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- Consider both the population shape and the sample size. - Separate an approximate result caused by averaging from an exact result inherited from the population. - Strong skewness matters more when only a few observations are averaged.

1. In a), the small sample does not provide enough averaging to overcome strong skewness, so a normal model is not justified from the information given. 2. In b), the large sample makes the sample mean approximately normal by the central limit theorem. 3. In c), the sample mean is normal because the population itself is normal; the central limit theorem is not needed.

a) A normal model is not justified from the information given. b) The sample mean is reasonably modeled as normal by the central limit theorem. c) The sample mean is normal because the population is normal; the central limit theorem is not needed.

- Determine whether the population shape already guarantees the shape of the sample mean. - Find the spread of the average for the small sample. - Standardize the requested cutoff.

1. Because the population is normal, the sample mean is exactly normal for any sample size. 2. Its mean is \(20\) and its standard error is \(\frac{6}{\sqrt{4}}=3\). 3. The standardized value is \(z=\frac{17-20}{3}=-1\). 4. Therefore, \(P(\bar X<17)=P(Z<-1)\approx0.1587\). The central limit theorem is not needed.

\(P(\bar X<17)\approx0.1587\). The normal model for \(\bar X\) is exact because the population is normal; the central limit theorem is unnecessary.

- Express the average as a scaled sum to determine its spread. - Compare the six-unit margin with the corrected sampling scale. - Use the central area between symmetric standardized bounds.

1. The standard deviation of the sample mean is \(\frac{18}{\sqrt{36}}=3\), not \(\frac{18}{36}\). 2. A distance of \(6\) units is \(\frac{6}{3}=2\) standard errors. 3. Therefore, \(P(|\bar X-\mu|<6)\approx P(-2<Z<2)\approx0.9545\).

a) The correct standard error is \(3\). b) The approximate probability is \(0.9545\).

- List the parameters needed to specify the approximate sampling distribution. - Check which of those parameters can be obtained from the problem statement. - Decide whether the cutoff can be expressed in standardized units.

1. The central limit theorem locates the sample-mean distribution at \(70\). 2. Its standard error would be \(\frac{\sigma}{\sqrt{100}}=\frac{\sigma}{10}\). 3. Without the population standard deviation \(\sigma\), the cutoff \(72\) cannot be standardized. 4. Therefore, the numerical probability cannot be determined; the population standard deviation or equivalent spread information is needed.

No. The population standard deviation \(\sigma\) is needed to obtain the standard error \(\sigma/10\) and standardize \(72\).

- Distinguish how means and variances scale in an independent sum. - Correct the total's spread before standardizing. - Use the upper-tail area beyond the corrected cutoff.

1. The student multiplied the standard deviation by \(100\). For an independent sum, variances add, so \(\sigma_S=2\sqrt{100}=20\). 2. The approximate mean is correctly \(100(4)=400\). 3. The standardized cutoff is \(z=\frac{430-400}{20}=1.5\). 4. Therefore, \(P(S>430)\approx P(Z>1.5)\approx0.0668\).

a) The correct standard deviation is \(20\), not \(200\). b) The approximate probability is \(0.0668\).

- Express each sampling spread using the square root of its sample size. - Compare how averaging more observations changes variability. - Apply the theorem's qualitative statement about shape as sample size grows.

1. The standard errors are \(\frac{\sigma}{4}\) for \(n=16\), \(\frac{\sigma}{8}\) for \(n=64\), and \(\frac{\sigma}{16}\) for \(n=256\). 2. Greatest to least standard error: \(n=16,64,256\). 3. As sample size increases, the central limit theorem gives a distribution generally closer to normal. Least to most nearly normal: \(n=16,64,256\).

a) Greatest to least standard error: \(16,64,256\). b) Least to most nearly normal: \(16,64,256\). c) \(\sigma/4,\sigma/8,\sigma/16\), respectively.

- Focus on the distribution of the average rather than the distribution of one call. - Determine the center and spread of the sample mean. - Assess whether the sample size supports an approximate bell-shaped model.

1. For \(n=64\), the central limit theorem gives \(\bar X\) an approximately normal distribution with mean \(8\) and standard error \(\frac{6}{\sqrt{64}}=0.75\). 2. The standardized value for \(9.5\) is \(z=\frac{9.5-8}{0.75}=2\). 3. Therefore, \(P(\bar X>9.5)\approx P(Z>2)\approx0.0228\). 4. The large random sample makes the distribution of the sample mean approximately normal despite the skewed population.

\(P(\bar X>9.5)\approx0.0228\). The approximation is reasonable because the sample size is large and the observations are randomly selected.

- Compute the spread of the average for the stated sample size. - Locate both dollar limits relative to the sampling center. - Consider how quickly an unusual population shape may be smoothed by averaging.

1. The sample mean has mean \(50\) dollars and standard error \(\frac{20}{\sqrt{25}}=4\) dollars. 2. The standardized bounds are \(z=\frac{45-50}{4}=-1.25\) and \(z=\frac{55-50}{4}=1.25\). 3. The approximate probability is \(P(-1.25<Z<1.25)\approx0.7887\). 4. Because \(n=25\) is only moderate and the population is irregular and multimodal, the normal approximation may be imperfect.

The approximate probability is \(0.7887\). The moderate sample size and irregular population shape make the approximation less secure than it would be for a larger sample.

- Determine the approximate distribution of the sample mean. - Use the standard-normal location that leaves five percent above it. - Transform that standardized location back to the original measurement scale.

1. The sample mean is approximately normal with mean \(72\) and standard error \(\frac{18}{\sqrt{81}}=2\). 2. The standard normal \(95\)th percentile is \(z\approx1.6449\). 3. The corresponding sample-mean value is \(72+1.6449(2)\approx75.29\).

The approximate \(95\)th percentile is \(75.29\).

- Express the same cutoff in standard-error units for each sample size. - Compare how the denominator changes when the sample size is quadrupled. - Interpret the probabilities through the changing spread of the average.

1. For \(n=16\), the standard error is \(\frac{\sigma}{4}\), so the cutoff is \(z=\frac{\sigma/2}{\sigma/4}=2\). The probability is approximately \(0.0228\). 2. For \(n=64\), the standard error is \(\frac{\sigma}{8}\), so the cutoff is \(z=\frac{\sigma/2}{\sigma/8}=4\). The probability is approximately \(0.000032\). 3. Because the population is strongly nonnormal, the approximation for \(n=16\) may be unreliable, while the approximation for \(n=64\) is more defensible. 4. Increasing the sample size makes the sample mean much less variable, so the same absolute distance from \(\mu\) becomes far less likely.

For \(n=16\), the central limit theorem approximation is \(0.0228\), though it may be unreliable because the population is strongly nonnormal and the sample is small. For \(n=64\), the approximation is \(0.000032\) and is more reliable. The larger sample produces a much tighter distribution of sample means.

- Relate the sample sum and sample mean algebraically. - Check whether the two inequalities describe the same sample outcomes. - Standardize using whichever of the equivalent variables is simpler.

1. Since \(S=36\bar X\), the event \(S>522\) is exactly the event \(\bar X>\frac{522}{36}=14.5\). 2. The sample mean has approximate mean \(14\) and standard error \(\frac{3}{\sqrt{36}}=0.5\). 3. The standardized cutoff is \(z=\frac{14.5-14}{0.5}=1\). 4. Both probabilities are therefore approximately \(P(Z>1)=0.1587\).

\(P(S>522)\approx0.1587\) and \(P(\bar X>14.5)\approx0.1587\). They are equal because the two events are equivalent.

- Find the standard error of the sample mean. - Convert the stated central percentage into equal tail areas. - Transform the resulting standardized endpoints back to the population scale.

1. The sample mean is approximately normal with mean \(100\) and standard error \(\frac{24}{\sqrt{64}}=3\). 2. A central area of \(0.80\) leaves \(0.10\) in each tail, so the positive cutoff is \(z\approx1.2816\). 3. The margin is \(1.2816(3)\approx3.84\). 4. The interval is approximately \((100-3.84,100+3.84)=(96.16,103.84)\).

Approximately \((96.16,103.84)\).

- Find the spread of the total without needing its numerical mean. - Express the symmetric distance from the expected value in standardized units. - Find the central area between the two standardized limits.

1. The standard deviation of the total is \(5\sqrt{225}=75\). 2. Being within \(15\) units of the expected total corresponds to \(-15<S-E(S)<15\). 3. The standardized bounds are \(-\frac{15}{75}=-0.2\) and \(\frac{15}{75}=0.2\). 4. Therefore, the approximate probability is \(P(-0.2<Z<0.2)\approx0.1585\).

The approximate probability is \(0.1585\).

- Build the approximate distribution of the full total. - Standardize each endpoint using the total's spread. - Find the area between the two standardized values.

1. The total \(S\) has mean \(64(4)=256\) minutes and standard deviation \(1.5\sqrt{64}=12\) minutes. 2. The standardized bounds are \(z=\frac{250-256}{12}=-0.5\) and \(z=\frac{270-256}{12}\approx1.167\). 3. Therefore, \(P(250<S<270)\approx P(-0.5<Z<1.167)\approx0.5698\).

The approximate probability is \(0.5698\).

- Determine the standard error for the large sample. - Standardize the two unequal distances from the population mean. - Find the normal area between the resulting endpoints.

1. The sample mean is approximately normal with mean \(6\) and standard error \(\frac{2.5}{\sqrt{121}}=\frac{2.5}{11}\approx0.2273\). 2. The lower standardized bound is \(z=\frac{5.5-6}{0.2273}=-2.2\). 3. The upper standardized bound is \(z=\frac{6.3-6}{0.2273}=1.32\). 4. Therefore, \(P(5.5<\bar X<6.3)\approx P(-2.2<Z<1.32)\approx0.8927\).

The approximate probability is \(0.8927\).

- The theorem describes convergence, not an identical rate for every population shape. - Consider how a rare extreme observation affects an average of only a few dozen values. - Compare the amount of smoothing each original shape needs.

1. The central limit theorem makes both sample-mean distributions move toward normality as sample size increases. 2. A symmetric, unimodal population generally requires less averaging before its sample mean is well approximated by a normal distribution. 3. Rare extreme values in Population B can continue to produce skewed sample means at \(n=30\). 4. Therefore, the approximation should generally be more reliable for Population A, while more information or a larger sample would be desirable for Population B.

Population A should generally have the more reliable normal approximation. Population B's rare extreme values can slow convergence to normality at \(n=30\).

- Form the center and spread of the multi-day total. - Use the number of days to assess whether averaging can overcome the daily skewness. - Standardize the total-rainfall threshold.

1. The total rainfall has mean \(81(0.12)=9.72\) inches and standard deviation \(0.40\sqrt{81}=3.6\) inches. 2. The standardized cutoff is \(z=\frac{15-9.72}{3.6}\approx1.467\). 3. The central limit theorem gives \(P(S>15)\approx P(Z>1.467)\approx0.0712\). 4. The large sample supports the approximation despite the zero-inflated, skewed daily distribution.

The approximate probability is \(0.0712\).

- Separate an asymptotic statement from an exact statement. - Consider how extreme skewness affects the amount of averaging needed. - Compare the two sample sizes without relying on a universal cutoff.

1. The central limit theorem gives an approximation that improves as sample size increases; it does not make every finite-sample distribution exactly normal. 2. With extreme skewness, \(n=20\) may be too small for a reliable normal approximation. 3. The distribution for \(n=200\) should generally be much closer to normal because substantially more observations are averaged. 4. Neither approximation is described as exact solely by the central limit theorem.

The statement is false. The \(n=200\) sample-mean distribution should generally be much closer to normal, while \(n=20\) may still show substantial skewness. The theorem provides approximation, not exact finite-sample normality.

- Convert the percentage tolerance into units on the original scale. - Find the sampling spread of the average. - Use a symmetric normal area around the center.

1. Five percent of \(100\) is \(5\), so the event is \(95<\bar X<105\). 2. The standard error is \(\frac{30}{\sqrt{100}}=3\). 3. The standardized bounds are \(z=\pm\frac{5}{3}\approx\pm1.667\). 4. Therefore, the approximate probability is \(P(-1.667<Z<1.667)\approx0.9044\).

The approximate probability is \(0.9044\).

- Find the standardized quartile location first. - Transform it to the sampling distribution of the mean. - Use the exact relationship between the sum and mean for the same sample.

1. The sample mean has approximate mean \(40\) and standard error \(\frac{8}{\sqrt{25}}=1.6\). 2. The standard normal \(75\)th percentile is \(z\approx0.6745\). 3. The sample-mean percentile is \(40+0.6745(1.6)\approx41.08\). 4. Since \(S=25\bar X\), use the unrounded percentile from step 3: \(25[40+0.6745(1.6)]\approx1026.98\).

The approximate \(75\)th percentile of \(\bar X\) is \(41.08\), and that of \(S\) is \(1026.98\).

- Find the population center and spread after the linear transformation. - Then determine the sampling spread for the average of transformed values. - Standardize the requested cutoff on that scale.

1. The transformed population has mean \(\mu_Y=3(20)-7=53\) and standard deviation \(\sigma_Y=3(4)=12\). 2. The sample mean \(\bar Y\) has approximate standard error \(\frac{12}{\sqrt{64}}=1.5\). 3. The standardized cutoff is \(z=\frac{55-53}{1.5}\approx1.333\). 4. Therefore, \(P(\bar Y>55)\approx P(Z>1.333)\approx0.0912\).

The approximate probability is \(0.0912\).

- Identify the statistic named in the theorem's standard classroom form. - Check whether the population mean must be the center of the statistic used here. - Ask whether the supplied population parameters are enough to determine this statistic's spread.

1. The standard central limit theorem result using only the population mean and standard deviation applies to the sample mean or sum. 2. A sample median has a different sampling distribution whose center and spread depend on additional features of the population distribution. 3. The population mean need not equal the population median, and \(12/\sqrt{64}\) is not generally the standard error of the sample median. 4. Therefore, the student's proposed model is not justified from the given information.

The claim is not justified. The usual central limit theorem model with mean \(\mu\) and standard error \(\sigma/\sqrt{n}\) applies to the sample mean, not automatically to the sample median.

- Find the sampling spread of the mean. - Convert the absolute-value event into two symmetric tails. - Standardize the positive boundary and double the corresponding tail area.

1. The sample mean has approximate mean \(0\) and standard error \(\frac{1}{\sqrt{49}}=\frac{1}{7}\). 2. The standardized magnitude is \(z=\frac{0.30}{1/7}=2.1\). 3. The event has two tails, so \(P(|\bar X|>0.30)\approx2P(Z>2.1)\approx0.0357\).

The approximate probability is \(0.0357\).

- Convert the stated cumulative probability to a standardized location. - Determine the sampling spread from the known population spread. - Use the cutoff equation to solve for the unknown center.

1. A cumulative probability of \(0.9772\) corresponds to \(z\approx2\). 2. The standard error is \(\frac{8}{\sqrt{64}}=1\). 3. Therefore, \(\frac{52-\mu}{1}=2\). 4. Solving gives \(\mu=50\).

\(\mu=50\).

- Compare how much influence one unusually large observation can have in each population. - Think about what averaging does when individual contributions are bounded. - The same sample size need not give the same approximation quality for different shapes.

1. Both sample means involve averages of \(40\) observations. 2. The original population's rare enormous values can dominate an average and preserve strong right skewness. 3. Capping the values limits the effect of any one observation and makes convergence toward a normal sampling shape faster. 4. Therefore, the sample mean from the capped population should generally be closer to normal.

The sample mean from the capped population should generally be closer to normal because no single observation can be arbitrarily dominant.

- Find the center and spread of the large sum. - Use symmetric normal cutoffs for the central percentage. - Add and subtract the resulting margin from the expected total.

1. The total has mean \(E(S)=225(7)=1575\) and standard deviation \(4\sqrt{225}=60\). 2. A central probability of \(0.95\) uses the cutoff \(z\approx1.96\). 3. The margin is \(1.96(60)=117.6\). 4. The interval is \((1575-117.6,1575+117.6)=(1457.4,1692.6)\).

Approximately \((1457.4,1692.6)\).

- Identify which sampling-distribution parameters depend only on the population mean, standard deviation, and sample size. - Separate an approximation from an exact distributional statement. - Consider whether different source populations can leave small finite-sample differences.

1. Both sample means have center \(\mu\) and standard error \(\frac{\sigma}{\sqrt{200}}\). 2. The large sample size makes both sampling distributions approximately normal. 3. Thus, their normal approximations use the same mean and standard deviation. 4. The actual finite-sample distributions need not be exactly identical because they arise from different population shapes.

Both sample means are approximately normal with mean \(\mu\) and standard deviation \(\sigma/\sqrt{200}\). Their approximations match, but their exact finite-sample distributions need not be identical.

- Find the center and spread of the multi-day total. - Adjust the boundary to represent the integer-valued event on a continuous model. - Use the upper-tail area beyond the standardized cutoff.

1. The total \(S\) has mean \(49(30)=1470\) and standard deviation \(8\sqrt{49}=56\). 2. For the integer event \(S\ge1550\), use the continuity-corrected boundary \(1549.5\). 3. The standardized value is \(z=\frac{1549.5-1470}{56}\approx1.420\). 4. Therefore, \(P(S\ge1550)\approx P(Z\ge1.420)\approx0.0779\).

The approximate probability is \(0.0779\).

- Match the stated central probability to symmetric standardized endpoints. - Express the three-unit margin in standard-error units. - Solve the resulting scale equation for the population spread.

1. A central normal probability of \(0.8664\) corresponds to approximately \(-1.5<Z<1.5\). 2. Therefore, \(\frac{3}{\sigma/\sqrt{64}}=1.5\). 3. Since \(\sqrt{64}=8\), \(\frac{24}{\sigma}=1.5\). 4. Solving gives \(\sigma=16\).

\(\sigma=16\).

- Determine the approximate distribution of the multi-day total. - Shift the integer boundary by one half before using a continuous model. - Find the lower-tail area at the standardized cutoff.

1. The total \(S\) has mean \(100(4)=400\) and standard deviation \(3\sqrt{100}=30\). 2. For \(S\le350\), the continuity-corrected boundary is \(350.5\). 3. The standardized value is \(z=\frac{350.5-400}{30}=-1.65\). 4. Therefore, \(P(S\le350)\approx P(Z\le-1.65)\approx0.0495\).

The approximate probability is \(0.0495\).

- Express the fixed three-unit difference in standard-error units. - Use the standardized cutoff associated with a one-percent upper tail. - Round the sample size in the direction that preserves the probability condition.

1. The cutoff \(\mu+3\) has standardized value \(z=\frac{3}{15/\sqrt{n}}=\frac{\sqrt{n}}{5}\). 2. An upper-tail probability of at most \(0.01\) in the normal model requires \(z\ge2.3263\). 3. Thus, \(\sqrt{n}\ge5(2.3263)\), so \(n\ge135.28\). 4. The smallest integer satisfying the normal-model probability condition is \(136\).

Under the stated normal approximation, \(n=136\).

- Identify which random component is shared by every observation. - Ask whether averaging can cancel a component that moves all readings together. - Check the independence condition before applying the usual sampling-scale rule.

1. The readings are not independent because all contain the same random calibration shift. 2. Averaging can reduce the independent measurement errors, but it does not reduce the shared shift. 3. Therefore, the standard deviation of the sample mean is not generally the one-reading standard deviation divided by \(\sqrt{100}\). 4. The usual independent-observation central limit theorem calculation is not justified without separately modeling the common shift.

The proposal is not justified. The common calibration shift creates dependence and remains in the average, so the usual \(1/\sqrt{n}\) spread reduction does not apply to the full reading variability.

- Build the approximate distribution of the multi-day total. - Expand both integer endpoints by one half for an inclusive interval. - Find the normal area between the corrected standardized bounds.

1. The total has mean \(50(6)=300\) and standard deviation \(2\sqrt{50}\approx14.14\). 2. The inclusive integer interval \(290\le S\le310\) becomes \(289.5<S<310.5\) under continuity correction. 3. The standardized bounds are approximately \(-0.742\) and \(0.742\). 4. Therefore, the approximate probability is \(P(-0.742<Z<0.742)\approx0.5422\).

The approximate probability is \(0.5422\).

- Convert the desired central probability into a symmetric standardized cutoff. - Relate the one-unit accuracy to the standard error. - Round upward so the final sample meets the requirement.

1. A central probability of \(0.99\) in the normal model uses the cutoff \(z\approx2.5758\). 2. Require \(2.5758\left(\frac{4}{\sqrt{n}}\right)\le1\). 3. This gives \(n\ge[2.5758(4)]^2\approx106.16\). 4. The smallest integer meeting the normal-model probability requirement is \(107\).

Under the stated normal approximation, \(n=107\).

- Combine the centers from both contribution types. - Combine independent variability using variances. - Assess whether the large collection of contributions supports a normal approximation before standardizing.

1. The total mean is \(50(2)+50(5)=350\). 2. The total variance is \(50(1^2)+50(3^2)=500\), so the standard deviation is \(\sqrt{500}\approx22.36\). 3. With many independent contributions and no single term dominating, the total is approximately normal. 4. The standardized cutoff is \(z=\frac{390-350}{22.36}\approx1.789\). 5. Therefore, \(P(S>390)\approx0.0368\).

The approximate probability is \(0.0368\).

- Compute the population parameters from the two possible values. - For the exact calculation, translate the average threshold into a required count of rare observations. - Compare the two results in light of how unusual values affect moderate samples.

1. The population mean is \(1\), and the variance is \(0.99(0-1)^2+0.01(100-1)^2=99\). Thus, the sample-mean standard error is \(\frac{\sqrt{99}}{\sqrt{30}}\approx1.817\). 2. The CLT standardized cutoff is \(z=\frac{5-1}{1.817}\approx2.202\), giving an approximate probability of \(0.0138\). 3. The sample mean exceeds \(5\) only when at least two of the \(30\) observations equal \(100\). The exact probability is \(1-(0.99)^{30}-30(0.01)(0.99)^{29}\approx0.0361\). 4. The approximation is poor because the population is extremely skewed and rare large values dominate the sample mean at this sample size.

a) CLT approximation: \(0.0138\). b) Exact probability: approximately \(0.0361\). c) The approximation substantially understates the probability; \(n=30\) is not enough for this extremely skewed population.