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44 points BBUnderStat1o008. 0/15 Submissions UsedMy NotesThe incubation time for Rhode Island Red chicks is normally distributed with mean of 29 days and standard d...

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44 points BBUnderStat1o008. 0/15 Submissions UsedMy NotesThe incubation time for Rhode Island Red chicks is normally distributed with mean of 29 days and standard deviation of approximately day_ Look at the figure below and answer the following questions_ If 1000 eggs are being incubated_ how many chicks do we expect will hatch in the following time periods? (Note: In this problem; let us agree to think of a single day succession of days continuous interval of time. Assume all eggs eventually ha

44 points BBUnderStat1o 008. 0/15 Submissions Used My Notes The incubation time for Rhode Island Red chicks is normally distributed with mean of 29 days and standard deviation of approximately day_ Look at the figure below and answer the following questions_ If 1000 eggs are being incubated_ how many chicks do we expect will hatch in the following time periods? (Note: In this problem; let us agree to think of a single day succession of days continuous interval of time. Assume all eggs eventually hatch:) Area Under Normal Curve 2.3590 13.590 3490 349 13.596 2.35% p -Zo K + 3 6890 9590 99.790 (a) in 27 to 31 days chicks (b) in 28 to 30 days chicks in 29 days or fewer chicks in 26 to 32 days chicks



Answers

The incubation time for Rhode Island Red chicks is normally distributed with a mean of 21 days and standard deviation of approximately 1 day (based on information from The Merck Veterinary Manual). Look at Figure $6-3$ and answer the following questions. If 1000 eggs are being incubated, how many chicks do we expect will hatch (a) in 19 to 23 days? (b) in 20 to 22 days? (c) in 21 days or fewer? (d) in 18 to 24 days? (Assume all eggs eventually hatch.) Note: In this problem, let us agree to think of a single day or a succession of days as a continuous interval of time.

Looking at incubation of these chicks, and we're told that the distribution of incubation is approximately normally distributed, with a mean of 21 days in the standard deviation of one. So this would be at 22 days, Here's 23 there's 24. So marking one within one standard deviation within two standard deviations and within three standard deviations. Okay, So the standard deviation is one, and we have 1000 eggs, and we want to know approximately how many of them will hatch between 19 2, 23 days, and we can see if we look at our DeGraff, 19 to 23 days is within two standard deviations, and we know that that is 95% of those chicks is within what would happen within two standard deviations. So 95% of that will give us 950 chicks. Part B. Yeah, we want 20 right, 2, 22 days, you can. Mhm. And 20 to 22 days is from here to hear, and that's within one standard deviation. We know that 68%. So that would be 680 chicks. Part C. Just put that over here, we want 21 days or less, and we can see that that's dealing with the mean, and that's 50% of the chicks, and 50% of this will be 500 chicks. And then the last question question. D between 18 2, 24 days, 18 to 24 days, let's go with blue is from here to hear. And that's three standard deviations, and that's 99.7% would go within three standard deviations, and 99.7% of the 1000 will end up being 997 6. Mhm.

47. The main incubation time of a fertilized chicken egg kept at 100.5 degrees Fahrenheit in a still air incubator is 21 days suppose that the incubation times are approximately normally distributed. With the standard deviation of one day. We're also told that we have a normal distribution. We want to a determined the 17th percentile for incubation times of fertilized chicken eggs, so the 17th percentile on a quick sketch or we have the mean of 21 center deviation of one. We want to find what value on this X axis gives us 17% of the data below it. So when you use our to 84 Yeah. Use the inverse norman command. It is going to ask for the area we want to type in the area to the left of our boundary value that we're looking for inserted in as a decimal 0.17 The mean is 21 the standard deviation is one. This gives us the value of 20.5 so 20.5 is at the 17th percentile B determine the incubation times that make up the middle 95% of fertilized chicken eggs. So we're looking for the middle 95%. Now we could do this by estimating using the empirical rule or you can use your calculator to find the actual values. We still have the sketch of 21 and this is spaced out by ones by our standard deviation. And we're looking for that value that gives us this middle 95%. Now. In order to create this middle 95% I have an X value here and next value here. I need to find both of those values. So we're gonna use our T A a t fours inverse norman command. And because our distribution is symmetrical, these two values will have the same exact Z score except one is negative, one is positive. We could convert all of these disease scores by standardizing and then found it that way and then convert them back into incubation times. Or we could treat them as two separate problems. I'm going to treat them as two separate problems. So in this picture I'm gonna find X one first. I want to know what boundary value has this blue region to the left of it. So we're going to think a little bit. Well at this boundary point, what percent of the curves to the left of it with 95 is in the center. That means that we have 5% left over that 5% is evenly divided amongst both tails. So this tale has 2.5% to the left of it. So we want to find this first value by playing in 2.5% as 0.25 as our area are mean 21 our standard deviation is one. We can do this and we get this X one value to be 19.4 and then we can also do it with the X two value inverse norm. So we're looking for the screen region, which is 95 plus two 0.5, so 97 a half percent of the entire curve lies to the left of X two for Putin 97 5 as our area, with our mean of 21 our standard deviation of one, and this gives us 22 point 96 So those two values are 19.4 22.96

So let us look at this question. Now, the incubation tank are approximately normally distributed with a standard deviation of one day and mean as 21 days. Okay, so the mean happens to be 21 days and standard deviation is one day. Right? So this is 21 Sigma Standard Division is one. All right. Now, what is the probability that a randomly selected fertilized chicken egg hatches in less than 20 days? Less than 20 days? Okay, so we know the formula Z is equal to X minus mu by sigma. What is X? In this case, this is part a 20. So this is 20 minus move, which is minus one upon one, which happens to be minus one. So my Z statistic, my that statistic is minus one. Okay, now what do I do? I find the p value so my Z score is minus one. Great. This is one tail. You can either use a set table or you can use an online package like or any other statistical package like our or excel, which will give you, uh, the value that is exact and quite quickly. So p values 0.15 86 Okay, so, Mike, P value for this corresponding. That statistic is point 1586 Okay, so what is the probability that a randomly selected fertilized chicken and hatches in less than 20 days it is going to be 200.1586? All right, moving on. But visas? What is the probability that randomly selected for daylight you can takes over 22 days to hatch. Oh, shape. Let us look at this. Now, my ex is 22. My ex is 20 to 22 now, this is movie moved on to part B 20 to minus 20 by one, which happens to be too. So why is that statistic now? Is to I find the p value or two. Okay, I hit on calculate and I get 0.2 275 My P value turns out to be 2750.2275 So what is the probability if we read the question again? Okay. What is the probability that are randomly selected? Fertilized chicken egg takes more than 22 days to hatch. It is going to be the answer that we found. That is 0.2275 And if I want to find the probability that the egg will hatch in less than 22 days, right, 22 or less number of days, it will be one minus 0.0 to 75. Okay, so if I use my calculator for this, this is going to be one minus 0.2275 which is 0.97725 0.97725 Alright. Now moving on to part C. What is part C have to say, What is the probability hatches that the egg hatches between 19 and 21 days. Okay, 19 and 21 days. So let's just look at the diagram. The mean happens to be 2019 is going to be somewhere around here, right? This is going to be 19 and 21 will be somewhere around here. What we want is the probability in between. Okay, so what I do is I find the P value for 19 and find the P value for 20. And that is the areas in the teens, right? These two areas and I subtract both of them from one. So this is going to be one minus the P value of 19 minus the P value of 21. Okay, so if I put 19, this is going to be my Zach statistic. Turns out to be minus one for 21. My, that statistic turns out to be one. Okay, so if I find the p value for this, the value for both of them is going to be the same, because both of them are one away. One standard deviation away. Right? So my p values 0.1586 So p value is 0.1586 So disk reason is 1586 and this region is also 0.1586 So which means I have to do one minus twice off 0.1586 which is 0.6828 So this 0.6828 This is my answer for part C, where my probability that my ex is between 19 and 21 days. Yeah. Okay. What is part D? Would it be unusual for an act to hatch in less than 18 days? Yes. The probability that an egg hatches in less than 18 days is unusual because this is less than two standard deviations away, right? 18 is less than two standard deviations away. If this is 18, this is 20. This is 18. So this less than 18 means this This is less than two standard deviations of it. So this is going to be less than 2.5%. Okay, between 2.5 to 5% of the area. So yeah, this is rare. This is unusual.

So we have two populations population A as 100 millions and isil Jim also saying it reared Assange bar relacion de he has a genetically variable was actually just a one today. Unbearable. So we're just like that g p and we have no mention of the environment. So average weight of the yes, the excess laid by a get the two volumes of the various of 3.5 square and number two we have ah, me with 52 grounds of GM. We're varies of 21. So the first thing we condemn the notices the bear is the variability of environment. Oh, now ekes But let's keep it should be is raising environment is the equivalent to publish in a So there you go. So that being said what, he'll stay environmental Various for equity was engineer various They quit well with the environment actually being equivalent, it doesn t equal. By the way, that's a great word. There will be an improvement. We do know what it was in an environment Environmental factors must be playing a rope special in a because his ice agents there was no genetic difference. So we know that environment various that Beaky, as we call it, is on equal that just embarrassed. Doesn't have anything todo due to Oh, jeez. Now Population B has a total area of 21. If you remember correctly, it is on the same environmental conditions. What's so this name? And it will be, because again, the genes there. So we had a little more United V X equals E plus de G. So environment fishnets and so arranged his home look. Since we know two of these and V G equals the X minus B, which would then give us 21 minus 3.5, which would give us 17.5 where Molly's elsewhere but basically 10 list that Genet Americans accounted for US 17.5 size difference in the I A square here would be for B B G, divided by V six, which is 17.5 about like 21 which gives the 210.83 everywhere we're tryingto broke someone, it's on. There we go


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