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A recent study by the Lembaga Lebuhraya Malaysia revealed that20 percent of Malaysian drivers do not use their seat belts. (a) A sample of 15 driversis select...

Question

A recent study by the Lembaga Lebuhraya Malaysia revealed that20 percent of Malaysian drivers do not use their seat belts. (a) A sample of 15 driversis selected. What is the probability that: (i) 4 or fewer of the drivers are not wearing seatbelts? (2 m)(ii) more than 6 of the drivers are not wearing seatbelts? (2 m)(iii) less than 4 or more than 10 are not wearing seatbelts?

A recent study by the Lembaga Lebuhraya Malaysia revealed that 20 percent of Malaysian drivers do not use their seat belts. (a) A sample of 15 drivers is selected. What is the probability that: (i) 4 or fewer of the drivers are not wearing seat belts? (2 m) (ii) more than 6 of the drivers are not wearing seat belts? (2 m) (iii) less than 4 or more than 10 are not wearing seat belts? (3 m) (b) Calculate the mean, variance and standard deviation of the number of drivers who do not wear seat belt in a sample of 1000 randomly selected drivers.



Answers

Police estimate that 80$\%$ of drivers now wear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use.
a) How many cars do they expect to stop before finding a driver whose seatbelt is not buckled?
b) What’s the probability that the first unbelted driver is in the 6th car stopped?
c) What’s the probability that the first 10 drivers are all wearing their seatbelts?
d) If they stop 30 cars during the first hour, find the mean and standard deviation of the number of drivers expected to be wearing seatbelts.
e) If they stop 120 cars during this safety check, what’s the probability they find at least 20 drivers not wearing their seatbelts?

Chancellor since or giving honest people to 500 p is equal to 0.70 The requirements are met and P is equal to 100 times 0.7 just 3 15 bigger equals 10 and then times one minus p. People to 500 turns one You minus 27 which is equal to 1 50 Just figured you'd want to 10. So we move on to parking. So we have seen is equal to x minus and p over a square root of NPD tons one minus people. This is equal to three inch of 19.5 minus 400 times 0.7 over a square It 0 500 times 0.7 times one minus points. Um, you might be asking us. So why are we using 319.5? This is because of the continuities correlation. We're gonna do the same thing for by using 370.5 minus 500 turns 5000.7 over a square of 500 terms. 0.7 turns one point throws one minus 0.7. Just put it to 0.0 And when we calculate this between we got this is equal to negative 2.90. When you calculate this between you get of are you. 0.97 59 For part, B C is equal to 324.5, minus 500 turns 5000.7 well over a squared off 500 tons, 5000.7 times one minus point. Stopping is equal to negative 2.49 And the probability? Um, Becks, this less spring 325. I was able to the probability the Texas less than 324.5 because of problem continually correlation, which is equal to probability. Mugsy is less than negative. 2.49 just pickle chip won't Sarah rose zero 639 fewer than 315 and C is equal to 314.5, minus 500 turns 0.7 over a square root of 500 times 0.7 turns one minus 10.7 just equal to negative 3.46 So the probability that fence was less than 315 equal to probability that excess less than 314 there in your 15 and three inch of 14.5 because of continued correlation again, a sequel to Probability that see it's less than negative three point forces just able to point their other of Sarah to seven.

In this exercise were given the scenario in which 75% of all drivers always wear the seat belts. And then we're investigating five cars to see if the drivers are wearing seat belts. The investigation of each car can be regarded as a Bernoulli trial because we can see that there is a success and a failure outcome and success would be wearing a seat belt. And we could also say that each trial is independent of each other because the likelihood of wearing a seat belt is not effective for anyone driver by the situation in the other cards on the highway. So if we're investigating five cars, we can say X is the number of successes in five cars in which Case X is distributed according to a binomial random variable based on five trials and a probability of success of 0.75 For part A were asked to describe how we would simulate the number of seat belt wearing drivers among the five cards, so we would need to find some type of random number generator that has a probability of success of 0.75 So one example might be that you could use a deck of cards and you randomly draw a card from the deck. And once you've taken that card, you replace it back into the deck and shuffle again, and you could designate three suits as being success. And then one of the suits is being failure. And so then, once you've randomly selected five cards from the deck one at a time, replacing and shuffling in between, you could look at how many of these success sits that you have. Alternatively, you could define a random number generator in in software, that it has a probability of success of 75% and run that five times and count up how many successes you have. But that would only be one experiment. You would then want to repeat that many times to see how the number of successes varies and how many year most likely to get. So to get an idea of the probability distribution for the number of successes out of five trials, so for Part B were asked to run at least 30 such trials. So I did mine in our software, and I'm not going to run through the code line by line because that's beyond the scope of this tutorial. But I basically just defined a random variable with probability success of 75% and I ran it five times and counted the number of successes. And then this gets repeated many times, and you can see done 50 trials in this simulation. And my results for these 50 trials are given in this line here, and you can see that it's mostly fours and fives and there's a few threes, a few twos, and there are no one's and No. Zero. Now we know there is a non zero probability of getting zero successes and one success, so you might look at this and think, OK, 50 simulations is a few too few for this type of scenario, but anyway, these air my results for Part B and for Part C, we're asked to estimate the probabilities for the different numbers of possible successes based on our simulation so we could make a Table X is the number of successes so have X equal to zero one to up to five. And to estimate these probabilities, we can use the relative frequencies from our simulation results to recall that the relative frequency for our simulation would be equal to the the number of trials, with I successes divided by the total number of trials. So, for example, for I equals two, we have one to four. Looks like we have four four trials that had that resulted in two successes. So in this case, the he estimated probability for X equals two would be 4/50 in that equals 0.8 So if we repeat this calculation for all possible values of successes based on our simulation, we get these results. So that concludes part. See, that's our estimate for the probability distribution. And now for party, we are asked to calculate the actual probability distribution. So the probability mass function forex, based on every newly random variable or IT binomial random variable of five trials and probability success of 0.75 is given by five Choose X time 0.75 to the exponents X time 0.25 to the exponents, five minus x. So we would repeat this calculation for all possible values of X from 035 And so we could add this to as another role in our table and we get the following results. And so our answers for part D are provided in this rule of the table. And finally for party were s to compare the distribution of outcomes from our simulation with the actual probability model. So we've already done that by putting them in the same table so you can easily make a 1 to 1 comparison for the different probabilities of the outcomes. And they're actually quite close. You could say that our simulation resemble is the actual problem probability model Quite closely, Any time you have actual probabilities that air this low, you know, for a 0.1 you would only expect to get one of these out of 1000 trials. So 50 trials is not likely to give you a good estimate of that value.

In this exercise were given that the probability is 0.4 that a traffic fatality will involve an intoxicated or alcohol impaired driver or non occupant in the first part of the problem. A We're supposed to find the probability that the number y of fatalities that involved into intoxicated or alcohol impaired drivers or non occupant is exactly three, at least three. And at most three. Now, before we can do that, uh we we would need to create the probability distribution for that uh random variable Y. And first, why can take on the value zero All the way to 8? Because we are looking at eight traffic fatalities. This make the numbers all the way 28 And then the next column will be for the probabilities. So six. See Yeah, you need to fill up the table with a different probabilities for different numbers. So we use a binomial distribution. Yeah. Number of fatalities is zero out of eight trouts. And the successful ability is 0.4. Mhm. Since and and it's a false and I yeah we copy the formula all the way through for the eight different values of y. Now we have our probabilities, the probability distribution. And in part a of the question We're looking at three different probabilities. The first one. The probability but why is exactly three? That means that we're looking at Y equals three six. Yeah. Oh And come up to the table. The value is zero points. So you just need to put that they're 0.278. Which you can round off to make it 0.279 67 Sure. Next looking at the probability That the number is at least three. So at least three means It could be three four or or four or 567 and eight. So the probability that y is at least three. So sad sleep. You obtained as follows. You have to get uh the probabilities for zero one and to then we add them up and from then and then we subtract from one. The some of these probabilities. So it will be equal to one minus Some of your abilities of 01 and two. See That gives us 0.684605 Michigan round off to 0.685. Next we're looking at the probability that the number is at most three. I believe that by he is at most three. Yeah this is Now for white we at most three it means why could be 012 or three but not any number Greater than three. So we need to get the some of these four probabilities. So we put the formula equals the sum first. For That will be 0.594-086. Which you can round up to zero five and four. So x. In part B. We're supposed to find the probability that the number is between two and four inclusive. That means they're going to be looking probability that it's two or 3 or four. So we focus on these three probabilities and get there some. So the some of those three possibilities is given by the formula equals some. It's three powerful. That's going to be 0.719954. Which can round off 0.7 20 But see we're supposed to find and interpret the mean of the random variable. Y. Now the mean of the random variable by meal is given by N. P. In this case n equals eight. So you put the formula equals eight Times here, which is 0.4 success probability. So the mean is 3.2 fatalities. So we can interpret it as follows that on average 3.2 of every eight traffic fatalities involved an intoxicated or alcohol impaired driver or non occupant. Yeah. And lastly, but d sorry, supposed to obtain the standard deviation of why? Now the standard deviation is given by the square root of n times p times one minus P. Which we can compute using the formulas with the equal sign and then square uh where it off? Yeah. Uh n which is eight times p which is 0.4 Times one may not be, which is 0.6 plus, And you have not at 1.3 uh which we can approximate it to the 1.4385641, which is 1.4 traffic fatalities. Mhm. Yeah.

Okay, so our problem is about licensed drivers. So let's go to party. Party is asking us about Ah, the probability of them being a male, right, a male and and being 19 years old or on there, Right? So I'm gonna put years, So let's see. So here regard the probability of being male and let's less than or equals 2 19 years of age. I'm gonna put age or years, right? So let's see. So that will be. When we look at the table, we're gonna end up with 100 MT. That is 4746 divided by the total amount of all the licences held. So 15,747. And that's gonna give this a probability of 0.301 thousands. Okay, so that's how you work on part A. Now let's go to part B. So for part B, this one is a little bit different. This one is either they are 20 years of age, right? 20 years of age or or they are female. They are female, okay? And I'm using this female symbol here, so let's see. So 20 years of age or female. Okay, so here we're gonna use the addition, Rule number two. So in order to get that, I'm gonna start with the 20 years of age, okay? 20 years of age plus bright, and I'm gonna add the probability of them being female. So all the female license driver's minus that. That is overlapping, right? The overlap, which is any female That it that that is 20 years, all 20 years old. So let's see how it looks like. So we're gonna take that first amount. So anybody that is 20 years of age. So they are 3178 plus right that, plus right here being the probability of being female or the amount of female. So 7697 minus that overlap of females that are 20 years of age, so minus 1553. And that's gonna give me 9000 322. And I'm gonna divide that by the total amount of licensed drivers. So 15,747. And that gives me a probability of 59 hundreds. And there you go. That's how you get for part B? Now that's going part C So poor for part C, they're asking us to find the probability of anybody that is greater. Ah, more than 20 years of age, right so greater or equals to 20 years of age. I'll put age here anytime I put age of years. We're talking about it cycle yet years of age. So here we're gonna take the probability of being 20 years of age female and plus 20 years of age males, right plus smells. And that's where I'm using both symbols and for us, the probability of being 21 female, right, and males. And that's gonna give me now let's look at the amount. So we got 3178. There's 3000 306 and that's gonna give me 6484 and I'm gonna divided by the total amount of license rivers of 15 1007 147 and the probability of being greater than 20 years of age. And remember, we used both. Um, sex is in this one or genders, and it will be zero point 412 thousands. And there you go. That's the probability off anybody that is greater than or equal to 20 years of age.


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