Free math worksheets

- The condition compares the sample size with the size of the population from which it is drawn. - Express the sample as a fraction of the population and impose the stated percentage limit. - Check the proposed population separately after finding the threshold.

1. The \(10\%\) condition requires the population size \(N\) to be at least \(10\) times the sample size. 2. With \(n=75\), the minimum population size is \(N=10(75)=750\). 3. A population of \(600\) does not satisfy the condition because \(75\) is \(12.5\%\) of \(600\), which exceeds \(10\%\).

The minimum population size is \(750\). A population of \(600\) does not satisfy the \(10\%\) condition because the sample would be more than \(10\%\) of the population.

- Use the same population spread for all three sampling plans. - Compare the square roots of the three sample sizes rather than the sample sizes directly. - Look for a simple multiplicative pattern in the resulting sampling spreads.

1. For \(n=16\), the standard deviation is \(\frac{24}{\sqrt{16}}=6\). 2. For \(n=64\), the standard deviation is \(\frac{24}{\sqrt{64}}=3\). 3. For \(n=144\), the standard deviation is \(\frac{24}{\sqrt{144}}=2\). 4. The spread decreases according to the inverse square root of sample size: larger samples produce more concentrated sample means.

The standard deviations are \(6\), \(3\), and \(2\), respectively. Sampling variability decreases as \(1/\sqrt{n}\).

- Write out what a sample mean contains when there is only one observation. - Compare that statistic directly with the sampled value. - Then use the usual center and spread relationships at \(n=1\).

1. When \(n=1\), the sample mean is just the single sampled observation. 2. Therefore, the sampling distribution of \(\bar{x}\) has the same shape as the population, so it is strongly right-skewed. 3. Its mean is the population mean \(\mu\). 4. Its standard deviation is the population standard deviation \(\sigma\), since \(\sigma/\sqrt{1}=\sigma\).

For \(n=1\), the sampling distribution of \(\bar{x}\) is the population distribution itself: it has the same strongly right-skewed shape, mean \(\mu\), and standard deviation \(\sigma\).

- Ask what values can appear in any sample from this population. - An average cannot vary if none of its component observations vary. - A distribution with only one possible value has no spread.

1. Every sampled observation equals \(7\). 2. Therefore, every sample mean also equals \(7\), regardless of the sample size. 3. The sampling distribution is concentrated entirely at \(7\), with mean \(7\) and standard deviation \(0\).

\(\bar{x}=7\) for every possible sample. Its sampling distribution has mean \(7\) and standard deviation \(0\).

- Think about what happens when several integer values are averaged. - The possible values of a statistic do not have to match the possible values of an individual observation. - Try a small sample of integer counts and compute its mean.

1. A sample mean is an average of several integer observations. 2. Averages of integers need not be integers; for example, values \(1,2,2,2\) have sample mean \(7/4=1.75\). 3. Therefore, the sampling distribution of \(\bar{x}\) can include noninteger values even though every individual population value is an integer.

The claim is false. Sample means are averages, so the sampling distribution can contain fractions or decimals even when all population observations are integers.

- Think about what happens to the center when many sample means are taken from the same population. - Consider how averaging \(64\) observations changes the spread compared with the spread of individual observations. - After describing the sampling distribution, translate the probability question into a standardized distance from its center.

1. The sampling distribution is normal because the population distribution is approximately normal. 2. Its mean is \(\mu_{\bar{x}}=18.4\,\text{s}\). 3. Its standard deviation is \(\sigma_{\bar{x}}=\frac{4.8}{\sqrt{64}}=0.6\,\text{s}\). 4. The standardized value for \(19.5\) is \(z=\frac{19.5-18.4}{0.6}\approx1.83\). 5. Therefore, \(P(\bar{x}>19.5)\approx0.0334\).

The sampling distribution has mean \(18.4\,\text{s}\) and standard deviation \(0.6\,\text{s}\), and \(P(\bar{x}>19.5)\approx0.0334\).

- The center of a sample-mean distribution does not depend on the sample size. - Focus on how the spread of sample means changes as the sample size grows. - Solve the spread requirement for the smallest allowable whole-number sample size.

1. The sampling distribution has mean \(\mu_{\bar{x}}=240\,\text{hours}\). 2. Require \(\frac{36}{\sqrt{n}}\le4.5\). 3. This gives \(\sqrt{n}\ge8\), so \(n\ge64\). 4. The smallest whole-number sample size is \(64\).

The smallest sample size is \(n=64\), and the sampling-distribution mean is \(240\,\text{hours}\).

- First decide whether the sample size is large enough to model sample means more simply than the original population. - Find the center and sampling spread before translating the two endpoints. - A between-probability requires the area bounded by both standardized endpoints.

1. Since \(n=49\ge30\), the sampling distribution of \(\bar{x}\) can be modeled as approximately normal unless the population is extremely skewed. 2. Its mean is \(12\,\text{min}\), and its standard deviation is \(\frac{7}{\sqrt{49}}=1\,\text{min}\). 3. The lower endpoint has \(z=\frac{10-12}{1}=-2\), and the upper endpoint has \(z=\frac{13.5-12}{1}=1.5\). 4. Therefore, \(P(10<\bar{x}<13.5)=P(-2<Z<1.5)\approx0.9104\).

\(P(10<\bar{x}<13.5)\approx0.9104\).

- Compare the rule for the variability of individual observations with the rule for an average of many independent observations. - Check whether increasing the sample size by a factor of \(36\) should reduce spread by the same factor or by a smaller factor. - Revisit the distinction between variance and standard deviation when combining independent observations.

1. The standard deviation of a sample mean is found by dividing the population standard deviation by the square root of the sample size, not by the sample size itself. 2. The correct standard deviation is \(\sigma_{\bar{x}}=\frac{18}{\sqrt{36}}=3\). 3. Averaging \(36\) observations reduces variability by a factor of \(\sqrt{36}=6\), not by a factor of \(36\). 4. Thus, \(0.5\) understates the sampling variability of \(\bar{x}\).

The student divided by \(n\) instead of \(\sqrt{n}\). The correct sampling-distribution standard deviation is \(3\), not \(0.5\).

- Compare the population spread with the square root of the sample size in each study. - The center of a sample-mean distribution follows the population mean. - Look for proportional changes in the numerator and denominator of the sampling-spread expression.

1. Because the populations have the same mean, the two sampling distributions have the same mean. 2. Study A has sampling-distribution standard deviation \(\frac{12}{\sqrt{25}}=2.4\). 3. Study B has sampling-distribution standard deviation \(\frac{24}{\sqrt{100}}=2.4\). 4. Study B doubles the population standard deviation but quadruples the sample size, so the square root of the sample size also doubles. The two effects cancel in the standard deviation of \(\bar{x}\).

The sampling distributions have the same mean and the same standard deviation, \(2.4\). In Study B, doubling the population standard deviation is exactly offset by doubling \(\sqrt{n}\) when the sample size is quadrupled.

- One sampling-distribution parameter is inherited directly from the population. - The other parameter changes by a square-root sample-size factor. - Work backward from the sample-mean spread to recover the population spread.

1. The mean of the sampling distribution of \(\bar{x}\) equals the population mean, so \(\mu=145\). 2. The sampling-distribution standard deviation satisfies \(2.2=\frac{\sigma}{\sqrt{81}}\). 3. Since \(\sqrt{81}=9\), \(\sigma=2.2(9)=19.8\).

The population mean is \(145\), and the population standard deviation is \(19.8\).

- Model the hourly sample mean, not the weight of a single unit. - The two alert limits are equally far from the sampling-distribution center. - Convert the alert rule into two tail events in the standardized sampling distribution.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(500\,\text{g}\) and standard deviation \(\frac{40}{\sqrt{100}}=4\,\text{g}\). 2. The alert limits are \(8\,\text{g}\) below and above the mean, corresponding to \(z=-2\) and \(z=2\). 3. The alert probability is \(P(Z<-2)+P(Z>2)\approx0.0455\).

The probability of triggering an alert is approximately \(0.0455\).

- Sample size is not the only route to a normal sampling distribution. - Determine the sampling-distribution spread before standardizing the interval endpoints. - The requested event is centered symmetrically around the population mean.

1. Because the population itself is normal, the sampling distribution of \(\bar{x}\) is normal for any sample size. 2. Its mean is \(10\), and its standard deviation is \(\frac{2}{\sqrt{5}}\approx0.894\). 3. The standardized endpoints are approximately \(-1.118\) and \(1.118\). 4. Therefore, \(P(9<\bar{x}<11)\approx0.7364\).

\(P(9<\bar{x}<11)\approx0.7364\). The normal model is exact here because the source population is normal, so a large-sample condition is not needed for the shape.

- Compare the sample size with a fixed percentage of the finite population. - Sampling without replacement creates dependence when the sampling fraction becomes large. - The issue here is the sampling design, not the shape of the population distribution.

1. Ten percent of the population is \(0.10(500)=50\). 2. The sample size is \(80\), which is greater than \(50\). 3. Equivalently, the sample is \(16\%\) of the population. 4. Therefore, the \(10\%\) condition fails, so this guideline does not support the usual independence approximation or the unadjusted standard-deviation formula for sampling without replacement.

The \(10\%\) condition fails because \(80>50\), so the sample is too large a fraction of the population for the usual independence approximation and unadjusted formula to be justified by that guideline.

- Treat common sample-size cutoffs as context-dependent guidelines, not automatic laws. - Consider how severe the source-population skewness is in this problem. - The central limit theorem describes convergence, but the speed of convergence depends on population shape.

1. A sample size of at least \(30\) often supports an approximately normal model for the sampling distribution of a sample mean when the population is moderately nonnormal. 2. However, an extremely skewed population may require a sample size much larger than \(30\) before the normal approximation is accurate. 3. Therefore, \(n=30\) does not guarantee approximate normality in this setting. 4. More information or a substantially larger sample would be needed before confidently using the normal approximation.

No. The \(n\ge30\) guideline is not an absolute guarantee. For an extremely skewed population, a much larger sample may be needed before the sampling distribution of \(\bar{x}\) is approximately normal.

- Find the spread of the sampling distribution separately for each sample size. - The interval “within \(2\)” is the same in original units but represents different standardized distances in the two distributions. - Compare how concentration around the population mean changes as \(n\) increases.

1. For \(n=25\), the standard deviation of \(\bar{x}\) is \(\frac{10}{\sqrt{25}}=2\). Being within \(2\) units of the mean corresponds to \(-1<Z<1\), so the probability is approximately \(0.6827\). 2. For \(n=100\), the standard deviation of \(\bar{x}\) is \(\frac{10}{\sqrt{100}}=1\). Being within \(2\) units corresponds to \(-2<Z<2\), so the probability is approximately \(0.9545\). 3. The larger sample size produces a narrower sampling distribution, so sample means are more likely to fall close to the population mean.

For \(n=25\), the probability is approximately \(0.6827\). For \(n=100\), it is approximately \(0.9545\). The larger sample gives the greater probability because its sample means vary less.

- Think about what adding the same constant to every observation does to a sample mean. - Separate the effect on the center from the effect on the spread. - The usual sample-size effect on the spread of a sample mean still applies after calibration.

1. Adding \(7\) to every observation adds \(7\) to every sample mean, so \(\overline{Y}=\overline{X}+7\). 2. Therefore, the sampling-distribution mean is \(\mu+7\). 3. Adding a constant does not change variability, so the population standard deviation remains \(\sigma\). 4. Thus, the sampling-distribution standard deviation is \(\frac{\sigma}{\sqrt{n}}\).

The sampling distribution of \(\overline{Y}\) has mean \(\mu+7\) and standard deviation \(\frac{\sigma}{\sqrt{n}}\).

- Do not confuse the shape of the population with the center of the sampling distribution. - Recall what determines the expected value of a sample mean. - Consider what a large sample size does to the shape of the sampling distribution of the mean.

1. The sampling distribution of the sample mean is centered at the population mean \(\mu\), not at a population mode. 2. Its standard deviation is \(\frac{\sigma}{\sqrt{64}}=\frac{\sigma}{8}\), assuming the sampling observations are independent. 3. With a sample size of \(64\), the central limit theorem generally supports an approximately normal sampling distribution of the sample mean even though the population itself is bimodal. 4. Therefore, the student's claims about both the center and the shape are incorrect.

The statement is incorrect. The sampling distribution is centered at \(\mu\), has standard deviation \(\frac{\sigma}{8}\), and is approximately normal for a sufficiently well-behaved finite-variance population at \(n=64\), despite the population's bimodality.

- Find the center and spread of the sample-mean distribution before working with the two bounds. - Standardize the lower and upper endpoints separately. - The requested probability is the area between the two standardized values.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(70\) minutes and standard deviation \(\frac{12}{\sqrt{36}}=2\) minutes. 2. The lower bound gives \(z=\frac{68-70}{2}=-1.00\). 3. The upper bound gives \(z=\frac{73-70}{2}=1.50\). 4. Therefore, \(P(68<\bar{x}<73)=P(-1<Z<1.5)\approx0.7745\).

The probability is approximately \(0.7745\).

- Separate random sampling variability from systematic selection bias. - Ask how participants enter each sample before comparing the sample sizes. - A larger sample can narrow a distribution without correcting a biased center.

1. A simple random sample gives each appropriate population member a chance to be selected through a random mechanism, so the sample mean is an unbiased estimator of the population mean under the sampling model. 2. Method B uses voluntary response, so people who choose to participate may differ systematically from the population. 3. Increasing the number of voluntary responses can reduce random sampling variability around the volunteer-group tendency, but it does not remove selection bias. 4. Therefore, the larger sample size in Method B does not guarantee that its sample-mean distribution is centered at the population mean.

Method A uses random sampling and is designed to produce a sample mean centered at the population mean. Method B can remain biased even with \(400\) responses because voluntary response affects the center, not just the spread.

- Distinguish the population standard deviation from the sample standard deviation. - The question asks for the theoretical spread of all possible sample means under the stated population. - Use the population quantity when it is given exactly.

1. The theoretical standard deviation of the sampling distribution uses the population standard deviation \(\sigma\), when it is known. 2. Therefore, \(\sigma_{\bar{x}}=\frac{15}{\sqrt{36}}=2.5\). 3. The value \(\frac{12}{\sqrt{36}}=2\) is an estimated standard error based on this particular sample, not the known theoretical standard deviation of the sampling distribution.

The correct sampling-distribution standard deviation is \(2.5\). The value \(2\) uses the sample standard deviation and is an estimate, not the known theoretical spread.

- Ask whether different repetitions of the stated sampling procedure can produce different samples. - Sampling variability comes from variation among possible samples. - Compare a census with the independence assumption behind the usual sample-mean spread formula.

1. Every repetition selects the entire population, so every sample contains exactly the same \(80\) values. 2. Therefore, every sample mean equals the population mean \(\mu\). 3. The sampling distribution is concentrated at the single value \(\mu\) and has standard deviation \(0\). 4. The usual \(\sigma/\sqrt{n}\) formula assumes approximately independent sampling observations; sampling the entire finite population without replacement violates that approximation completely.

The sampling distribution consists only of \(\mu\), so its mean is \(\mu\) and its standard deviation is \(0\). The usual \(\sigma/\sqrt{n}\) formula does not apply to a census drawn without replacement.

- Express the distance from the population mean in units of the sampling-distribution standard deviation. - Once standardized, the problem becomes a fixed probability under the standard normal distribution. - The original measurement scale disappears after standardization.

1. Standardizing gives \(Z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}\). 2. Being within \(2\sigma_{\bar{x}}\) of \(\mu\) is equivalent to \(-2<Z<2\). 3. For a standard normal distribution, \(P(-2<Z<2)\approx0.9545\). 4. Standardization removes the original center and scale, so the probability depends only on the standardized cutoff \(2\).

The probability is approximately \(0.9545\), regardless of the particular values of \(\mu\), \(\sigma\), and \(n\), as long as the sampling distribution is normal.

- First find the center and spread of the sample mean on the original measurement scale. - Think about how multiplying every data value by the same constant changes a mean. - The same scale factor also changes measures of spread.

1. Before conversion, the sampling distribution of \(\bar{x}\) has mean \(12\) miles and standard deviation \(\frac{3}{\sqrt{25}}=0.6\) mile. 2. Multiplying every observation by \(1.6\) multiplies both the sample mean and its sampling-distribution standard deviation by \(1.6\). 3. The converted mean is \(12\cdot1.6=19.2\) km. 4. The converted standard deviation is \(0.6\cdot1.6=0.96\) km.

The sampling distribution has mean \(19.2\,\text{km}\) and standard deviation \(0.96\,\text{km}\).

- Separate the ideas of center and spread in a sampling distribution. - Ask whether the sample mean systematically overestimates or underestimates the population mean. - Larger samples affect how much sample means vary from sample to sample.

1. The sample mean is an unbiased estimator of the population mean, so its sampling distribution is centered at \(48\) for both sample sizes. 2. Increasing the sample size does not move the center toward \(0\). 3. The sampling-distribution standard deviation decreases as the sample size increases, so the sample means for \(n=100\) are less variable than those for \(n=25\).

The claim is false. Both sampling distributions are centered at \(48\). Increasing the sample size reduces the spread of \(\bar{x}\); it does not shrink the center toward \(0\).

- Ask whether the added rows contain new random outcomes. - The usual sample-size effect on spread assumes independent observations. - Reformatting or duplicating data cannot create new information about the population mean.

1. Duplicating each measurement creates exact copies, not new independent observations from the population. 2. The duplicated dataset contains the same information about the population mean as the original \(30\) observations. 3. Therefore, the effective sample size remains \(30\); using a standard deviation proportional to \(1/\sqrt{60}\) would falsely claim more precision. 4. The sampling variability is determined by the original data-collection process, not by the number of rows after duplication.

The effective sample size is still \(30\). Duplicating rows does not create independent information and does not reduce the sampling-distribution standard deviation as a genuine sample of size \(60\) would.

- First identify the center and spread of a standardized individual value. - Then apply the sampling-distribution relationships for a sample mean. - The sample size affects the spread but not the center.

1. Standardizing the population gives \(Y\) mean \(0\) and standard deviation \(1\). 2. The sample mean \(\bar{Y}\) is therefore centered at \(0\). 3. Its sampling-distribution standard deviation is \(1/\sqrt{16}=0.25\).

The sampling distribution of \(\bar{Y}\) has mean \(0\) and standard deviation \(0.25\).

- Use the fact that an average must lie between the smallest and largest values being averaged. - Compare the true support of the statistic with the unbounded support of a normal approximation. - An approximation does not override an exact logical constraint.

1. A sample mean is an average of sampled population values. 2. If every sampled value is at most \(10\), their average cannot exceed \(10\). 3. Therefore, \(P(\bar{x}>10)=0\) exactly. 4. A normal approximation can assign extremely small probability outside the true range because a normal distribution is unbounded; such tail probability is an approximation artifact.

The exact probability is \(0\). An average of values all between \(0\) and \(10\) cannot be greater than \(10\), even if a normal approximation assigns a tiny impossible tail probability there.

- Keep variance and standard deviation formulas distinct. - You can convert the population variance to a population standard deviation first. - Check the result by squaring the sampling-distribution standard deviation.

1. The population standard deviation is \(\sqrt{81}=9\). 2. The standard deviation of \(\bar{x}\) is \(9/\sqrt{9}=3\). 3. Therefore, the variance of \(\bar{x}\) is \(3^2=9\), equivalently \(81/9=9\). 4. The student incorrectly divided the variance by \(\sqrt{n}\); variances scale by \(1/n\), while standard deviations scale by \(1/\sqrt{n}\).

The sampling-distribution variance is \(9\), and the sampling-distribution standard deviation is \(3\).

- Distinguish a statement about one realized sample from a statement about repeated sampling. - Larger samples reduce sampling variability. - Lower variability increases concentration around the population mean but does not eliminate randomness.

1. Random samples vary, so a particular sample of size \(10\) can happen to have a mean very close to \(\mu\), while a particular sample of size \(100\) can be farther away. 2. Increasing the sample size reduces the standard deviation of the sampling distribution of \(\bar{x}\). 3. Therefore, larger samples make sample means more tightly concentrated around \(\mu\) and increase the probability of being within any fixed reasonable distance of \(\mu\). 4. This is a probabilistic tendency, not a guarantee comparing every pair of samples.

A size-\(100\) sample mean is not guaranteed to be closer to \(\mu\) than a size-\(10\) sample mean. It is more likely to be close because the sampling distribution is less variable.

- The population shape affects whether a small-sample mean can be modeled normally. - Separate the questions of center, spread, and shape; sample size influences them differently. - A large random sample can make the distribution of sample means much more regular than the original population distribution.

1. For \(n=16\), the sampling-distribution mean is \(\$48\) and the standard deviation is \(\frac{30}{\sqrt{16}}=\$7.50\). 2. Because the population is strongly right-skewed and \(n=16\) is small, an approximately normal sampling distribution is not guaranteed. 3. For \(n=100\), the sampling-distribution mean is \(\$48\) and the standard deviation is \(\frac{30}{\sqrt{100}}=\$3.00\). 4. With \(n=100\), the central limit theorem supports modeling the sampling distribution of \(\bar{x}\) as approximately normal, provided the population is not so extremely skewed that an even larger sample is needed. 5. Both sample sizes satisfy the \(10\%\) condition because the population is much larger than either sample.

For \(n=16\): mean \(\$48\), standard deviation \(\$7.50\), and normality is not guaranteed because the population is strongly skewed and the sample is small. For \(n=100\): mean \(\$48\), standard deviation \(\$3.00\), and the sampling distribution is approximately normal by the central limit theorem, barring extreme skewness.

- Describe the sampling distribution before looking for the requested cutoff. - The question asks for a percentile of sample means, not a percentile of individual job sizes. - Convert the desired cumulative probability into a standardized location and then return to page units.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(120\) pages and standard deviation \(\frac{24}{\sqrt{36}}=4\) pages. 2. The \(95\)th percentile of the standard normal distribution is approximately \(1.645\). 3. Therefore, \(c=120+1.645(4)\approx126.58\) pages. 4. About \(95\%\) of random samples of \(36\) jobs have a sample mean print-job size below about \(126.58\) pages.

\(c\approx126.58\) pages. About \(95\%\) of random samples of \(36\) print jobs have \(\bar{x}<126.58\) pages.

- Do not use the same spread for an individual observation and for an average of \(25\) observations. - Compare how many standard deviations \(74\) lies above the center in each distribution. - The shared mean does not imply the two distributions have the same concentration around that mean.

1. For an individual value, \(z=\frac{74-70}{10}=0.4\), so \(P(X>74)\approx0.3446\). 2. For the sample mean, the standard deviation is \(\frac{10}{\sqrt{25}}=2\). 3. Thus, \(z=\frac{74-70}{2}=2\) for \(\bar{x}\), so \(P(\bar{x}>74)\approx0.0228\). 4. Both distributions are centered at \(70\), but sample means are much less variable than individual observations, making a mean above \(74\) much less common.

\(P(X>74)\approx0.3446\), while \(P(\bar{x}>74)\approx0.0228\). Sample means have the same center as individual values but a much smaller standard deviation, so values far from the mean are less likely for \(\bar{x}\).

- Express the spread of possible sample means in terms of the unknown sample size. - Think about the standardized cutoff that captures the middle \(95\%\) of a normal distribution. - After solving for the sample size, check which way a noninteger result must be rounded to preserve the probability requirement.

1. Because the population distribution is approximately normal, the sampling distribution of \(\bar{x}\) is normal with standard deviation \(\frac{4.2}{\sqrt{n}}\). 2. To place the middle \(95\%\) of sample means within \(0.7\,\text{min}\) of the population mean, require \(0.7\ge 1.960\left(\frac{4.2}{\sqrt{n}}\right)\). 3. Solving gives \(n\ge\left(\frac{1.960\cdot4.2}{0.7}\right)^2\approx138.30\). 4. The sample size must be a whole number and must meet the requirement, so round up to \(n=139\). 5. For \(n=139\), the probability is approximately \(0.9506\), while \(n=138\) gives approximately \(0.9498\).

The minimum sample size is \(139\).

- First describe the distribution of a sample mean for samples of size \(25\). - Convert the cutoff of \(55\,\text{s}\) into a probability for one sample mean. - Use that probability to find an expected count out of \(300\) repeated samples.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(52\,\text{s}\) and standard deviation \(\frac{10}{\sqrt{25}}=2\,\text{s}\). 2. For \(55\,\text{s}\), the standardized value is \(z=\frac{55-52}{2}=1.50\). 3. The probability that one sample mean exceeds \(55\,\text{s}\) is \(P(Z>1.50)\approx0.0668\). 4. Across \(300\) independent samples, the expected number is \(300\cdot0.0668\approx20.0\).

About \(20\) of the \(300\) sample means would be expected to exceed \(55\,\text{s}\).

- The center is the same for individual observations and sample means, but their spreads are not. - Standardize \(118\,\text{s}\) once using the population standard deviation and once using the standard deviation of \(\bar{x}\). - Compare the resulting standardized distances before comparing the tail probabilities.

1. For one individual time, \(z=\frac{118-100}{12}=1.50\), so \(P(X>118)\approx0.0668\). 2. For samples of size \(16\), the sampling distribution of \(\bar{x}\) is normal with mean \(100\,\text{s}\) and standard deviation \(\frac{12}{\sqrt{16}}=3\,\text{s}\). 3. For the sample mean, \(z=\frac{118-100}{3}=6\), so \(P(\bar{x}>118)\approx9.87\times10^{-10}\). 4. The same numerical cutoff is far more extreme for a sample mean because sample means vary much less than individual observations.

For an individual time, the flag probability is approximately \(0.0668\). For a sample mean of \(16\) times, the flag probability is approximately \(9.87\times10^{-10}\). The cutoff is therefore vastly more extreme for the sample mean.

- Convert the condition on the total into an equivalent condition on the sample mean. - The sample is large enough to use the sampling distribution of the mean even though individual amounts are skewed. - Standardize the mean cutoff using the spread of sample means, not the spread of individual households.

1. A total above \(2000\,\text{lb}\) for \(100\) households is equivalent to a sample mean above \(\frac{2000}{100}=20\,\text{lb}\). 2. Since \(n=100\), the sampling distribution of \(\bar{x}\) is approximately normal despite the right-skewed population. 3. The sampling distribution has mean \(18\,\text{lb}\) and standard deviation \(\frac{12}{\sqrt{100}}=1.2\,\text{lb}\). 4. The standardized value is \(z=\frac{20-18}{1.2}\approx1.67\). 5. Therefore, \(P(\bar{x}>20)\approx0.0478\).

The probability that the sampled households total more than \(2000\,\text{lb}\) is approximately \(0.0478\).

- Look at how the observations were selected, not only how many were recorded. - Ask whether two observations from the same selected unit could be related. - The usual sample-mean spread formula relies on independence among the sampled values.

1. The formula \(\sigma/\sqrt{n}\) assumes the sampled observations are independent or approximately independent. 2. People from the same household may have related measurements, so the \(40\) observations are clustered rather than independently sampled individuals. 3. Randomly selecting households does not make the four observations within a household independent of one another. 4. Therefore, treating the data as \(40\) independent observations can understate the sampling variability of the mean.

The formula \(\sigma/\sqrt{40}\) is not justified from the stated design because observations within the same household may be dependent. The effective sampling structure is clustered, not a simple random sample of \(40\) independent people.

- Find the center and standard deviation of the uniform population first. - Use the sample size to determine the spread of sample means. - The requested interval is symmetric around the population mean, so the standardized bounds have equal magnitude.

1. For a uniform distribution from \(58\) to \(82\), the population mean is \(\mu=70\), and the population standard deviation is \(\sigma=\frac{82-58}{\sqrt{12}}=4\sqrt{3}\approx6.928\). 2. Since \(n=36\), the sampling distribution of \(\bar{x}\) is approximately normal with standard deviation \(\frac{4\sqrt{3}}{\sqrt{36}}=\frac{2\sqrt{3}}{3}\approx1.155\). 3. The endpoints \(68\) and \(72\) are each \(2\) units from the mean, corresponding to standardized values approximately \(-1.732\) and \(1.732\). 4. Therefore, \(P(68<\bar{x}<72)\approx0.9167\).

The probability is approximately \(0.9167\).

- Decide which plan produces independent draws and therefore uses the unadjusted sample-mean spread. - For the without-replacement plan, apply the supplied finite-population factor after computing \(\sigma/\sqrt{n}\). - Compare the two correction factors before interpreting which plan has less sampling variability.

1. With replacement, the draws are independent, so the sampling-distribution standard deviation is \(\frac{10}{\sqrt{30}}\approx1.826\). 2. Without replacement, the sampling fraction is large and the finite-population factor is \(\sqrt{\frac{50-30}{50-1}}=\sqrt{\frac{20}{49}}\). 3. The without-replacement standard deviation is \(\frac{10}{\sqrt{30}}\sqrt{\frac{20}{49}}\approx1.166\). 4. Sampling without replacement produces less variability because each selected value reduces the remaining population variation available to later draws.

a) With replacement: standard deviation \(\approx1.826\). b) Without replacement: standard deviation \(\approx1.166\). The without-replacement sampling distribution is less variable.

- First find the probability for a single sample mean. - It may be easier to work with the complementary event that every sample mean stays at or below the cutoff. - The repeated samples are independent.

1. For one sample, \(\bar{x}\) is normal with mean \(50\) and standard deviation \(10/\sqrt{25}=2\). 2. The cutoff \(54\) corresponds to \(z=(54-50)/2=2\), so \(P(\bar{x}>54)\approx0.02275\). 3. The probability that none of the \(20\) independent sample means exceeds \(54\) is \((1-0.02275)^{20}\approx0.63112\). 4. Therefore, the probability that at least one exceeds \(54\) is \(1-0.63112\approx0.36888\).

The probability is approximately \(0.3689\).

- First find the sampling distribution of one sample mean. - Then treat the independent sample means as the values being averaged at the second stage. - Averaging independent quantities reduces spread while preserving their common center.

1. Each sample mean is centered at \(80\) and has standard deviation \(12/\sqrt{25}=2.4\). 2. The \(16\) sample means are independent because they come from independent random samples. 3. Averaging the \(16\) sample means keeps the center at \(80\). 4. The standard deviation of their average is \(2.4/\sqrt{16}=0.6\), equivalent to \(12/\sqrt{400}\).

The sampling distribution of \(G\) has mean \(80\) and standard deviation \(0.6\).

- List the possible numbers of \(10\)s that can appear in a sample of size \(3\). - Convert each count into a possible sample mean. - The expected value of a distribution does not have to be one of its possible outcomes.

1. If the sample contains \(k\) values equal to \(10\), then \(\bar{x}=10k/3\), so the possible sample means are \(0,\frac{10}{3},\frac{20}{3},10\). 2. Thus, \(5\) is not a possible realized value of \(\bar{x}\). 3. The population mean is \(0(0.5)+10(0.5)=5\). 4. The sample mean is unbiased, so the mean of its sampling distribution is \(5\), even though that center need not itself be a possible outcome.

The possible sample means are \(0,\frac{10}{3},\frac{20}{3},10\), so \(5\) cannot occur. Nevertheless, the sampling distribution is centered at \(E(\bar{x})=\mu=5\).

- Find the expected value of one observation under the actual selection mechanism. - The sample mean reflects the distribution being sampled by the procedure, not merely the listed population composition. - Unbiasedness depends on how the sample is selected.

1. Under the biased sampling method, one selected observation has expected value \(0(0.2)+10(0.8)=8\). 2. The mean of a fixed number of independent draws from this selection mechanism also has expected value \(8\). 3. Therefore, the sampling distribution of \(\bar{x}\) is centered at \(8\), not at the population mean \(5\). 4. The usual unbiasedness of the sample mean depends on a sampling process that represents the population rather than systematically oversampling high values.

The sampling distribution is centered at \(8\). The biased sampling method oversamples the value \(10\), so \(\bar{x}\) is not centered at the population mean \(5\).

- Separate the center of a sampling distribution from its spread. - A simple random sample can remain representative even when it is a large fraction of the population. - The \(10\%\) condition is tied to dependence from sampling without replacement.

1. A simple random sample without replacement still gives a sample mean whose expected value is the population mean. 2. Therefore, \(\bar{x}\) remains an unbiased estimator of \(\mu\) even though the sampling fraction is large. 3. The failed \(10\%\) condition affects the approximation that treats sampled observations as independent and uses the usual \(\sigma/\sqrt{n}\) spread formula without a finite-population adjustment. 4. Thus, the issue is sampling variability and dependence, not bias in the center.

The sample mean is still centered at the population mean. Failing the \(10\%\) condition affects the usual independence and standard-error approximation, not the unbiasedness of \(\bar{x}\).

- List the ordered pairs that can occur when sampling with replacement. - Different ordered pairs can produce the same sample mean, so combine equal outcomes when assigning probabilities. - Check the sampling-distribution center and spread against the population mean and the sample-mean spread rule.

1. There are \(9\) equally likely ordered samples of size \(2\). 2. The possible sample means and probabilities are \(P(\bar{x}=2)=\frac{1}{9}\), \(P(\bar{x}=4)=\frac{2}{9}\), \(P(\bar{x}=6)=\frac{3}{9}\), \(P(\bar{x}=8)=\frac{2}{9}\), and \(P(\bar{x}=10)=\frac{1}{9}\). 3. The mean of this sampling distribution is \(6\), equal to the population mean. 4. The population variance is \(\frac{(2-6)^2+(6-6)^2+(10-6)^2}{3}=\frac{32}{3}\), so the standard deviation of \(\bar{x}\) is \(\sqrt{\frac{32/3}{2}}=\frac{4}{\sqrt{3}}\approx2.309\).

\(P(\bar{x}=2)=\frac{1}{9}\), \(P(\bar{x}=4)=\frac{2}{9}\), \(P(\bar{x}=6)=\frac{3}{9}\), \(P(\bar{x}=8)=\frac{2}{9}\), and \(P(\bar{x}=10)=\frac{1}{9}\). The sampling distribution has mean \(6\) and standard deviation \(\frac{4}{\sqrt{3}}\approx2.309\).

- List all equally likely subsets of two values from the population. - Several different samples can have the same sample mean, so combine their probabilities. - Because the sample is a large fraction of this tiny population, sampling without replacement reduces the spread compared with independent draws.

1. The \(6\) equally likely samples are the unordered pairs from the four population values. 2. Their sample means are \(3,5,7,7,9,11\). 3. Thus, \(P(\bar{x}=3)=\frac{1}{6}\), \(P(\bar{x}=5)=\frac{1}{6}\), \(P(\bar{x}=7)=\frac{2}{6}\), \(P(\bar{x}=9)=\frac{1}{6}\), and \(P(\bar{x}=11)=\frac{1}{6}\). 4. The sampling-distribution mean is \(7\), equal to the population mean. 5. The sampling-distribution standard deviation is \(\sqrt{\frac{20}{2}}\sqrt{\frac{4-2}{4-1}}=\sqrt{\frac{20}{3}}\approx2.582\), including the finite-population correction for sampling without replacement.

\(P(\bar{x}=3)=\frac{1}{6}\), \(P(\bar{x}=5)=\frac{1}{6}\), \(P(\bar{x}=7)=\frac{1}{3}\), \(P(\bar{x}=9)=\frac{1}{6}\), and \(P(\bar{x}=11)=\frac{1}{6}\). The mean is \(7\), and the standard deviation is approximately \(2.582\).

- Treat the two possible sample sizes as two different sampling distributions first. - Their centers are the same, but their spreads are not. - When equally mixing distributions with the same center, compare their variances rather than averaging their sample sizes.

1. The sampling-distribution standard deviation must be calculated for each actual sample size, not from the average sample size. 2. For \(n=25\), the standard deviation is \(20/\sqrt{25}=4\). For \(n=100\), it is \(20/\sqrt{100}=2\). 3. Both component distributions have the same mean \(100\), so the variance of the equally weighted mixture is the average of their variances: \(\frac{4^2+2^2}{2}=10\). 4. The combined distribution therefore has standard deviation \(\sqrt{10}\approx3.162\), not \(20/\sqrt{62.5}\approx2.530\).

Averaging the sample sizes does not produce the correct sampling variability. The combined distribution has standard deviation \(\sqrt{10}\approx3.162\).

- Each sample mean should contribute in proportion to how many observations it represents. - After writing the combined statistic as a weighted sum, track how independent sources of variability combine. - Check whether the result agrees with what you would expect from one sample containing all \(100\) observations.

1. The two sample means should not receive equal weight because the second sample contains three times as many observations as the first. 2. The combined mean is \(\bar{x}=\frac{25}{100}\bar{x}_1+\frac{75}{100}\bar{x}_2=0.25\bar{x}_1+0.75\bar{x}_2\). 3. Its sampling-distribution mean is \(0.25(60)+0.75(60)=60\). 4. Because the two samples are independent, its variance is \((0.25)^2\frac{10^2}{25}+(0.75)^2\frac{10^2}{75}=1\). 5. Therefore, the sampling-distribution standard deviation is \(1\), the same as for a single random sample of size \(100\).

The correct combined mean is \(0.25\bar{x}_1+0.75\bar{x}_2\). Its sampling distribution has mean \(60\) and standard deviation \(1\).

- First identify the sampling-distribution spread for samples of size \(25\). - Convert the stated lower-tail probability to the supplied standardized location. - Work backward from the standardized sample-mean value to the unknown center.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(\mu\) and standard deviation \(10/\sqrt{25}=2\). 2. The given probability means that \(47\) has standardized value approximately \(-0.842\). 3. Therefore, \(\frac{47-\mu}{2}=-0.842\). 4. Solving gives \(47-\mu=-1.684\), so \(\mu\approx48.684\).

The population mean is approximately \(48.68\).