- Recall where the sampling distribution of a sample mean is centered. - Ask whether repeated sample means are centered at the population mean.

1. The sample mean is an unbiased estimator of the population mean. 2. Therefore, \(E(\bar X)=\mu=50.\)

\(50\).

- Distinguish the distribution of individual data values from the distribution of a statistic across repeated samples. - Ask which random quantity the central limit theorem describes.

1. The statement is not correct. 2. Increasing the sample size does not change the population distribution or make the individual observations normal. 3. Under suitable conditions, the central limit theorem says that the sampling distribution of the sample mean becomes approximately normal as the sample size grows.

No. The individual observations do not become normal. The sampling distribution of the sample mean becomes approximately normal as the sample size grows.

- Which mean describes the whole library, and which describes one group of songs? - Can one unusually long or short sample change the long-run center of repeated sample means?

1. The \(4.1\,\text{min}\) mean came from just one sample, so it need not equal the population mean. 2. Across repeated random samples of the same size, the mean of the sample means equals the population mean, \(3.5\,\text{min}\).

No. The sample means are centered at \(3.5\,\text{minutes}\), the population mean, not the mean from one sample.

- Count what the researcher records after taking one sample. - The number of students measured is not necessarily the number of plotted values.

1. The researcher records one number—the mean—for each of the \(50\) samples. 2. The histogram therefore contains \(50\) sample means, not \(500\) individual study times.

\(50\) values. Each is the mean study time for a group of \(10\) students.

- Focus on variation among averages, not among individual observations. - What usually happens to the stability of an average when more observations contribute to it?

1. The standard deviation of a sample mean is \(\sigma/\sqrt{n}\) for independent observations. 2. A larger sample size gives a smaller standard deviation of the average. 3. Means of samples of size \(16\) are therefore less spread out than means of samples of size \(4\).

Sample size \(16\). More observations per average make the sampling distribution less spread out.

- Compare the widths of the two distributions on the same horizontal scale. - Ask whether the larger sample size should make averages more or less variable. - A similar center does not imply a similar spread.

1. Both histograms are centered near \(30\), so the center alone does not identify sample size. 2. Panel b) has its sample means concentrated more closely around \(30\), showing a smaller spread. 3. Larger samples tend to have more stable averages, so panel b) most likely comes from samples of size \(36\).

Panel b). Its sample means are less spread out, as expected for samples of size \(36\).

- 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.

- Decide whether normality of the sample mean is approximate or exact. - Find the standard error of the sample mean. - Standardize the 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 \(\frac{6}{\sqrt4}=3.\) 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 sample mean is exactly normal because the population is normal, so the central limit theorem is unnecessary.

- The spread of a sample mean decreases with the square root of sample size. - Convert the six-unit distance into standard-error units. - Use a central standard-normal probability.

1. The standard deviation of the sample mean is \(\sigma_{\bar X}=\frac{18}{\sqrt{36}}=3\), not \(18/36\). 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) Approximately \(0.9545\).

- 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 \(72\).

- For a sum, compare how means and variances scale with \(n\). - Correct the spread before standardizing the total. - Use the CLT approximation only after fixing the model.

1. Soo-jin multiplied the standard deviation by \(100\). For an independent sum, variances add, so \(\sigma_S=2\sqrt{100}=20\). 2. The mean is correctly \(\mu_S=100(4)=400\). 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) Approximately \(0.0668\).

- 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.

- Separate what happens to the center from what happens to the spread of a sample mean. - Compare the square roots of the two sample sizes. - For shape, consider how much averaging occurs in each statistic.

1. Both sampling distributions have mean \(\mu\). 2. For \(n=4\), the standard deviation is \(\frac{\sigma}{\sqrt4}=\frac{\sigma}{2}\). 3. For \(n=36\), the standard deviation is \(\frac{\sigma}{\sqrt{36}}=\frac{\sigma}{6}\). 4. The first standard deviation is \(3\) times the second. 5. The \(n=36\) sampling distribution would generally be more nearly normal because it averages more independent observations, so the central limit theorem gives a better approximation.

a) Both means are \(\mu\). The standard deviations are \(\frac{\sigma}{2}\) and \(\frac{\sigma}{6}\). b) The \(n=4\) standard deviation is \(3\) times as large. c) The \(n=36\) distribution would generally be more nearly normal.

- Use the standard-error formula for a sample mean. - Compare the square roots of \(25\) and \(100\). - The standard error changes with \(\sqrt n\), not directly with \(n\).

1. For \(n=25\), \(\sigma_{\bar X}=\frac{\sigma}{\sqrt{25}}=\frac{\sigma}{5}.\) 2. For \(n=100\), \(\sigma_{\bar X}=\frac{\sigma}{\sqrt{100}}=\frac{\sigma}{10}.\) 3. The sample size is multiplied by \(4\), while the standard error is divided by \(2\).

a) For \(n=25\), \(\sigma_{\bar X}=\frac{\sigma}{5}\). For \(n=100\), \(\sigma_{\bar X}=\frac{\sigma}{10}\). b) The standard error is cut in half.

- Compare both shape and spread across the three panels. - Larger samples make sample means less variable. - The central limit theorem concerns the shape of the sampling distribution of the sample mean, not the shape of the original population.

1. Panel 2 is the widest and most strongly right-skewed, so it corresponds to \(n=1\). 2. Panel 3 is less spread out and more symmetric, so it corresponds to \(n=4\). 3. Panel 1 is the narrowest and most nearly symmetric, so it corresponds to \(n=25\). 4. As \(n\) increases, the sampling distribution becomes less spread out and more nearly normal while staying centered near the same population mean. 5. The increasing symmetry and bell-shaped appearance of the sample-mean distribution as \(n\) grows illustrates the central limit theorem.

a) Panel 1: \(n=25\); Panel 2: \(n=1\); Panel 3: \(n=4\). b) The spread decreases, and the shape becomes more nearly normal while the center stays near \(10\). c) The progression toward a more nearly normal shape as \(n\) increases illustrates the central limit theorem.

- Keep the population spread fixed and think about what changes when the sample size changes. - Standard error depends on the square root of sample size. - Check whether the sample size should increase or decrease.

1. With the same population standard deviation, standard error is inversely proportional to \(\sqrt{n}\). 2. To make the standard error one-third as large, the sample size must be \(3^2=9\) times as large. 3. Thus the needed sample size is \(25\cdot9=225\).

\(225\) observations.

- Write the shared part of two or three readings before averaging them. - Check whether a value added to every reading cancels when you compute a mean. - Separate the shared error from the independent errors.

1. Each of the \(100\) readings includes the same offset \(C\). 2. When the readings are averaged, the offsets contribute \(100C/100=C\). 3. Averaging reduces independent measurement noise, but it does not remove the shared offset.

No. The average still includes the full offset \(C\); only the independent noise is reduced by averaging.

- Compare where each histogram is centered before comparing their shapes. - Which display has a longer tail toward unusually large times? - Think about how averaging many independent times changes extreme values.

1. Panel a) has a long right tail and a wide spread; it is plausible for individual commute times. 2. Panel b) is more concentrated and roughly symmetric around the same center. 3. Averages of independent groups of \(30\) tend to have less spread and a more nearly Normal shape than the original individual times, so panel b) represents group averages.

Panel b). The sample means have a smaller spread and a more symmetric shape than the right-skewed individual commute times in panel a).

- Find the mean and standard deviation of the sample mean. - Standardize the sample-mean cutoff, not an individual call length. - Separate the shape of the population from the approximate shape of the sampling distribution.

1. For the sample mean, \(\mu_{\bar X}=8,\qquad \sigma_{\bar X}=\frac{6}{\sqrt{64}}=0.75.\) 2. Standardize \(9.5\): \(z=\frac{9.5-8}{0.75}=2.\) 3. Therefore, \(P(\bar X>9.5)\approx P(Z>2)\approx0.0228.\) 4. Although the population is strongly right-skewed, the observations are independent and the sample size \(n=64\) is large enough for the central limit theorem to make the sampling distribution of \(\bar X\) approximately normal.

\(P(\bar X>9.5)\approx0.0228\). The approximation is reasonable because the observations are independent and the sample size is large, so the central limit theorem applies despite the skewed population.

- Use the standard deviation of the sample mean rather than the population standard deviation directly. - Standardize both interval endpoints. - Consider both sample size and population shape when judging the quality of the approximation.

1. The sample mean has \(\mu_{\bar X}=50,\qquad \sigma_{\bar X}=\frac{20}{\sqrt{25}}=4.\) 2. The endpoints have standard scores \(z_1=\frac{45-50}{4}=-1.25,\qquad z_2=\frac{55-50}{4}=1.25.\) 3. Thus, \(P(45\le\bar X\le55) \approx\Phi(1.25)-\Phi(-1.25) \approx0.7887.\) 4. The approximation should be treated cautiously because \(n=25\) is only moderate while the population shape is irregular and multimodal; a larger sample would provide stronger CLT support.

\(P(45\le\bar X\le55)\approx0.7887\). A caution is that the sample size is only moderate for such an irregular population, so the normal approximation may be imperfect.

- Find the standard deviation of the sample mean. - Use the standard-normal \(95\)th percentile. - Transform the standard-normal percentile back to the sample-mean scale.

1. The sample mean has \(\mu_{\bar X}=72,\qquad \sigma_{\bar X}=\frac{18}{\sqrt{81}}=2.\) 2. The \(95\)th standard-normal percentile is \(z_{0.95}\approx1.6449.\) 3. Therefore, the corresponding sample-mean percentile is \(72+1.6449(2)\approx75.29.\) 4. The central limit theorem supports the normal approximation because the observations are independent and \(n=81\) is large.

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

- Express the sample-mean standard error in terms of \(\sigma\) for each sample size. - Measure the same absolute cutoff in standard-error units. - Separate the numerical normal approximation from the question of whether that approximation is trustworthy.

1. For \(n=16\), the standard error is \(\frac{\sigma}{\sqrt{16}}=\frac{\sigma}{4}.\) The cutoff has standard score \(z=\frac{\sigma/2}{\sigma/4}=2.\) Thus, \(P\left(\bar X>\mu+\frac{\sigma}{2}\right)\approx0.0228.\) 2. For \(n=64\), the standard error is \(\frac{\sigma}{8},\) so the cutoff has \(z=\frac{\sigma/2}{\sigma/8}=4.\) Therefore, \(P\left(\bar X>\mu+\frac{\sigma}{2}\right) \approx3.17\times10^{-5}.\) 3. Because the population is strongly nonnormal, the approximation for \(n=16\) may be unreliable. The larger sample provides stronger CLT support. 4. Increasing \(n\) reduces the standard error, so the same absolute distance \(\sigma/2\) from \(\mu\) becomes many more standard errors away and therefore much less likely.

For \(n=16\): approximately \(0.0228\), with caution because the sample is small relative to the strong nonnormality. For \(n=64\): approximately \(3.17\times10^{-5}\). The larger sample has a much tighter sampling distribution and provides stronger CLT justification.

- Express the sample sum in terms of the sample mean. - Check whether the two stated events are actually different events. - Standardize the sample-mean cutoff.

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. Thus, both probabilities are approximately \(P(Z>1)\approx0.1587.\)

\(P(S>522)\approx0.1587\) and \(P(\bar X>14.5)\approx0.1587.\) They are equal because the two events are equivalent through \(S=36\bar X\).

- Find the standard error of the sample mean. - Split the probability outside a central \(80\%\) interval equally between the two tails. - Convert the standard-normal critical value back to the sample-mean scale.

1. The sample mean is approximately normal with mean \(\mu_{\bar X}=100\) and standard error \(\sigma_{\bar X}=\frac{24}{\sqrt{64}}=3\). 2. A central area of \(0.80\) leaves \(0.10\) in each tail. The positive standard-normal cutoff is \(z^{*}=\Phi^{-1}(0.90)\approx1.2816\). 3. The margin is approximately \(1.2816\cdot3=3.8448\). Keeping full precision until rounding, the endpoints are approximately \(96.15535\) and \(103.84465\). 4. Rounded to two decimal places, the interval is \([96.16, 103.84]\).

Approximately \([96.16, 103.84]\).

- Find the standard deviation of the sum. - Center the event at the expected total. - Convert the two deviations from the mean into standard scores.

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

Approximately \(0.1585\).

- Find the center and spread of the total. - Standardize both total-value boundaries. - Use the central limit theorem to justify the approximate normal shape of the total.

1. Let \(S\) be the total. Then \(\mu_S=64(4)=256\) minutes and \(\sigma_S=1.5\sqrt{64}=12\) minutes. 2. The standardized bounds are \(z_1=\frac{250-256}{12}=-0.5,\) and \(z_2=\frac{270-256}{12}\approx1.1667.\) 3. Therefore, \(P(250<S<270) \approx P(-0.5<Z<1.1667) \approx0.5698.\)

Approximately \(0.5698\).

- Find the standard error of the sample mean. - Standardize both endpoints using the same standard error. - Use the CLT approximation for the sampling distribution of \(\bar X\).

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 \(z_1=\frac{5.5-6}{2.5/11}=-2.2.\) 3. The upper standardized bound is \(z_2=\frac{6.3-6}{2.5/11}=1.32.\) 4. Therefore, \(P(5.5<\bar X<6.3) \approx P(-2.2<Z<1.32) \approx0.8927.\)

Approximately \(0.8927\).

- 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 \(n=30\).

- Find the center and spread of the total. - Standardize the total-rainfall cutoff. - Separate the skewness of daily rainfall from the approximate shape of the large-sample total.

1. Let \(S\) be the total rainfall. Its mean is \(\mu_S=81(0.12)=9.72\) inches and its standard deviation is \(\sigma_S=0.40\sqrt{81}=3.6\) inches. 2. The standardized cutoff is \(z=\frac{15-9.72}{3.6}\approx1.4667.\) 3. Therefore, \(P(S>15)\approx P(Z>1.4667)\approx0.0712.\) 4. The large sample supports the CLT approximation despite the zero-inflated, highly skewed daily distribution.

Approximately \(0.0712\).

- 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 index-value units. - Find the standard error of the sample mean. - Standardize both endpoints symmetrically about the mean.

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 \(z=\pm\frac{5}{3}\approx\pm1.6667.\) 4. Therefore, \(P(95<\bar X<105) \approx P(-1.6667<Z<1.6667) \approx0.9044.\)

Approximately \(0.9044\).

- Find the standard error of the sample mean. - Transform the standard-normal percentile to the \(\bar X\) scale. - Use the deterministic relationship \(S=n\bar X\) to transform the percentile to the sum scale.

1. The sample mean has approximate mean \(\mu_{\bar X}=40\) and standard error \(\sigma_{\bar X}=\frac{8}{\sqrt{25}}=1.6.\) 2. The standard-normal \(75\)th percentile is \(z_{0.75}\approx0.6745.\) 3. The sample-mean percentile is \(40+0.6745(1.6)\approx41.08.\) 4. Since \(S=25\bar X,\) use the unrounded sample-mean percentile: \(25[40+0.6745(1.6)]\approx1026.98.\)

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

- Transform the population center and spread before working with the sample mean. - Use the standard error for \(\bar Y\). - Standardize the cutoff on the transformed scale.

1. The transformed population has \(\mu_Y=3(20)-7=53\) and \(\sigma_Y=3(4)=12.\) 2. The sample mean has approximate standard error \(\frac{12}{\sqrt{64}}=1.5.\) 3. The standardized cutoff is \(z=\frac{55-53}{1.5}\approx1.3333.\) 4. Therefore, \(P(\bar Y>55)\approx P(Z>1.3333)\approx0.0912.\)

Approximately \(0.0912\).

- 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, Tariq'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 standard error of the sample mean. - Convert the absolute deviation to a standard score. - Include both tails.

1. The sample mean has approximate mean \(0\) and standard error \(\frac{1}{\sqrt{49}}=\frac17.\) 2. The standardized magnitude is \(z=\frac{0.30}{1/7}=2.1.\) 3. The event has two symmetric tails, so \(P(|\bar X|>0.30) \approx2P(Z>2.1) \approx0.0357.\)

Approximately \(0.0357\).

- Convert the given cumulative probability into a standard-normal score. - Find the spread of the sample mean from the population standard deviation. - Work backward from the cutoff to the population mean.

1. An approximate cumulative probability of \(0.9772\) corresponds to a standard score of about \(2\). 2. The standard error is \(\frac{8}{\sqrt{64}}=1\). 3. Thus \(\frac{52-\mu}{1}\approx2\), giving \(\mu\approx50\). 4. To the nearest whole number, the estimated population mean is \(50\).

\(\mu\approx50\), to the nearest whole number.

- 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 mean and standard deviation of the sum. - Use the central-\(95\%\) standard-normal critical value. - Apply the margin symmetrically around the expected total.

1. The total has mean \(E(S)=225(7)=1575\) and standard deviation \(\sigma_S=4\sqrt{225}=60.\) 2. A central probability of \(0.95\) uses \(z^*\approx1.96.\) 3. The margin is \(1.96(60)=117.6.\) 4. Therefore, 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 mean and standard deviation of the sum. - Because the total is integer-valued, translate the lower boundary using a half-unit continuity correction. - Standardize the corrected boundary.

1. Let \(S\) be the total over \(49\) days. Then \(\mu_S=49(30)=1470,\) and \(\sigma_S=8\sqrt{49}=56.\) 2. For the integer event \(S\ge1550\), use the continuity-corrected boundary \(1549.5\). 3. The standardized boundary is \(z=\frac{1549.5-1470}{56}\approx1.420.\) 4. Therefore, \(P(S\ge1550)\approx P(Z\ge1.420)\approx0.0779.\)

\(P(S\ge1550)\approx0.0779\).

- A symmetric central probability leaves equal probabilities in the two tails. - Express the three-unit margin in units of the sample mean's standard error. - Use the corresponding normal cutoff to solve for the population standard deviation.

1. The central probability \(0.8664\) gives an upper cumulative probability of approximately \(\frac{1+0.8664}{2}=0.9332\). 2. Its positive standard-normal cutoff is \(z=\Phi^{-1}(0.9332)\approx1.5001\). 3. The three-unit margin corresponds to \(\frac{3}{\sigma/\sqrt{64}}\approx1.5001\). 4. Solving gives \(\sigma\approx\frac{24}{1.5001}\approx16.0\), so the nearest whole number is \(16\).

\(\sigma\approx16\).

- Find the mean and standard deviation of the total. - Translate the discrete upper endpoint with a half-unit continuity correction. - Standardize the corrected boundary.

1. Let \(S\) be the total. Then \(\mu_S=100(4)=400\) and \(\sigma_S=3\sqrt{100}=30.\) 2. For the integer event \(S\le350\), use the continuity-corrected boundary \(350.5\). 3. The standardized boundary is \(z=\frac{350.5-400}{30}=-1.65.\) 4. Therefore, \(P(S\le350)\approx P(Z\le-1.65)\approx0.0495.\)

Approximately \(0.0495\).

- 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.

- Find the mean and standard deviation of the total. - Extend the inclusive integer interval by half a unit at each end. - Standardize the corrected boundaries.

1. Let \(S\) be the total. Then \(\mu_S=50(6)=300\) and \(\sigma_S=2\sqrt{50}\approx14.142.\) 2. The inclusive integer event \(290\le S\le310\) becomes \(289.5<S<310.5\) under continuity correction. 3. The standardized bounds are approximately \(-0.7425\quad\text{and}\quad0.7425.\) 4. Therefore, \(P(290\le S\le310) \approx P(-0.7425<Z<0.7425) \approx0.5422.\)

Approximately \(0.5422\).

- 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, \(n=107\).

- Add the two groups' mean contributions. - Add variances, not standard deviations, for independent contributions. - Standardize the total using its combined standard deviation.

1. The total mean is \(\mu_S=50(2)+50(5)=350.\) 2. Since all contributions are independent, the total variance is \(\sigma_S^2=50(1^2)+50(3^2)=500,\) so \(\sigma_S=\sqrt{500}=10\sqrt5\approx22.36.\) 3. With many independent contributions and no single term dominating the total variance, a normal approximation is reasonable. 4. The standardized cutoff is \(z=\frac{390-350}{\sqrt{500}}\approx1.7889.\) 5. Therefore, \(P(S>390)\approx0.0368.\)

Approximately \(0.0368\).

- Compute the population mean and variance from the two possible values. - For the exact calculation, translate the sample-mean cutoff into a required count of \(100\) values. - Compare the approximate and exact probabilities numerically before judging the approximation.

1. The population mean is \(\mu=0(0.99)+100(0.01)=1.\) The variance is \(0.99(0-1)^2+0.01(100-1)^2=99.\) Thus, the sample-mean standard error is \(\frac{\sqrt{99}}{\sqrt{30}}\approx1.8166.\) 2. The CLT standardized cutoff is \(z=\frac{5-1}{1.8166}\approx2.202.\) Therefore, \(P(\bar X>5)\approx0.0138.\) 3. The sample mean exceeds \(5\) exactly when at least two of the \(30\) observations equal \(100\). Hence, \(P(\bar X>5) =1-(0.99)^{30}-30(0.01)(0.99)^{29} \approx0.0361.\) 4. The CLT approximation substantially understates the exact probability. The population is extremely skewed, with rare values that dominate the sample mean, so \(n=30\) is not large enough for a good normal approximation here.

a) CLT approximation: \(0.0138\). b) Exact probability: \(0.0361\). c) The approximation is poor and substantially understates the true probability.