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Suppose the lengths of the pregnancies of certain animal are approximately normally distributed with mean pl = 194 days and standard deviation 0 = 19 dayClick here ...

Question

Suppose the lengths of the pregnancies of certain animal are approximately normally distributed with mean pl = 194 days and standard deviation 0 = 19 dayClick here to view the standard normal distrbution table (page lick here to_view the standard normal distribution table (page 2)_(a) What is the probability that a randomly selected pregnancy asts ess than 188 days?The probability that a randomly selected pregnancy lasts less than 188 days is approximately 0.3761 (Round to four decimal places a

Suppose the lengths of the pregnancies of certain animal are approximately normally distributed with mean pl = 194 days and standard deviation 0 = 19 day Click here to view the standard normal distrbution table (page lick here to_view the standard normal distribution table (page 2)_ (a) What is the probability that a randomly selected pregnancy asts ess than 188 days? The probability that a randomly selected pregnancy lasts less than 188 days is approximately 0.3761 (Round to four decimal places as needed:) Interpret this probability: Select the correct choice below and fill in the answer box within your choice _ (Round to the nearest integer as needed ) CA If 100 pregnant individuals were selected independently from this population, we would expect pregnancies to last more than 188 days: If 100 pregnant individuals were selected independently from this population, we would expect pregnancies t0 ast exactly 188 days_ If 100 pregnant individuals were selected independently from this population, we would expect 38 pregnancies to last less than 188 days_ (b) Suppose random sample of 23 pregnancies is obtained. Describe the sampling distribution of the sample mean length of pregnancies_ The sampling distribution of x is normal (Round to four decimal places as needed:) with 194 and Gx 3.9618 (c) What is the probability that random sample of 23 pregnancies has mean gestation period of 188 days or less? The probability that the mean of a random sample of 23 pregnancies is less than 188 days approximately (Round to four decimal places as needed:)



Answers

The length of human pregnancies is approximately normally distributed with mean $\mu=266$ days and standard deviation $\sigma=16$ days. (a) What is the probability a randomly selected pregnancy lasts less than 260 days? (b) Suppose a random sample of 20 pregnancies is obtained. Describe the sampling distribution of the sample mean length of human pregnancies. (c) What is the probability that a random sample of 20 pregnancies has a mean gestation period of 260 days or less? (d) What is the probability that a random sample of 50 pregnancies has a mean gestation period of 260 days or less? (e) What might you conclude if a random sample of 50 pregnancies resulted in a mean gestation period of 260 days or less? (f) What is the probability a random sample of size 15 will have a mean gestation period within 10 days of the mean?

All right. So, we're looking at the length of human pregnancies, which is approximately normally distributed with a mean 266 days days. And then the standard deviation of the population 16. And were asked some questions about what's the probability randomly selected pregnancy uh lasts less than This many days. 260 days. Well, so they were basically asking basically we are asking the problem a probability that a random sample two away. Mhm. A selected pregnancy. So this one we're just gonna take X and want that to be less than 260 days. So for this one, what we're gonna need is the Z score, which is calculated by x minus mu over sigma, Which for our purposes here is to 60 -2, 66, all over 16, which is going to give us -375. And then we do a look up table and that would tell us that the probability is equal to 0.34, looks 354. Excuse me. Mhm. And then, just as a little picture, here's your normal distribution This .354. So, here's your mean Of to 66- 60 is somewhere over here. We'll call that to 60. And then it's this area, This number .354 is this area. And then be we have the probability that a random sample with 20 as a mean of 26 year last. So we have the probability that the sampling distribution has a mean less than 260. So this one we're gonna apply the we use the central limit theorem which says that the sample distribution has a mean that is approximately normal or that excuse me. That is that the sampling distribution is approximately normal given a and we as long as the problem the population distribution is normal, which this is we can use this. He was essential limit room which says the Z scores was given this way with the sample mean minus mu over the standard error of the mean. Which is given as sigma over route. And um If it's not normally if the population distribution is not normal then we have to have the sample size of and greater than 30 but it's our population is already normal. So we can just apply this. So we're gonna do Z score. But we want the X. Bar which in this case is to the mean to be to 60, That is to 66 all over the The standard error of the mean, which is in this case is 16 over and then N is 20 in this case. So route N if you do that, you end up with the negative 1677. That's the Z score. Mhm. Okay. And then this is going to be equal to You. Do a little table look up or some sort of calculating device to get 0.047. Yeah. And again that would be a the area over here. Although the the sampling distribution Oh and the sampling distribution of a population that's approximately normal would have a mean, That is the population means so we can do that then for c it's similar. Uh except the sample size is now 50, so we do the same kind of thing. Mm The probability of the X. Bar is less than 2 60. However, the mean is different. So just to Put that all together here, this is where N20 20 and this is where and is 50. So it's a larger sample. And that means we're gonna do the same thing with the Z score to 16 -2 66. of Make that six. All over 16 over route 50. That is now the standard deviation of the mean. And then this is going to give us -2 651. So that's all good. And that's the Z. Value. And then we do the table look up And this is .004, I'll draw a picture of it because I was like pictures, pictures and statistics, and pictures and math. Just go together. Like just so well, Peanut Butter and Jelly. Here we go. So here's here's where the mean was to 66. Popular the uh to 60. So There were the means to 66 right in the middle, Normal distribution is symmetric. And so here's to 60. And then this .04.2.004 is this area right here. My graph is my drawing not to scale, but just to give you a picture of it, where is here, I'll do it in red. It's the same kind of thing here is to 66, but the standard deviation or the standard error of the mean changed um And to 66 is right here to 60. Excuse me, is right here And .04 is right here. So this .47 is right here. So this area, this red area is a bit bigger than this one. And that's because when we increase the sample size, we reduced, we decrease the variability. So the probability of this happening is pretty, pretty rare because with size 50 we're getting a better picture of what the, the mean is and that's actually what the next question is. He says, what can we conclude if this sample of 50 pregnancies gave us the mean gestation of Of 260 days or less? Well, that my friends would be very rare, that means one of two things. one we either picked a very unique, So we either picked a very unique set of 50 people who have a, I mean that's 260 or less, we impact a we randomly selected ah a unique group Who had the mean of 200 of less than 260. Well, when you put sample mean two days, the gestation period, uh there's that or you know what, that could, that could mean that the population is wrong or that the population mean is wrong. So this could mean this could be a telling sign for us to say, you know what? We should reevaluate this, this uh This assumption that the mean is 266 days. So we, this could also tell us that perhaps the The population mean is not to 66. Let's use symbols perhaps. Oh my gosh, having trouble writing today. She really, perhaps new is not 66. You know, I love symbols of And the last party. What the probability that Sample size of 15, we'll have a gestation within 10 days of the mean. So um like the calculations will be the same here. Um and but what we're looking for, basically the probability That the sample mean is within 10 of the population. So the sample mean, the population mean is to 66 within 10 would be like, Right, if here's to 66 Shortly. Here's to 66, you could go 10 this way, which take you 2 to 76 or 10 this way Be right plus 10 or -10. It should be 2:56. And what's the probability of landing within here. And so if we grab this out, just kind of super do that. Here's to 66 and we want Here's one standard deviation two Standard deviations three, Sandy aviation's government standing ovations where it is, which is actually the Z score, we would want this total area right here, So we want the Z score for 2 56 and Z score for 2 76 which this is a beautiful thing about normal distribution. It's symmetric people, so this is pretty, pretty cool. Make that better. Mhm Yeah, so once we get the Z scores with the mean, it's the standard normal. So The mean is zero You get plus 24, two -2.42, which hopefully That shouldn't seem random to that should seem pretty understanding because you're 10 away on the left and 10 away on the right, so that you're going to be the same distance away on either side, we're going to have the same standard deviations away, positive. End of the negative. And then you do your table look up and then basically what you end up doing is the probability of X less than two 0.42 minus. But Princess, usually 1 -1 X are less than To -2.42. And this is going to end up giving us .984 and what we're doing, so we're doing this, we're doing the Z look up and you're doing the same table look up. But just to get you that the concept here, this portion here, X bar minus, Excuse me, the probability that export is less than 2.42 is all this stuff and then probably the X bar is less than point Excuse me. The probability that export is less than negative, 2.42 is this stuff? So we need to subtract this magenta portion away from the whole green thing, which will give us This magical air we want. So that's why we do one that probability. Um So you go. And if you're just to make sure you're getting that, this ends up being .992 and you're gonna subtract that from 1 -1. Oh no, sorry, this ends up, this whole thing ends up being mhm .008. That's that's where you get. So there you go.

It's Claire's. When you read, here's a party. We have a graph in a ladle, the bottom horizontal in hundreds. So, for example, 17 it's gonna be equal to 1700 23 is 2300. So for Part B, we're just finding the probability that access between 3000 and 4500 we noticed that included, whether included or not included noticing burnt See, we have to probably get X is less than 2500 which is equal to the probability that c'est les than negative 1.67 tickles 0.4 So in five no for party. We're finding a probability that X is bigger than 6000 which is equal to the probability that sea is bigger than 4.17 and you answer her part. E. The most extreme point is there of 1%. Values are the highest 0.5% in the lowest 0.5%. So the the scores that separate them are, see uh, through a 0.5 and Z of 0.999 by meaning greater than sea of throw up Windsor of there are five or last thing. So you're terribly 9995 so we can save by of C 0.5 equal to 0.9995 So we get around 3.3. Then, using symmetry, we get sea of 0.9995 to be around negative 3.3 know solving. For this, we get excellent 0.0 You get X of 0.5 minus 3500 over 600 to be C of 0.5 which we get to be 3.3. So for X value 0.5 we get it for 80. We're gonna use the same thing for next of nine. 0.9995 we get 15 20. So are most extreme values are either less than 15 20 grams or more than 5000 for injured 80 grams. Report. Uh, you know, for our shape, it does not change. So it's Bill Kirk relatively, and it has a normal distribution. It's not really effective. Well, just just for sure, then, for the meat and standard deviation. We got 2500 minus 3500 over 600 Logistical too negative. 1.67 when we get a value, is there a 0.475 which is the same as part?

In question. 17. We have a situation in which we're looking at a deer in the weight of deer in kilograms were given some information. Uh, that says that we have a a distribution that we're gonna call X, and the distribution is normally distributed with mean of 63 kg and a standard deviation of 7.1 kg. Way. Have a few questions based off of that support. A. Ask what is the probability that a single dough captured Wade and released at random in December is undernourished. We also get additional information that says undernourished is less than 54 kg for a single dough. So a What's the probability that a single dough is undernourished? So, basically, what's the probability that a single dose less than 54 kg we can use our normal CDF command in our T 84 calculator? Our picture would be a pretty simple, normal curve because it tells us it's normally distributed with me and 63 it's based out by 7.1 on each side. Ultimately, we're looking for what's the probability of being less than 54? So I draw the picture to show the shading we start shading at negative infinity. That's our lower bound or upper bound goes all the way up to that 54 with our mean of 63 a standard deviation of 7.1. Just gonna label those here and we can get 0.102 and be it says, if the park has about 2200 does. So we know that the park has 2200 Does. What number do you expect to be undernourished in December? Um, we confined the expected number to be undernourished by basically taking the size that we have multiplied by the probability, um, of undernourished, which was found in part A so we could take the 2200. Multiply by 0.102 and we can expect 224.4 to be undernourished. Or we could just say that's 2 24. See to estimate the health of the December dough population, park rangers use the rule that the average weight of 50 does should be more than 60 kg. What is the probability that the average weight X bar for a random sample of 50 does is less than 60 kg So this time we want to find the probability that the average of 50 does is less than 60 kg. Eso We're looking at a sampling distribution here. We're not dealing with just one singular dough. We're looking at multiple, um and then average them together. So in this case, we can still use a normal distribution. Use the CDF command in your TI 83 or 84 calculator again, just kind of drawing the picture to see where our shading is. We have our center at 63 and we want to know what's the probability of being 60 or less so you can see the shading is gonna be on the left side, which means our lower bound is going to start at negative infinity and we're going to go all the way up to 60 are mean is still at 63. But this time, our standard deviation is gonna change just a bit. Standard deviation for a sampling distribution is gonna be whatever the standard deviation of the population is. But divided by our sample size in this case is 50. So you can type this in and you'll find that the probability is 0.0 14 Last one in d compute the probability that X bar is less than 64.2 eso on the probability that X bar is less than 64.2. We get additional information that says I'm supposed park ranger captures ways and released 50 does in December and the average weight was 64.2. Do you think the dough population is undernourished? I was told earlier that when the average of 50 does are basically when the averages less than 60 kg, we consider the entire population to be undernourished. So basically we're looking at these 50 does finding their average and want to know if this sample, which had an average of 64.2 We want to know if that is odd enough to consider the entire population to be undernourished. So we can use our normal CDF command here are lower bound again. We're gonna draw the picture to see where shading is. We have 63 we're marking the 64.2 is what we want. Notice the shape here. We want to know what's the probability of less than that value. So we're gonna shade into the left. So which tells me that my lower bound is going to start a negative. Infinity, which is just 999 on the calculator upper bound at 64.2 mean is 63. And because we're dealing with 50 does that were averaged them? We're gonna do standard deviation about about Route 50. That gives us a pretty large probability of 0.84 And we wanna answer the question. Do we think this population of dough is undernourished? So I'm going to say, since the probability founding an average weight of 64.2 kg or less ISS so high we know that it's so hard because it was pointed out for the dough population is most likely not undernourished.

All right. So babies typically have a mean Mean gestation period of 266 days, with the standard deviation being 16 days. All right. So, in all of these uh problems, we're going to be using this formula for finding uh the probabilities of a normal distribution. Alright. Your your input is goes on a the number given in the problem goes in a You subtract the mean and divide by the standard deviation. All right. And that gives you a Z value. That you can then use a table to look up and answer to. All right. So, for part a we want to know the probability that a gestation period lasts longer than 270 days. Okay, so 2 70 -266 over 16. Now, the key to this one is it is greater than it is a greater than 270. So, whatever answer we get, we have to remember we have to subtract it from one. All right. So this is 4/16, which is a Z value of 0.25 Okay. Then use a Z table and find 0.25. And you see that it is 59.87%. Yeah, It's going All right. But remember this must be subtracted from one so -1987. That gives us .4013 or 40.13%. All right. So, there's a 40.13% chance. A gestation period lasts longer than 270 days. Mhm. Alright, let her be wants to know about less than 250 days. Okay. Okay. So, 2 50 -2/16. This is negative 16 over positive 16. Which is just negative one. Mhm. Yeah. And a z value of negative one Is 15.87%. All right. And this is a less than so, that is your answer. 15.87% of all births uh Are less than 250 days in length. Alright. Part C Wants to know about what percentage of births are between 240 and 280 days long gestation periods. All right. So start with the largest to 80 -2, minus 2, 40 -266. Okay. All right. This is 14/16, which is 0.88 This is negative one. For a Z value. These are Z values. So now you need to look up the Z value for 10.88 which is 81.6%. And for negative one. Just as in the last problem is 15.87% we're subtracting needs. So 81 point Those 6 -15.87 gives us a 65.19 Of all births being between 240 880 days long. All right. And then we have part d Greater than 280. Okay. So we already calculated that in the last problem. uh this part right here, this was actually less than 280. This is actually less than So, if we want greater than then we take one minus the 81 point oh six and that gives us 18.94%. Alright. Letter e. less than 246 246, -2, Over 16. That is negative 20/16, which is negative 1.25. All right. And again looking up that z value negative 1.3 is 15 0.15%. Mhm. And since this is a less than this is our answer. All right. And then for part f wants to know about less than 224. Alright, so to 20 for -2, over 16. Mhm. That's 42/16. Which is a Z value of 2.6 three. Alright. On your table, 2.63 is 99.57 mm. All right. Okay. Excuse me. This is a negative. That didn't make sense. Phative. is 0.43% on the table. That makes much more sense. Only .3% of all gestation periods are 224 days are shorter.


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