May/Jun 2023 WST01 (Q4 - 7) - IAL Statistics 1

Ajouté : 2026-05-20

[00:00:00]Hi, welcome to my channel PI byM Maths.

[00:00:03]This is by Jasan.

[00:00:05]Here we are solving May 2023 statistics one paper. In part one of this video, we looked at question number 1, 2, and 3. Let's jump into question number four. A bag contains a large number of colored counters. Each counter is labeled A, B, or C. 30%age of the counters are A labeled as A. 45 are B.

[00:00:30]The rest of them are C. That means this is 75 already. So this will be 25%age.

[00:00:38]It is known that 2 percentage of the counters labeled as A are red. 4 percentage of B are red. 6 percentage of C are red. One counter is selected at random from the back. Complete the tree diagram on the opposite page to illustrate this information. We have a tree diagram given and in the question it says A is 30%age that's why.3 B is 45%age so it's 0.45 so the balance the total should be 1. So C is going to be 1 minus 75.25.

[00:01:15]Now in a two percentage of M are red right that's what given here right for B 4 C 6. So for B it's 4 C it's 6 but this total should always be this total should be 1.

[00:01:33]So if this is 0.02 02. This is no brainer. Just subtract from 1 one and write it here. 0.98 here. 0.96 0.94.

[00:01:47]That's all. We have completed the tree diagram already. That's the two more question. Now we are going to solve next part using this tree diagram.

[00:01:56]Part B. Calculate the probability that the counter is labeled A and is not red.

[00:02:03]So they want you to find P of A and not red.

[00:02:11]So you have to take P of A multiply by P of A and not red.

[00:02:21]So we cannot just put P of not red because it can be B and not red also. So we what we can do it's a two more question right?

[00:02:30]A and not ready. You just need to multiply these two.3 and 98.

[00:02:37]So 0.3 * 0.98 0.3 * 0.98 0.294 is the answer. It's a part it's a two more question.

[00:02:56]Let's move on to part C. Now calculate the probability that the counter is red.

[00:03:01]They want you to find P of red. It can be P of A and red plus P of B and red plus P of C and red.

[00:03:18]We just calculated A and red. Oh, sorry.

[00:03:21]So what we calculated here is A and not red. So here we need to multiply a 100 b 100 c.

[00:03:32]So 3 02.3 * 02 plus b is450 red is 04 plus c.25 25 * 06.25 * 06 Let's calculate this now.

[00:04:02]0.3 * 0.02 + 0.45 * 0.04 + 0.25 * 0.06 0.039.

[00:04:26]That's the answer. Let's move on to part D. Now, usually you will always see a conditional probability in the uh tree diagram questions here. Given that the counter is red, that's a condition.

[00:04:43]Find the probability that it is labeled C. So, they want you to find this.

[00:04:49]So use the formula for P of A given B is P of A intersection B divided by P of B.

[00:04:59]C is this the one circuit.25 * 06 0.25 * 0.06 06 divided by P of red which is what we calculated here P of red 0.039 0.039 don't calculate the values again you should be able to use the answers from the previous parts actually 0.25 * 0.06 divided by 0.039 039.

[00:05:45]Okay, that's a recurring decimal.

[00:05:48]5 over3. I'm going to keep it as 5 over 13. That's all. Question number five.

[00:05:55]This is from chapter 6. Discrete random variable. A discrete random variable has probability function. This show that k is 1 / 30. Okay. Sub in y is equal to 1 here, 1 and two here.

[00:06:10]And then these values here when it's six it's just k and if you add all the probabilities it should always be equal to 1 right that's what we are going to do when you sub in one here 3 - 1 is 2k plus when you sub in 2 3 3 - 2 is just 1 1k and then when you sub in 3 here 3² 9 - 8 is 1k 4 square - 8 is 8k 5² 25 - 8 is 17k and then when y = 6 it's just k.

[00:06:52]This total probability should always be 1. So if you add them you get 10 27 30 k is 1.

[00:07:01]K is 1 / 30. Let's move on to part B.

[00:07:07]Find the exact value of P of 1 to 4.

[00:07:12]Let's draw the probability distribution table first.

[00:07:16]Okay, I have drawn the table here. If you want, you can replace K by 1 / 30 and write it this way. Instead of 2K, put it as 2 over 30, 1 / 30.

[00:07:27]This also 1 / 30 8 over 30 17 / 30 1 / 30. Now part B they want you to find P of 1 is less than Y is less than or equal to 4. So we don't take one we take the values more than one including four till four. So y = 2, y = 3 and p of y = 4. Take the corresponding probabilities 2 3 and 4.

[00:08:06]So you get 1 / 30 + 1 / 30 + 8 / 30 which is 10 / 30 1 / 3. Part C, find E of Y. The formula to find the expected value of y is sigma y * p of y.

[00:08:28]That mean basically you need to multiply the y with the probability and then sum up find the summation of so there is 1 / 30 is a common factor.

[00:08:42]I'm going to multiply 1 by 2 + 2 by 1 + 3 by 1. If you don't understand this, do not worry about the common factor. Just multiply 1 by 2 over 30 + 2 by 1 / 30 3 * 1 / 30 4 * 8 / 30 5 * 17 / 30 6 * 1 / 30.

[00:09:16]You can keep 1 / 30 as a common factor outside also.

[00:09:22]So what I'm going to do, I'm going to put 1 / 30 outside.

[00:09:26]Yeah. Okay.

[00:09:28]And then bracket 1 * 2 + 2 * 1 + 3 * 1 + 48 are 32 + 5 * 17 is 50 85 + 6 * 1 close the bracket 13 / 3 is the answer for D. Now the random variable X is this. Calculate P of Y is greater than or equal to X.

[00:10:03]So first write it as P of Y is greater than or equal to replace X by the expression given. Make Y as a subject.

[00:10:12]So you'll get 3 Y is greater than or equal to 15ide by 3 Y is greater than or = 5.

[00:10:21]So you need to take P of Y = 5 + P of Y = 6 from the table. It's 17 / 30 + 1 / 30.

[00:10:37]18 / 30 18 / 30 is 3 over 5. That's the answer.

[00:10:49]Part E. Now variance of x is same as variance of 15 - 2 y and we have learned this rule actually e of a x + b is a e of x + b but if it's variance of a x + b we ignore b square the cofficient of x a square variance of x. So in this case the cofficient of x is -2.

[00:11:21]So it's -2² variance of y which is 4 * the formula for e of variance of y is e of y² minus e of y square.

[00:11:36]So we are going to calculate this actually. First we need to find e of y².

[00:11:43]E of y square is sigma y² * the probability.

[00:11:51]So from the table we need to square these values square these values and then multiply by the probability and then find the summation of it.

[00:12:05]That's what e of y square. So, it's going to be 1 2 * 2 over 30 + 2² 2 * 1 / 30 + 3² * 1 / 30 + 4² * 8 / 30 5 2 * 17 / 30 6 * 1 / 30 We are going to calculate this now.

[00:12:41]So I'm going to put 1 / 30 as a common factor.

[00:12:47]1 2 * 2 is 2. 2 * 1 is 4. 3 2 * 1 + 4 2 * 8. Okay, I'm going to put it as 4² * 8 + 5² * 17 + 6² * 1 is 36.

[00:13:14]So I get 302 over 152 over 15. So here we sub in the value 302 over 15 minus we calculated e of y here right which is 13 / 3 so square of it 13 / 3 square of it calculate this now so this answer - 13 / 3 square of * 4 24 / 45 is answer.

[00:13:58]Question number six. Now there are three events A, B, C such that P of A is given. B given A is given. The conditional probability P of A union B is this. C is this. Given that A and C are mutually exclusive. The formula for mutually exclusive is because there will be no intersection between these two set. So A union C is simply P of A plus P of C. So take the values at them.1 +.5 is equal to 6.

[00:14:33]It's just one more question.

[00:14:36]Okay, let's look at part B. Now I have kept all the values here. So to find part B I'm going to start with this actually the formula for P of B given A is equal to the formula for conditional probability P of B intersection A over P of A right so this value is.3 B intersection A is same as A intersection B divided by.1 so we know P of A intersection B is.1 * it's going to be 0.0 03.

[00:15:11]Now I'm going to use this probability addition formula. P of A intersection B is equal to P of A plus P of B minus P of A union B. Okay.

[00:15:25]Now sub in all the known quantities here.

[00:15:30]A intersection B is P of A. We don't know P of B. Keep it as it is. A union B is 0.25.

[00:15:39]So bring everything this side you'll get P of B is 03.28 - 1 is 0.18.

[00:15:50]That's all. Three more question.

[00:15:53]Given also that B and C are independent.

[00:15:57]If B and C are independent the formula is P of B intersection C is P of B * P of C. So take P of B times P of C. P of C is 0.5.

[00:16:16]So the intersection part of B and C would be 0.18 * 0.5 0.09.

[00:16:30]Now I'm going to draw the V diagram. You cannot just put three circles this way.

[00:16:35]That's wrong because you need to read the question from the beginning carefully. In the beginning, it says A and C are mutually exclusive. That means there is no intersection between A and C. So I'm going to put a A here and I'm going to draw C here without intersecting A and then B will have intersection with these two.

[00:17:04]So this is a C b and then let's put a box around the circles.

[00:17:15]Okay. Now B intersection C is 0.09.

[00:17:20]We know A intersection B we just calculated here 0.03.

[00:17:28]We know P of A 01. So we already put 0.03 here. So the balance 0.07 and we know P of C P of C is.5 and the intersection is 09.

[00:17:44]So the balance will be 5 minus 09.41 here.

[00:17:52]And we know P of B is 0.18.

[00:17:56]That's P of B. The whole circle we need to subtract 03.09.

[00:18:04]So.12 will be gone. So you'll get 0.06.

[00:18:10]Now sum of all the probabilities should be equal to 1. If not put the balance outside the circle.

[00:18:18]So 0.07 + 0.03 + 0.06 + 0.0 09 + 0.41 41 66 - 1 will be.34 0.34 here. That's all.

[00:18:45]You will get five marks for this.

[00:18:48]Let's look at question number seven. A machine squeezes apples to extract their juice. The volume of juice jl extracted from 1 kilogram apple apple is modeled by normal distribution with mean mu sigma standard deviation sigma. So you can write j follows normal distribution of mu comma sigma squ.

[00:19:14]Show that part a i show that p of j is greater than 510. Is this okay?

[00:19:24]Standardize the normal distribution using this code. Zed is X - mu / sigma.

[00:19:31]So Z is greater than X is 510. Mu is 500. Sigma is 25.

[00:19:40]So when you calculate this you'll get 10 / 25. 10 / 25.

[00:19:48]P of Z is greater than 0.4. and four.

[00:19:53]Finding the area on the right hand side of a positive number is same as 1 minus the area on the left hand side of the positive number. So you can use the table value of 0.4 for 0.4 5 is here 0.4 4 is this 654 0 1 - 0.654 and 654 is 0.3446.

[00:20:28]That's what they want us to prove.

[00:20:30]Easy. Two more question. We are going to move on to double I. Now we proved I already double I. Find the value of D.

[00:20:38]Calculate the value of D such that P of J greater than D is this. Okay.

[00:20:43]Standardize it. Z greater than X - mu / sigma is equal to 0.9192 PF. So what I notice when I give this question to students right they immediately open the table look for this value. You shouldn't be doing that actually because you use the table only if the if this value is a positive value. So you need to know whether this is positive or negative. So I'm going to draw a build curve first standard normal distribution. Let's say mu is here. Okay. Now zed is greater than means the area on the right hand side of a point and this value is more than 0.5. We know this is 0.5. This also 0.5. So you need to draw a line somewhere. You need to take this value somewhere. It can be here on the negative side or it can be on the positive side. But the condition is the area on the right hand side of the point is more than 0.5.

[00:21:59]If you take the value positive, this area is not more than 0.5. Right?

[00:22:05]So definitely the value is negative. I hope you understand this because if you take this negative value definitely this area is more than 0.5 right so this value is negative now we cannot use negative value for the table we need to change it to positive how do we change a negative value to positive we multiply by minus one right so it becomes 500 minus du over 25 and then the area on the right hand side of a negative number is same as area on the left hand side of a positive number.

[00:22:44]So zed is less than is equal to 0.9192.

[00:22:53]Now this is a positive number. You have a four decimal place a digit a number with four dp here. So we are going to use the table the big table and look for the value 9192 91 92 is here 1.4 1.40 is the corresponding value. So equate this to 1.40.

[00:23:22]So 500 minus d / 25 is 1.4. make D as a subject.

[00:23:33]So 1.4 * 25 and then subtract it from 500 we'll get 465.

[00:23:45]465 is the answer.

[00:23:48]Now I'll give you a shortcut here if you don't understand how to change this to this way and changing the symbol and all what you can do first definitely you need to figure out whether this number is positive or negative they're talking about area on the right hand side this value is more than 0.5 so definitely this value is a negative number so what I want you to do you can skip this step if you want you can Skip it straight away. Take the table value of this. You'll get 1.40. Right? So write it as dus 500 over 25 is - 1.40. That's one that's a shortcut actually instead of writing this changing the negative number to positive and then find the value of d you can do this. Actually you can straight away get the value put a negative sign and then make d as a subject you will get the same answer.

[00:24:51]But it's better to understand this part why why this number is a negative number and how do we change it to positive and flip the sign. You can watch my uh remote learning series for normal distribution where I have explained this very clearly.

[00:25:08]Now part B. Zen randomly selects five bags each containing 1 kilogram of apple and records the volume of juice extracted from each bag of apple.

[00:25:19]Calculate the probability that each five of them produce less than 500 ml.

[00:25:25]The machines that produce the bags of apple that produce less than 500 ml is the probability of success. P of J is less than 510 is the probability of success. We have calculated J is greater than 510 as this one. So less than 510 will be 1 minus this 1 - 0.346 you will get 0.6554.

[00:25:59]That's a probability of success here.

[00:26:02]And then they want you to find five out of five. It's binomial distribution.

[00:26:08]The binomial distribution usually they ask this type of question actually.

[00:26:13]That means let's say you have five items.

[00:26:17]They want you to pick two items and the probability of success is given.

[00:26:25]Probability of failure is 1 minus P. You can the formula is 5 C2 P power success power 2 failure power 5 minus 2. In general this is the rule N C R P power R Q power N minus R.

[00:26:46]Actually there is a separate chapter in statistics 2 exclusively for this binomial distribution. But in S1 they add a two or three mark question involving this. How do we figure out this is binomial distribution? Usually the question will be like this five out of some number. Sorry this five out of this or two out of five. There are questions where they say there are five letters delivered and find the probability that two of them delivered by RA and all. If you go through the chapter wise question for normal distribution and then you need to look for probability of success. This is the probability of success here P. So immediately you can calculate Q which is 1 minus P is going to be 34 6.

[00:27:42]Now they want you to find 5 out of five.

[00:27:44]So we are going to use this rule. So P of J is = 5 is oh sorry not we cannot write J is equal to 5. The notation we are going to use another variable. Let's use X. X follows binomial distribution of 5 comma the probability of success.

[00:28:05]So you don't need to write this step actually you can straight away write x = 5 is 5 c 5 p ^ 5 q ^ 5 - 5 which is zero 5 c 5 is just 1 this power 0 is also one you just need to calculate 65 4 to the^ 5 0.654 to the power 5 0.1209 dot dot dot and round it to three significant figure that's all. So watch out for this binomial distribution. Mostly they add such questions in normal distribution question. Actually you'll see one of the part involving binomial distribution.

[00:29:01]Let's move on to part C. It's a seven mark question. We need to find the value of R and K. There is K here. The sigma R is this here also there is R. So basically we are going to frame two equations and solve the simultaneous equation. The first one P of R is less than R is 0.15.

[00:29:26]Definitely in the bill curve r is a positive number.

[00:29:30]Why I'm saying that?

[00:29:37]Let's say this is um the mu 520.

[00:29:42]We know this area is 0.5. This area is 0.5. Right? So the total area below the curve is one. First I'm going to standardize this. Z is less than X - mu over sigma is 0.15.

[00:30:04]So I can use a b curve for standard normal distribution. It's zero here.

[00:30:08]Okay. So when they say zed is less than they're talking about the area on the right left hand side of a point which is smaller than 0.5. This value is smaller than 0.5. Right?

[00:30:22]So definitely the number should be here on the negative side. It should be a negative number. This should be a negative number because the area on the left hand side it's given as less than.5.

[00:30:37]So we cannot use a negative number a negative number to when it comes to table right using the table. So I'm going to change it to positive here. So this left hand side area is same as the right hand side area of a positive number. So I can put it as P of Z is greater than how do we change this number to positive? Multiply by minus 0.15.

[00:31:05]This is like a rounded figure. This is a positive number. So we can use a smaller table. Actually look for 0.15.

[00:31:16]0.15 is here and the corresponding value is 1.0364.

[00:31:24]520 - R / K is 1.0364.

[00:31:32]Frame equation.

[00:31:35]Bring the K this side. Put all this side. You'll get 1.0364. 03 64 K + R is 520 take it as equation one now using this we are going to frame equation two now and solve the simultaneous equation okay I'm going to standardize this first zed is greater than X is this whole thing minus mu over sigma is 0.005.

[00:32:13]If you draw a bell curve to understand it whether to check whether it's a positive or negative number, the zero is here. The area on the right hand side is this number is smaller than 0.5. So definitely the value should be here. So that the area on the right hand side is smaller than 0.5. Right? So we know this is a positive number. So we can straight away use a smaller table because this is a positive number. This is like a a rounded figure, right? 0.5.

[00:32:46]So 0.5 is here. So the corresponding value is 2.5758.

[00:32:55]2.5758 is equal to this is 3 R -,320 over K.

[00:33:08]So bring the K here 3 R here you'll get 2.5758 K - 3R is -320.

[00:33:22]That's equation two.

[00:33:25]So we have two linear equations.

[00:33:28]Equation one here, equation two here.

[00:33:31]Solve the simultaneous equations.

[00:33:35]So I have both the equations here.

[00:33:37]Equation 1 and two. I'm going to multiply equation 1 by 3. First 1.0364 1.0364 364 * 3 3 1092 31092 + 3 R is equal to,560.

[00:34:06]Now add these two equations. The 3 R 3 R will be cancelled.

[00:34:12]This plus 2.5758 5.685 5.685.

[00:34:23]Sorry, there was a K.

[00:34:27]K is equal to 1560 - 1320 which is 240. So find the value of k 240 divided by 5.685 42.216 etc etc that's the value of k so what I'm going to do I'm going to sub in the value of k here to find the value of r.

[00:34:59]So multiply the k by this value and subtract from 520 1.0364.

[00:35:06]So times 1.0364 - 520 476 476 2469 etc. Now let's go back and read the question for the level of accuracy extracted from 1 kilogram. Nothing is mentioned. So we'll round it to three significant figure. So we can write therefore K is 42.2 R is 476.

[00:35:46]That's all.