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In biological treatment of wastewater; the time it takes & cell to grow (divide) is normally distributed with an average time of one hour and & standard dev...

Question

In biological treatment of wastewater; the time it takes & cell to grow (divide) is normally distributed with an average time of one hour and & standard deviation Of 5 minutes The probability that a cell divides in less that 45 minutes is: Points)0001350.0062005850.12500458702634Aeteeehetnee Nenbened ansuer

In biological treatment of wastewater; the time it takes & cell to grow (divide) is normally distributed with an average time of one hour and & standard deviation Of 5 minutes The probability that a cell divides in less that 45 minutes is: Points) 000135 0.0062 00585 0.1250 04587 02634 Aeteeehetnee Nenbened ansuer



Answers

Cell Division Let the expected number of cells in a culture that have an $x$ percent probability of undergoing cell division during the next hour be denoted by $n(x)$ .
(a) Explain why $\int_{20}^{30} n(x) d x$ approximates the total number of cells with a 20$\%$ to 30$\%$ chance of dividing during the next hour.
(b) Give an integral representing the number of cells that have less than a 60$\%$ chance of dividing during the next hour.
(c) Let $n(x)=\sqrt{5 x+1}$ give the expected number of cells (in millions) with $x$ percent probability of dividing during the next hour. Find the number of cells with a 5 to 10$\%$ chance of dividing.

It's clear someone named Reed here, so we know the density function for a normal distribution of a random variable X, with the main on standard deviation. So we're given that the mean is equal to 55 and the standard deviation is equal to four. So we have the equation of F of X is equal to one for its square root of two pi. Eat the negative X minus 55 square over 32. So the probability that a component will be made in less than 60 minutes, which is a Knauer to get X is less than 60 which is equal to but then to go from 0 to 60 of this equation. One over for a square root of two pi e the negative x minus 55 square over 32 d x, which gives us about 0.8 94 35

Okay, so this question is given a normal distribution of times it takes for this guy can to get to work. It's a normal distribution with mean of 25 standard deviation of 4.5. So yep. Stand deviation of 4.5 and he has 35 minutes to get to work before he is considered late. So and we want to see what percent of times, if it does indeed follow Roman distribution. What percent of times is this guy can going to be late for work. So basically 35 minutes or more is equal to late. And so there are two ways to approach this. One way is basically the second way is basically the first way, but with one extra step. So I'm gonna explain both ways, but I'm only going to do the first way on this whiteboard. Um, okay, So the way that I'm thinking is just calculate the percent off. Um, so let's draw this bell curve real quick with the 25 in the middle. Just calculate the percent of times it's 35 minutes or onward. So, like this, using the normal CDF function on your graphing calculator And if you don't have a graphing calculator, you can just search search a calculator up online. There are tons of them. So, yes, he did normal cdf um, remember to get to normal cdf at least 70 84. You go two seconds, and then you click the bars button, so get normal CDF of our minimum, which is 35 our maximum, which could just be any really large numbers so we can do 10,000 on that are mean andr. Standard deviation. And we should get using this method. We should get around 0.13 um, which goes to 1.3%. So that means that Ken is one late 1.3% of the time, which is actually pretty impressive everything about it. But also, you want to be late 0% of the time in most jobs. So, yeah, that's there's a little food for thought. The second way to calculate it is just take the easy score of 35 in relation to those normal distribution and then just do normal CDF off the Z score 1000 0 and one, and then you should get the same answer mainly because taking these quarters skills everything down to like our normal distribution was your own one. So, yeah, 1.3% is the answer because that's the percent of data points above 35 and that's the percent of times that can will be late.

In 15 minutes are bacterium is going to divide in half and give us to bacteria. Then in 15 minutes they're going to divide in half. Let's write this as a sequence to tell us the number of bacteria we have after a certain amount of minutes. So let's just assume that 15 minutes is one unit. So we can say that the lengthy unit is the set of 15 minutes will be the previous one A and minus one times to or more specifically, we could say that at any end moment, that is the end moment. There's going to be, too, to the end bacteria. So that is in the first recorded minute or set of 15 minutes. Rather, we're going to have to bacteria because we started with one, and then after 15 minutes, we and now have to. So let's write out the first couple terms for if we left this here for four hours, four hours times four sets of 15 minutes an hour, it means that we're going to do this 16 times using this formula where the term is just to the end. So that means that a two is going to be two squared or four. A three is going to be two cubed or eight a four. That's going to be two to the fourth power 16. And as you can see where this multiplying by two every single time, a five will be 32. Then we'll have 64 1 28 and eight. Halfway through. The experiment is to 56. These numbers are growing pretty fast, All right. Next up, we have a nine that's going to be 512. A 10 will be 10 24. A 11 will be 2048 and a 12 will be 40 96. We're almost done now. There's not quite enough room here in this last column, so I'm going to move down for the last couple terms. A 13. We're starting the last hour of the experiment now, and we're starting it with the term of 8192. Then we've got a 14 which is going to be equal to 16,384 then a 15 which is 32 7 68 And finally a 16 at the end of the fourth hour is the extraordinarily large 65,536. So, as you can see, if you have just one bacterium, you're going to end up with over 65,000 after just four hours. That's an insane amount of growth. But that is, in fact, what is going to look like as it is modeled by this sequence. So this is an accurate sequence for the modeling of what happens to these bacteria.

So to solve this problem, we're going to need to calculate four different T scores. One for each, um, specific number of minutes that we're looking at stone part. They were looking at the probability that somebody does less than 17 minutes. I used X to represent the number of minutes that they dio or do use the stair climber. So probability that it's less than 17 minutes. So we need a Z score for 17 minutes. Part B. We're looking for between 20 and 28. We need to one for 20 and one for 28 for part C, we're gonna need one for 30. I get dizzy score. We can use this formula. So Z score for 17 minutes, be given as X minus mu. So exes are 17 minus mu, which is the population, I mean, which is given us 20 just We divide all of that by the standard deviation, which is five Consol. This to find that disease scored 17 is negative 170.6. Now we do the same thing for a next see score. So the Z score for 20 minutes would be using the same formula. We're just gonna plug in 20 instead of 17. So right in that red bubble we're gonna put 20 instead. And then we calculated again we should get zero z score for 28 plugging in 28th this time. Okay, calculate that and you should get a C score of 1.6 and then our Z score for 30 should be too. So here's our Z scores. Oops, These air Z scores and now we can use them to find these probabilities. So each of these three scores represents of Mark on our standard normal curve. So our first the score of negative 0.6, we fall somewhere like here and then the corresponding value on RZ table represents the area to the left or the probability of a value being less than this value. So we look up negative 0.6 on RZ table. You should get points to seven. So 0.27 is the probability that somebody does less than 17 minutes. That's our answer for a part B is a little bit more complicated. Now we're looking for what's between two values. We have our first one, which again the corresponding values could be everything to its left. We also have a second one. And again, the corresponding value on RZ table will be everything to the left of that. Since we just want this middle section, we're going to take the corresponding value for our higher value. Sorry. 28. We're gonna take the course the Z score for the 28. So 1.6 and its corresponding probability, which again is going to cover this entire area to the left, all that green, and then we're subtracting off the same thing. But for are last year values. So the Z score of 20 everything to the left of that and that should just give us a middle region. This is just a generalized picture because the Z score of zero would actually line up or like here. But this gives you the general concept. So we look up our 1.6 on our, um, de table. You should get a probability of 0.945 and then we look up, are devalue for 20 on the Z table, and you should get a probability of 200.5 source subtracting 0.5. Then you can compute that, and you should get a probability of 0.445 and that's our answer for Part B. And then for part C, this is gonna be similar to part A. Except this time we want the area to the right instead of to the left, because we're looking for things that are greater than rather than less than because this represents a probability we're still gonna have our line. And normally the corresponding value gives us the area to the left. The probability that that it's less than that because we're working with probability, we can just do one minus this value, and that should give us the opposite, which is everything to the right instead, and that's what we're looking for. So we have our one minus, and then we look up our Z score or Z score for thirties to look up to on RZ table or again. You can use a graphing utility. You should get 0.977 We have one minus 10.977 gives us 0.0 to 3, and that's the probability that somebody's on the climber for more than 30 minutes.


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