Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
STAB22 section 5.1 ual observations, but the approximation will be better if the distribution of the observations is closer to normal. If you take a simple random sample of size 144 from this population, there is an approximate 95% chance that the sample mean will be between 240 − 2(1.5) = 237 and 240 + 2(1.5) = 243. 5.2 The sample mean has a sampling distribution with √ mean µ (the population mean) and SD σ/ n (smaller than the population SD). Here the √ sampling distribution has mean 240 and SD 18/ 36 = 3. That is, if you take a sample of 36 observations, the sample mean is likely to be close to 240 (because the SD of the sampling dis- 5.5 Replace 144 by 1296 in 5.3. The sample mean again has an approximate normal distribution tribution is a relatively small 3), even though the (the approximation will be better because the individual observations could be quite a bit larger sample √ size is bigger) with mean 240 and now or smaller than 240. (The high and low values will SD 18/ 1296 = 0.5. Now there is about a 95% tend to cancel each other out. If you only take chance that the sample mean will be between a small sample, you may be unlucky enough to 240 − 2(0.5) = 239 and 240 + 2(0.5) = 241. Beget, say, mostly large values, but the larger the cause of the larger sample, we can make a more sample you take, the more likely the large and precise statement about what the sample mean small values are to be more or less 50-50.) might be. (With a sample size this large, it pretty 5.3 The sampling distribution still has mean 240, but much doesn’t matter what the distribution of the √ now has SD 18/ 144 = 1.5, only half as big. original observations looks like; the sampling disThat is, if you take a bigger sample, the sample tribution of the sample mean will be pretty much mean is more likely to be even closer to the popnormal. See example 5.6 for an example of this. ulation mean (here 240) than it would be if you 5.6 The sampling distribution of the√sample mean took a smaller sample. has mean 25 mins and SD 25/ 50 = 3.5355 5.4 According to the central limit theorem, the samminutes (as in Example 5.7). It also has apple mean has an approximate normal distribution proximately a normal shape, so we can go ahead √ with mean µ = 240 and SD σ/ n = 18/12 = 1.5. and use the normal distribution to figure out the chance that the mean time between text mes(These are the same numbers as 5.3.) This is sages is less than 28 minutes, even though we true regardless of the distribution of the individ1 know that times between individual messages are not at all normal (see Figure 5.3 in the text). z = (28 − 25)/3.5355 = 0.85, so the probability that the mean time between text messages is less than 28 minutes is (Table A) approximately 0.8023. Note that times between individual text messages are all over the place, but the mean time between 50 of them is very likely to be less than 28 minutes, which is only a little bigger than the mean. This is because the mean time between 28 is quite predictable; a sample size of 28 is on the large side for this kind of calculation. mation required, regardless of the size of the sample. If you’re thinking about one can, it has a SD of 3 ml. If you look at 6 cans √ the sam√ and take ple mean x̄, x̄ has SD σ/ n = 3/ 6 = 1.225, smaller. The mean in each case is 250 ml, from the last sentence of 5.13. So your picture should look like Figure 1, with the solid line showing the curve for a single measurement and the dashed line showing the curve for the mean of 6. (Note that the dashed curve is less than half as wide, so it goes more than twice as high, roughly speaking.) 5.7 (a) would be correct if “variance” were replaced by “SD”. (“Sampling distribution of sample mean” would be clearer than just “mean”, too.) In (b), larger sample sizes will have sampling distributions for the sample mean with smaller SDs (which is a bit clearer than the statement given in the question). For (c), the mean of the sampling distribution of the sample mean stays where it is (it is equal to the population mean µ no matter how big the sample size is); it is the SD of the sampling distribution that changes when the sample size changes (gets smaller as the sample size gets bigger). In (b), the can has to contain less than 250 − 1 = 249 or more than 250 + 1. These two values correspond to z = (249 − 250)/3 = −0.33 and z = 0.33. The chance of either of those is 0.3707 (Table A), so 0.7414 of the time a single can will be outside these limits. In (c), because the SD is now less than half as big, the z values are now more than twice as big. z = ±0.82; the chance of being “outside” either of those is 0.2061 (Table A), so 0.4122 of the time, the mean of 6 cans will be outside those limits. The results of (b) and (c) make sense according to the picture. The moral of the story is that means of several measurements are more accurate than values of individual measurements, so the mean 5.15 First off, if individual observations have a normal distribution, the sampling distribution of a sample mean is exactly normal, with no approxi2 of 6 measurements is more likely to be inside a “tolerance” than an individual measurement is. This is why, in science class, you made several measurements and averaged them. 5.17 The sample mean that you actually observe might be smaller or bigger than the population mean, but on average (that is, over the many possible samples you might take), the sample mean will be the same as the population mean. (The sample mean is neither systematically too small, nor systematically too high. The distribution of the sample mean might look like Figure 3.14(b) on page 207 of the text, but at least it doesn’t look like (a) or (c).) The other side of the issue is of how close the sample mean might be to the population mean; this is summarized by the SD of the sampling distribution of the sample mean (in other words, how much sampling variability there is), and a larger sample will have less sampling variability, so it will produce a picture more like 3.14(d) than 3.14(b). 5.18 (a) z = (23 − 19.2)/5.1 = 0.75, so the chance of a value higher than this is 0.2266. (b) √ The mean 25 = 0.96. is still 20.8, but the SD is now 5.1/ √ (c) z is now (23 − 19.2)/(5.1/ 25) = (23 − 19.2)/0.96 = 3.73, and the probability of exceeding this is close to zero (since 3.73 is off the end of Table A). My software gives 0.000096. (d) The Figure 1: Normal curves for volume and mean volume of cans 3 0.5)/0.099 = 1.01 and is 1 − 0.8438 = 0.1562. Because the number of moths trapped is highly variable, the sample mean could easily be bigger than 0.6 even though the population mean is only 0.5. scores were only “roughly normal”, so the answer to (a) may not be very accurate. The answer to (c), however, is based on the mean of a number of observations, and (by the Central Limit Theorem) the sampling distribution of the sample mean is approximately normal even if the individual observations are not. So (c) is more accurate than (a). (25 is probably a large enough sample size to make the Central Limit Theorem work if the population is “roughly normal”, though of course this is guesswork.) A hint that that the normal approximation is going to be OK here is to look at whether the normal approximation to the sampling distribution of the sample mean could go down below zero (which we know to be impossible for these data). The 68–95–99.7 rule says that almost all of the sampling distribution should be between 0.5 − 3(0.099) ' 0.2 and 0.5 + 3(0.099) ' 0.8. This doesn’t get too close to zero, so we are all right. (0 is actually five SDs below the mean of the sampling distribution, so the normal approximation giving us a value below zero is as good as impossible. 5.19 When your observations have to be zero or positive (as here, they are numbers of moths trapped), and the SD is bigger than the mean, that’s a sign that the distribution is skewed to the right. (Try this for yourself with your software: make some lists of numbers that are zero or positive, and see what it takes to get the SD bigger than the mean.) My guess here is that 5.20 To find the mean of this discrete distribution, there are a lot of traps with no moths at all. But use “value times probability added up”, to get because of the central limit theorem, we can do a mean of µ = 4(0.33) + 3(0.24) + 2(0.18) + calculations for mean numbers of moths trapped 1(0.16) + 0(0.09) = 2.56. Then find the variance in a sample of traps with some confidence that as σ 2 = (4 − 2.56)2 (0.33) + (3 − 2.56)2 (0.24) + the answers will be accurate enough. · · · +√(0 − 2.56)2 (0.09) = 1.7664, and so the SD is The sampling distribution of the mean number σ = 1.7664 = 1.329. of moths trapped in 50 traps has mean µ = 0.5 √ √ The mean √ and SD of the sample mean are 2.56 and SD σ/ n = 0.7/ 50 = 0.099, so that the and 1.329/ 50 = 0.1880. chance of finding a sample mean number of moths trapped that is bigger than 0.6 uses z = (0.6 − For a randomly chosen student, just add up the 4 √ probabilities for “B or better”: 0.24+0.33 = 0.57. 125 and SD 10/ 3 = 5.773. The chance of this The average grade for the class as a whole was sample mean coming out above 140 now uses around B- (a grade point value of 2.56), and z = (140 − 125)/5.773 = 2.60, so the probabilso the chance that the 50 students will have a ity is 1 − 0.9953 = 0.0047, which is much smaller. mean grade point value of 3 or more ought to The chance of a wrong diagnosis has been much be less than 0.57. Using the answers for (b), reduced by using 3 measurements rather than 1. z = (3 − 2.56)/0.1880 = 2.34, and the chance of being higher than this is 1 − 0.9904 = 0.0096, 5.22 (a) is the same idea as 5.20: mean is 0(0.999) + 2 500(0.001) = 0.5, variance is (0 − 0.5)√ (0.999) + which is indeed (a lot) smaller than 0.57. (The 2 (500 − 0.5) (0.001) = 249.75, so SD is 249.75 = grade point mean for a sample of 50 students is 15.80. The law of large numbers says that Joe’s more likely to be close to 2.56 than the grade average winnings per play will be very close to point score for a single student.) 0.5 over a large number of plays (such as a year’s The mean, by the way, is less than 3 even though worth). The Central Limit Theorem says that the median is greater than 3 (more than 50% of the distribution of average winnings per play will students got a B or better), because the distribube close to normal √ over the year, with mean 0.5 tion is strongly left-skewed: you can’t do better and SD 15.80/ 104 = 1.549. (Actually, the Law than A, but there were some students who got a of Large Numbers is kind of optimistic here: beD or F and they pulled the mean downwards. cause the SD of winnings if you play once is so large, Joe’s average winnings over the year is still 5.21 Sheila ought not to get diagnosed with gestaquite variable even though he’s played 104 times. tional diabetes because her average glucose level This is saying that 104 isn’t really “large” in this is 125, a lot less than 140. But if only a single case. We’ll have to keep a close eye on the normal measurement is made, she might get diagnosed as approximation, too.) having gestational diabetes by chance. The probUsing the normal approx., find the chance that ability of this comes from z = (140 − 125)/10 = Joe comes out ahead for the year as P (x̄ ≥ 1) = 1.5, and is 1 − 0.9332 = 0.0668. P (Z ≥ (1 − 0.5)/1.549) = P (z ≥ 0.32). This, Measurements made on 3 separate days ought from Table A, is 1 − 0.6255 = 0.3745. to be independent. Sheila’s sample mean meaAlternatively, Joe will be ahead for the year if he surement has a sampling distribution with mean 5 wins once or more, because the $500 (or more) should be very close to 1.3. The 99.7 part of the he wins will pay for the whole year’s tickets, with rule says that the mean number of flaws is almost a profit left over. The probability that he fails certainly going to be between 1.0 and 1.6, and 2 104 to win at all over the year is 0.999 = 0.9012, is way above that.) This time, the Central Limit so the probability that he wins at least once is Theorem should work all right because the to1 − 0.9012 = 0.0988. This answer is quite a bit tal number of flaws found should be around 300, different from the one obtained using the norand there are many possible different numbers of mal approximation, which means that the norflaws that might be observed. mal approximation isn’t very good, even though the sample size (104) appears to be large. This is 5.25 If the weights of individual passengers are “not very non-normal”, the mean weight of 25 pasbecause the distribution of single observations is sengers should be pretty close√to normal, with very skewed, and n = 104 is not yet big enough mean 195 pounds and SD 35/ 25 = 7 pounds. for the central limit theorem to take hold. So If the total weight of passengers is going to exbeware: the normal approximation is only an apceed 5200 pounds, the mean weight has to exproximation, and sometimes it’s not very good! ceed 5200/25 = 208 pounds. The chance of this The clue here is that Joe’s number of wins is is figured using z = (208 − 195)/7 = 1.86, and likely to be very small, 0,1 or 2 say, whereas if is 1 − 0.9686 = 0.0314. (This kind of calculation the normal approx. is going to work well there is routinely done by airlines when deciding how needs to be a large number of possible outcomes. much fuel to put in: if the plane (and passen5.24 This is like a sample of 200 individual square gers) are heavier, it will need more fuel to get to yards of carpet, so the mean number of flaws has its destination, but if the airline puts in too much approx. a√normal distribution with mean 1.3 and fuel, it will be more costly because the fuel has SD 1.5/ 200 = 0.1061 (by the Central Limit weight too (and it takes more fuel to carry more Theorem). The chance that the mean number fuel). of flaws is more than 2 per square yard uses z = (2 − 1.3)/0.1061 = 6.60; the chance of ex- 5.26 According to the law of large numbers, the averceeding this is more or less 0. (2 isn’t much bigage loss to the company will be close to $250 per ger than 1.3, but the law of large numbers says house, if the number of policies sold is large. 12 that the mean number of flaws per square yard is probably not large enough; if one of those 12 6 mean 57 and variance 2.22 /6 = 0.8067; the sample mean of 6 treated specimens has a normal distribution with mean 30 and variance 1.62 /6 = 0.4267. The difference also has a normal distribution. The mean is the difference of means, 57 − 30 = 27; the variance is the sum of variances (remember the rules for means and variances?), which is 0.8067 + 0.4267 = 1.2333. For the chance that this difference is 25 or bigger, √ find z = (25 − 27)/ 1.2333 = −1.80; table A gives probability 0.0359 of being below, so 1 − 0.0359 = 0.9641 of being above. houses burns down, the insurance company will have to pay out a lot more than they took in in premiums. But with a large number of policies sold, the amount taken in in premiums will be enough to cover even a few houses burning down, almost certainly. If n = 10000, the average √ loss per policy sold has mean 250, SD 1000/ 10000 = 10 and approx. a normal distribution. Thus the chance that the average loss per year exceeds $275 uses z = (275 − 250)/10 = 2.5 and is approx. 1 − 0.9938 = 0.0062. Insurance, in sufficient volume, is a very safe business (from the point of view of an insurance company), just as owning a big casino is a safe business (from the point of view of the casino). Probabilistically speaking, insurance and gambling are really the same thing. If the individual values have a normal distribution, the sample mean has a normal distribution exactly; there’s no need to rely on the Central Limit Theorem. In this case, that’s probably a good thing, since the samples aren’t very big. 5.27 The sample mean strength of 6 untreated spec- 5.28 (a) The Central Limit Theorem says that samimens has a √ normal distribution with mean 57 ple means have approximately a normal distriand SD 2.2/ 6 = 0.8981 pounds. To find the bution in large samples. The individual scores probability that this exceeds 50 pounds, calcuwon’t be too non-normal (they are scores between late z = (50 − 57)/0.8981 = −7.79. The chance 1 and 7, so there can’t be any extreme outliers), of getting a value smaller than this is basically 0, so a sample of 28 scores will probably be large so the chance of exceeding it is basically 1. (The enough. sample mean will be close to 57; specifically, alThe sample mean of Journal scores has mean 4.8 √ most certainly closer than 50 is.) and SD 1.5/ 28 = 0.2835; the sample mean √ of Part (b) is tricky. The sample mean of 6 unEnquirer scores has mean 2.4 and SD 1.6/ 28 = treated specimens has a normal distribution with 0.3024. 7 The difference ȳ − x̄ has a normal distribution with mean the difference of means 4.8 − 2.4 = 2.4 and variance the sum of variances (like part (b) of the previous question): the variance of the difference √ is 0.28352 + 0.30242 = 0.1718 and so the SD is 0.1718 = 0.4145. the wall’s height differs from the design height by more than half an inch is 1−0.9876 = 0.0124: not very likely. If you want to, you can figure out the mean and SD of the total height of the wall by taking the mean and SD for the mean height of a row and multiplying them by 4. Then you can work with these values and find the chance that the total height is between 31.5 and 32.5. You’ll get the same z-values (±2.5) that we got above, and therefore the same answers. For the chance that the difference in sample means is greater than 1, calculate z = (1 − 2.4)/0.4145 = −3.38, so the chance of the difference being less than 1 is (Table A) 0.0004, and the chance of the difference being 1 or bigger is 1 − 0.0004 = 0.9996: almost a sure thing. 5.31 Like the question about the airline passengers, this question about totals can be translated into a question about means (by dividing by 4, there being 4 rows of blocks). We can figure out the probability that the wall does not differ from the design height by more than half an inch, and then subtract the answer from 1. We are going, therefore, to figure out the probability that the total height is between 31.5 and 32.5 inches, which (dividing by 4) is the same as saying that the mean height of a row is between 7.875 and 8.125 inches. The mean height of 4 rows has a sampling√distribution with mean 32/4 = 8 and SD 0.1/ 4 = 0.05 inches. So we need the chance of being between z = (7.875 − 8)/0.05 = −2.5 and z = (8.125 − 8)/0.05 = 2.5. This probability is 0.9938 − 0.0062 = 0.9876, so the probability that 8