3 and 18. So it's not outside the critical region. Therefore, there is not enough evidence to reject HO. Ho is the null hypothesis.

[00:11:43]There is not enough evidence to suggest that the rate of customers entering the shop has changed. In question three, we are told that a sweet shop produces different colored sweets and mixes them to sell in small, medium, or large bags.

[00:11:57]The proportion of sweets that are blue is P. Each small bag is filled with a random sample of N sweets. The mean number of blue sweets in a small bag is three, and the variance of the number of blue sweets in a small bag is 2.55.

[00:12:14]Now we know that in a binomial distribution the mean is equal to the number of sweets times the probability.

[00:12:23]So NP should be equal to 3. And we also know that the variance is is given by n * by the probability times by 1 minus the probability. So np * 1 - p should be equal to 2.55.

[00:12:41]Now we need to rearrange this one. So we're dealing with simultaneous equations. So we need to rearrange this one giving n= to 3 over p. If we play put instead of n here 3 over p, we're left with 3 / p * p * 1 - p is equal to 2.55.

[00:12:58]Now this p and this p cancel out and we're left with 3 * 1 - p, which is equal to 3 - 3 p is equal to 2.55.

[00:13:08]So 0.45 45 is equal to 3 p and p is equal to 0.15.

[00:13:16]And if we place 0.15 in here, so 3 / by 0.15, we are left with n = 20.

[00:13:25]Now in question b, in part B, we're told that the proportion of sweets that are green is 0.2. Each medium bag contains a random sample of 40 sweets. For medium bag, find the probability that the number of green sweets exceeds the expected number of green sweets. So as we said before, the expected value is just the mean. So expected value should be np.

[00:13:49]So the n is 40 in this case and the probability is 0.2. So 40 * 0.2 should be equal to 8. So this is the expected number of sweets. And we're told to find the value to find the probability that it exceeds the expected value. So it's equal to probability of X bigger than 8.

[00:14:13]Now notice that it's only bigger and not bigger or equal to 8 because it says exceeds. So probability of X is bigger to than 8.

[00:14:23]So that's equal to 1 minus probability of X smaller or equal to 8 which is equal to 0.469. 469.

[00:14:34]In part C, we are told that the proportion of sweets that are yellow is 0.1 and each large bag contains a random sample of 80 sweets. And using a plus on approximation, we need to find the probability that a large bag of sweets contains at least 12 yellow sweets. Now again, we're dealing with a binomial distribution here with n to be 80 in this case and p to be 0.1.

[00:15:00]So if we say that y follows a binomial distribution 80 and 0.1. Now if we use an approximation for a passon distribution we say that a follows a pluson of 8 because lambda in a pluson approximation is equal to n * p. So 80 * 0.1 is just 8. And then we told we need to find the probability that the large bag contains at least 12 yellow sweets.

[00:15:28]So in this case we include the 12. So the probability of X is should be bigger or equal to 12. Therefore the probability of A should be bigger or equal to 12 which is equal to 1 minus the probability of A being smaller or equal to 11 which we find from the table to be 0.8881.

[00:15:52]If we take one away from that, we're left with 0.1119.

[00:16:00]And in the final part, we are told to justify why a pass on approximation is suitable in this case. Well, it is suitable because the n is large and p is small, which should give a probability a lambda to be smaller than 10. That's why it is suitable.

[00:16:20]In question four, we are told that a bag contains a large number of beads of the same size and shape. The beads are either black or white. Black beads and white beads occur in the ratio of two to five, respectively. And in a game, a player takes a random sample of three beads from the bag. The player scores four points for every black bead he takes and one point for every white bead he takes. The random variable X represents the total number of points for the three beads. and we need to find the sampling distribution of X.

[00:16:54]Now, first we should write down to so it's clear what combinations we have. So the player can either pull black then black again then white. But that can happen in three different orders. So it can be either black first then black second and white then black white black or white black.

[00:17:17]So this this this can happen three times.

[00:17:23]So then what else we have? We have black, white, white. So one black and two whites that can also occur three times. Then we have black, black, black.

[00:17:33]That can happen only once. And white white that can also happen only once.

[00:17:39]And then if we write down the possible scores. So for black, black, white, it's 4 + 4 + 1, which is 9. For black, white, white, it's 4 + 1 + 1, so six. Black, black, black is 12. And white, white, white is three.

[00:17:57]Now we are told that the beads occur in a ratio of 2 to 5 respectively. So two for black and five for white. So if we write down the probability that a black bead can be is going to be taken out. So it's 2 over 7 and white is 5 over 7.

[00:18:16]Then if we write the sampling distribution we have 3 6 9 and 12 in ascending order. And first we can find the probability of white happening.

[00:18:29]So first we can figure out the probability of white white happening which is three. So that will just be 5 / 7 * 5 over 7 * 57. So 5 over 7 cubed and we're left with 125 over 343 for the probability. Then we can go ahead and find 6. So 6 can be found by black then white then white. So 2 over 7 * 5 over 7 * 5 over 7. But remember we have three possible combinations. So 27 * 5 over 7 * 57 * 3 leaving us with a probability of 150 over 343.

[00:19:14]Then if we do the same thing for the 9, it's 27^ 2 * 57 * 3 giving us a probability of 60 over 343.

[00:19:27]And then for 12, we can either find the probability by doing black * black * black. So 2 over 7 cubed. Or we can do one minus all of these probabilities like that together. Whatever uh benefits you, whatever helps you do it more easily. And we're left with an a probability of 8 over 343.

[00:19:54]Then we told that a random sample of n sets of three beads is taken. The random variable y represents the number of these n sets that have a total of exactly nine points. And we need to calculate the minimum value of n sets such that the probability of y being bigger or equal to 1 is bigger than 0.95.

[00:20:16]So we're told that y follows a binomial distribution of number of sets n and a probability that the there is exactly nine points which we found earlier to be 60 over 343.

[00:20:34]So then we are told that the probability of y being bigger or equal to 1 should be bigger than 0.95 which is equal to 1 minus the probability of y is equal to 0 is bigger than 0.95.

[00:20:47]So probability of y is equal to 0 should be smaller than 0.05.

[00:20:53]Then we need to find the probability of y equals to 0. How we do that is we take the n the number of terms choose the zero times 60 over 343. So the probability to the power of the y that we need. So to the power of 0. Then we do 1 minus the probability. So 283 over 343 to the power of n minus the y. So in this case n minus 0 remains n and that should be smaller than 0.05.

[00:21:30]All this remains one. So we're only left with 283 over 343 ^ of n should be smaller than 0.05.

[00:21:42]Now if we log both sides, we're left with n * log 283 over 343 should be smaller than log 0.05. 05.

[00:21:55]Now if we take this to the other side, so we leave n alone, we will have to reverse the sign because log of 283 over 343 is negative. So we're left with n should be bigger than log 0.05 over log of 283 over 343.

[00:22:16]So n is bigger than 15.6.

[00:22:19]So if we round that up, n is equal to 16. In question five, we are told that the continuous random variable x has a probability density function given by f ofx is equal to 2 over 21x from 0 to x to k and 2 over 15 * x - x * 6 - x from k to x to 6. And we need to show that k is equal to 3.5.

[00:22:48]We must make our meth method clear and show all stages of working.

[00:22:53]So what we can do is if we integrate this and this and add them two add those two together and equate them to one we can find the value of k. So first we will integrate the first equation 2 over 21 * x from limit k to limit zero.

[00:23:17]So if we integrate it from k to 0 2 over 21x that becomes + 1 ^ 2 / 2. So 2x^2 over 42 from k to 0. So that is just 2k ^ 2 over 42.

[00:23:36]Now if we do the same thing for the second equation. So 2 over 15 * 6 - x.

[00:23:44]But this is from the limit from 6 to k.

[00:23:46]So if we integrate it from 6 to k the equation 2 over 15 * 6 - x, we get 25 * 6x - x^2 / 2 from limit 6 to limit k.

[00:24:03]Now if we substitute 6, we get 2 over 15 * 6 * 6 - 6 ^ 2 / 2us the k limit. So 2 over 15 * 6 k - k ^ 2 / 2 and if we simplify this we get 2.4 - 0.8k + 2 over 30 k^ 2. Now what we said before is we will add these two together and equate them to one because when we integrate two equations so we should get the area underneath the line. So the area if we integrate it should be equal to 1 because that's the maximum probability 1 is 100%.

[00:24:47]So if we go ahead and do that we're left with 2k ^2 over 42 + 2.4 - 0.8k + 2k^ 2 over 30 should be equal to 1.

[00:25:04]Now adding up all the k^ 2 terms we're left with 4 over 35 k^ 2 - 0.8k + 2.4 is equal to 1. If we take the one to the other side we're left with + 1.4.

[00:25:18]To get rid of this uh denominator and to get rid of the decimals we multiply everything by 35 and we're left with 4k ^ 2 - 28 + 49 is equal to 0. And we get two values for k. One is k is equal to 3.5 and the other is k is equal to zero.

[00:25:39]Zero is rejected because as we see here k should be bigger than zero. So the only other valid answer is 3.5.

[00:25:49]Then in part b we're told using algebraic integration to find expected value of x and we must show all stages of our working. So to find the expected value of x we need to integrate again integrate these equations but this time these equations multiplied by x. So if we show that so it becomes more clear.

[00:26:16]So the first equation we integrated from k to 0. We found k to be 3.5.

[00:26:22]So we integrate the equation 2 over 21x but we said we multiply by x first. So in this case we are integrating 2 over 21x^2.

[00:26:34]So if we integrate 2 over 21x^2 add one to the power it becomes 3 and divide everything by the new power. So 3 that should become 2 x cubed over 63 from the interval 3.5 to 0. That gives a value of 49 over 36. If we do the same thing for the second equation from 6 to 3.5.

[00:27:00]So 2 over 15 * 6 - x * x. Remember we get an integration of 25 * 3x^2 - x cubed over 3. From the boundaries of 6 to 3.5. Putting that into the calculator, we get a value of 65 over 36. If we add this and this we get 65 over 36 + 49 over 36 which simplified is 19 over 6.

[00:27:31]And then we told given that the expected value of x^2 is equal to 277 / 24 we need to find find the variance of x.

[00:27:41]So we know that the variance of x is equal to the expected value of x^2 plus the expected value of x all to the squared. So if we know the expected value of x^2 which was given to be 277 / 24 and we take away the expected value of x all to the^ 2 which we found to be 19 / 6. So - 19 / 6^ 2.

[00:28:14]We get a value of variance of x to be 109 over 72. In question six, we're told that the continuous random variable x is uniformly distributed over the interval a b. And given that probability of 2 to x to 5 is equal to 36 and the expected value of x is equal to 6, we need to find the value of a and the value of b.

[00:28:36]So we will need to use these two equations for the first part.

[00:28:42]So first of all we have a uniform distribution that goes like this.

[00:28:50]This is the uniform distribution. This is the line and we don't know the values. So this is B and this is A. But we know that from 5 to 2, this area right here is equal to 3 over 16. And we know that the height is equal to 1 / b - a.

[00:29:16]So the area right here the probability the probability should be equal to 5 - 2 / b minus a which we know is equal to 3 / 16.

[00:29:32]So and we also know that the expected value of x is 6. So how we find the expected value of a continuous uniform distribution? Well, it's b + a over 2 and we know that it's equal to 6. So b + a is equal to 12. And from this equation here, we can also uh find that b is equal to 16 + a. If we put instead of b 16 + a into this equation, we're left with 16 + a + a is equal to 12. So 16 + 2 a is equal to 12. 2 a is -4 and a is minus2. If we put a is minus2 in here, we get that b is equal to 14.

[00:30:20]And in part b, we are told find the value of the constant c such that expected value of c * x + 1 is equal to 3. So the expected value of C * X + 1 is equal to 3 can be written as C * expected value of X + 1 should be equal to 3. And we we know already we were given that expected value of X is equal to 6. So C * 6 + 1 is equal to 3. 6 C is equal to 2 and C is equal to 1/3. And then in part C we told the exact value of variance of X. So in a continuous uniform distribution how we find variance of x well it's b minus a all to the^ 2 / 12 which is equal to we found b minus a to be equal to 16. So 16^2 / 12 is just equal to 64 over 3.

[00:31:20]Then in part d we're told to find the exact value of expected value of x^2. As we said in the previous question, variance of x is equal to expected value of x^2 minus expected value of x all to the^ 2. We found the variance to be 64 over 3. And we found we know that the expected value of x is 6. So 64 over 3 + 6^2 should be equal to expected value of x^2. So expected value of x^2 is equal to 172 over3.

[00:31:53]And in part D we are told to find the probability of 3 / 2x - b is bigger than a. So if we first rearrange that so probability of 3 over 2x - b is bigger than a is equal to probability of 3 / 2x bigger than a + b.

[00:32:14]So we know already that a + b is equal to 12.

[00:32:18]And we're left with probability of x is bigger than 8.

[00:32:23]So how we find the probability of that?

[00:32:25]So we get the upper bound which is 14 minus this value this x value here. So - 8 / 16 which is just equal to 3 over8.

[00:32:38]In question seven we told that when purchasing a pair of glasses from an optician, customers are offered insurance to cover accidental damage.

[00:32:47]Past records from the opetition show that 30% of customers buy insurance when per when they purchase a pair of glasses. In a random sample of 40 customers who have purchased a pair of glasses, X represents the number who buy insurance. And we need to state a suitable model for the distribution of X. Since this has an N and the probability, the N is 40 and the probability is 0.3.

[00:33:13]Th this follows a binomial distribution.

[00:33:17]So we can say that X follows a binomial distribution of 40 and 0.3.

[00:33:24]Then in part B we are told that to state a necessary assumption for the model in part A to be valid. Whenever we have a binomial distribution we need to say that things happen independently and at a constant rate. So customers buy insurance independently and the probability of buying insurance is constant.

[00:33:48]And then we're told that the probability that fewer than our customers who have purchased a pair of glasses by insurance is less than 0.05.

[00:33:57]And we need to find the largest probabil the largest possible value of R.

[00:34:02]So how we do that? probability of X is uh to be smaller than R is smaller than 0.05.

[00:34:10]We were told because you see uh we need fewer than our customers. So we do not include R and the probability is less than 0.05.

[00:34:20]So this is equal to probability of X smaller or equal to R -1 is smaller than 0.05.

[00:34:27]And from the table I found that the probability of X smaller or equal to 6 is 0.023. 0238.

[00:34:34]This is the closest value to 0.05 which is smaller than it. So r -1 is equal to 6. So r is equal to 7. Then we're told that a second random sample of 200 customers who have purchased a pair of glasses is taken. And using a normal approximation, the probability that at least t of these 200 customers buy insurance is 0.9474.

[00:34:58]Correct to the four decimal places.

[00:35:02]And we need to find the value of t. We must show our working. So in this case the n changed. It's 200.

[00:35:11]Well the probability stays the same 0.3.

[00:35:17]And if we approximate this to a normal distribution the mean is given by n * p.

[00:35:24]So 200 * 0.3 is 60. And the variance is the standard deviation is given by roo<unk> of np * 1 minus p.

[00:35:36]So the root the standard deviation is roo<unk> of 42 and the variance is roo<unk> of 42^ squared. So a follows a normal distribution of 60 and roo<unk>42.

[00:35:50]Now we are told that the probability of being at least t of these customers is 0.9474.

[00:35:59]So that is equal to probability of y bigger or equal to t. So probability of y bigger than t minus one is the same as that. Then then using normal correction we get that probability of a bigger than t minus 0.5 is is the same as this before.

[00:36:18]So using correction we get probability of a is bigger than t minus 0.5 minus the mean which is 60 divided by the standard deviation which is roo<unk>42 should be equal to 0.9474.

[00:36:36]Then if we flip this these two around so it becomes 60.5 minus t then we can flip over this sign as well. So it becomes probability of a smaller than 60.5 minus t / <unk>42 is equal to 0.9474.

[00:36:57]From the table we can find that the pro that the probability of x being smaller than 1.62 is equal to 0.9474.

[00:37:07]So 1.62 is equal to all of this. So rearranging that we get that t is equal to 50.00001.

[00:37:18]So rounding that up t is equal to 50.

[00:37:24]Then in part we told that the optitian decided to introduce a special offer in which customers who purchase a pair of glasses get the first three months of insurance free if they buy insurance.

[00:37:35]Following this, a random sample of 25 customers who purchase a pair of glasses is taken and 11 of them buy insurance.

[00:37:43]And we need to test at the 10% significance level whether or not there is evidence that the proportion of customers who buy insurance when purchasing a pair of glasses has increased. And we need to state our hypothesis clearly.

[00:37:58]Well, first we need to state our null hypothesis. So, Ho Ho is that the probability is the same. So, probability is 0.3. And the alternative hypothesis is that the probability changed. In this case, we told that he thinks that the probability increased.

[00:38:18]So, we say probability is bigger than 0.3.

[00:38:23]We are told that the number of customers observed is 11. And we told to test at the 10% significance level. So 0.1. So in this case, X follows a binomial distribution of 25 because 25 customers were uh analyzed and 0.3 probability.

[00:38:46]Since we are testing if probability increased, we will check for probability of X bigger or equal to X observations.

[00:38:54]So 11 which is equal to 1 minus probability of x smaller or equals to 10 which I found from the table to be 0.0978 which is smaller than 0.1.

[00:39:08]Since it's smaller than 0.1 it means that it falls within the critical region. So HO is immediately rejected.

[00:39:15]Therefore, there is enough evidence to suggest that the proportion of customers, this is important, proportion of customers buying insurance increased.

[00:39:25]So, the main words that you needed to say here, HO is rejected, proportion, and increased.

[00:39:33]This was a June 2025 math Alevel statistics 2 paper.