Free math worksheets

- Find the balance-point summary and the middle-position summary separately. - For the mean, divide the sum by the number of observations; for the median, use the ordered middle position.

1. The sum is \(60\), so the mean is \(\frac{60}{5}=12\). 2. The ordered middle value is \(12\).

Mean: \(12\). Median: \(12\).

- Interpret each statistic according to what it measures. - Equal means imply equal centers under that measure; compare standard deviations for consistency.

1. The equal means show that the samples have the same measured center. 2. The larger standard deviation for Sample B indicates greater typical distance from the mean. 3. Sample B is more variable.

The samples have the same mean, but Sample B has greater variability.

- Use the defining relationship between the two spread measures. - Variance is the square of standard deviation, so keep the units squared.

1. The sample variance is the square of the sample standard deviation. 2. \(s^2=(7.4)^2=54.76\).

The sample variance is \(54.76\).

- A percentile describes relative position within a distribution. - Separate rank information from the measurement scale of the score. - Ask what numerical details would require the original score distribution.

1. A \(72\)nd-percentile result means approximately \(72\%\) of the observations are at or below the student's score. 2. It does not mean the student answered \(72\%\) of the questions correctly. 3. It does not identify the numerical score or the number of points separating the student from another percentile.

Approximately \(72\%\) of scores are at or below the student's score. The percentile rank does not give the percent correct or the numerical score.

- Standard deviation measures distance from the mean. - Determine when a sum of nonnegative squared distances can equal zero. - Use the stated mean to identify the common value.

1. Standard deviation \(0\) means every squared deviation from the mean is \(0\). 2. Therefore every observation has deviation \(0\) from \(7\). 3. Every observation must equal \(7\).

Every observation equals \(7\).

- Order the values and separate the lower and upper halves. - After finding the quartiles, compute range from the extremes and IQR from the quartile difference.

1. The median is \(\frac{9+10}{2}=9.5\). 2. The lower-half median is \(Q_1=7\), and the upper-half median is \(Q_3=13\). 3. The range is \(17-4=13\). 4. The IQR is \(13-7=6\).

\(Q_1=7\), median \(=9.5\), \(Q_3=13\), range \(=13\), IQR \(=6\).

- Track deviations from the mean and remember that the requested spread is for a sample. - Square each deviation, divide their sum by \(n-1\), and take the square root only at the end.

1. The sample mean is \(\bar x=7.0\). 2. The sum of squared deviations is \(46.0\). 3. The sample variance is \(\frac{46.0}{5-1}=11.500\). 4. The sample standard deviation is \(s\approx 3.39\).

\(s\approx 3.39\).

- Work backward from the total implied by the mean. - Use \(\text{total}=n\bar{x}\), then compare that required total with the sum of the known values.

1. The required total is \(5\cdot 16=80\). 2. The known values total \(63\). 3. The missing value is \(80-63=17\).

The missing value is \(17\).

- Consider how scaling all data values changes both positions and distances. - A positive scale factor multiplies both locations and distances by that factor.

1. Multiplying every value by a positive constant multiplies all location and spread measures by that constant. 2. Therefore, the mean, median, range, IQR, and standard deviation are each multiplied by \(2.54\).

Each listed statistic is multiplied by \(2.54\).

- Translate the distance condition into lower and upper boundaries. - Build the interval \(\mu\pm2\sigma\) and classify values by whether they fall outside it.

1. Two standard deviations is \(2\cdot 6=12\). 2. The interval within two standard deviations is \([38,62]\). 3. Values \(37\) and \(64\) lie outside; \(41\) and \(62\) do not.

Potential outliers: \(37\) and \(64\).

- Identify the skewness and the presence of extreme values. - Compare the resistant pair median and IQR with the nonresistant pair mean and standard deviation.

1. Extreme high prices can greatly affect the mean and standard deviation. 2. The median and IQR are resistant to those extreme values. 3. Therefore, median and IQR are more appropriate.

Median and IQR, because they are resistant to the high-price outliers.

- Recover the original total from the old mean and sample size. - After adding the new observation, update both the total and the number of observations.

1. The original sum is \(9\cdot14=126\). 2. The new sum is \(126+24=150\). 3. The new mean is \(\frac{150}{10}=15\).

The new mean is \(15\).

- Convert the original mean to a total, adjust the total, and update the sample size. - Remove the specified observation from the total and reduce the number of observations by one.

1. The original sum is \(12\cdot18.5=222\). 2. The remaining sum is \(222-24=198\). 3. The new mean is \(\frac{198}{11}=18\).

The new mean is \(18\).

- Use the number of ordered values to locate the middle position or positions. - Because the list has an even number of terms, average the two central ordered values.

1. There are \(20\) values, so the median is the mean of the \(10\)th and \(11\)th values. 2. Those values are \(20\) and \(22\). 3. The median is \(\frac{20+22}{2}=21\).

The median is \(21\).

- Identify which data positions each spread measure uses. - Range depends on the extremes, while IQR depends only on the two quartiles.

1. The range depends directly on the maximum and minimum, so a much larger maximum increases the range because the minimum is the same. 2. The IQR depends only on \(Q_1\) and \(Q_3\). 3. Since the quartiles are the same, the IQR remains unchanged.

The range must increase; the IQR remains unchanged.

- Relate the direction of the mean–median difference to the likely tail. - An extreme high tail pulls the mean upward more strongly than it moves the median.

1. The mean is substantially greater than the median, consistent with right skew caused by high values. 2. The median is resistant to the extreme maximum, while the mean is not. 3. The median is the more resistant center.

The data are likely right-skewed, and the median is the more resistant measure of center.

- Convert the percentile to a decimal and apply the stated nearest-rank rule to the sample size. - Use the resulting ordered position without interpolating between observations.

1. Here, \(p=0.65\) and \(n=20\), so the rank is \(r=\lceil0.65\cdot20\rceil=13\). 2. The \(13\)th ordered value is \(16\). 3. Therefore, the \(65\)th percentile is \(16\,\text{minutes}\). In this sample, \(65\%\) of the delivery times are at or below \(16\,\text{minutes}\).

The \(65\)th percentile is \(16\,\text{minutes}\); \(65\%\) of the sampled delivery times are at or below this value.

- Identify whether the data are described as a sample or a population. - Use the appropriate degrees of freedom in the variance calculation. - Standard deviation is expressed in the original unit, not squared units.

1. The sample variance is \(s^2=\frac{180}{10-1}=20\). 2. The sample standard deviation is \(s=\sqrt{20}\approx4.472\).

The sample variance is \(20\), and the sample standard deviation is \(\sqrt{20}\approx4.472\).

- Use the stated percentile convention rather than a different software rule. - Calculate the position before reading the ordered list. - An integer position points directly to one observation.

1. Here \(n=11\), so \(L=0.75(11+1)=9\). 2. The ninth ordered value is \(15\).

The \(75\)th percentile is \(15\).

- Compute the full-data average first. - Remove exactly the observations specified before recomputing. - Compare which data values have the greatest influence on the two results.

1. The ordinary mean is \(\frac{2+3+4+5+6+7+50}{7}=\frac{77}{7}=11\). 2. After removing \(2\) and \(50\), the remaining total is \(3+4+5+6+7=25\), so the trimmed mean is \(\frac{25}{5}=5\). 3. The ordinary mean is much larger because the high value \(50\) has strong influence; trimming reduces that influence.

The ordinary mean is \(11\), and the trimmed mean is \(5\).

- Use the defining relationship among mean, sample size, and sum. - Compute the implied total from one statement. - Compare it with the independently reported total.

1. A mean of \(8\) for \(6\) observations requires a total of \(6\cdot8=48\). 2. The stated total \(43\) would instead give mean \(\frac{43}{6}\approx7.167\). 3. Therefore the two reported summaries are inconsistent.

No. A mean of \(8\) requires sum \(48\), not \(43\).

- Match the summary pair to the distribution's shape and unusual features. - Consider which center is naturally linked to standard deviation. - Resistant summaries are most valuable when extremes or skew are important.

1. For an approximately symmetric distribution without strong outliers, the mean represents the center well. 2. Standard deviation summarizes typical distance from the mean and uses all observations. 3. Therefore mean with standard deviation is the more natural pair; median with interquartile range would also be valid but is especially useful for skewed data or data with outliers.

Use the mean and standard deviation because the distribution is approximately symmetric and has no strong outliers.

- Separate measures of location from measures based on differences between values. - Adding a constant shifts every ordered position by the same amount. - Measures built from differences between data values are unaffected by a common shift.

1. Adding the same constant to every value increases the mean and median by \(273.15\). 2. Differences between values do not change. 3. Therefore, the IQR, range, and standard deviation remain unchanged.

Mean and median each increase by \(273.15\); IQR, range, and standard deviation do not change.

- Find the middle half first, then compare every extreme value with the resulting boundaries. - Use the stated quartile convention before calculating the IQR. - Form the lower and upper fences, then compare each extreme value using strict inequalities.

1. Excluding the overall median, \(Q_1=7.5\) and \(Q_3=12.5\). 2. The IQR is \(5\). 3. The fences are \(0\) and \(20\). 4. Only \(30>20\), so \(30\) is a potential outlier.

The fences are \(0\) and \(20\). The value \(30\) is a potential outlier.

- Use each class midpoint as the representative value for that interval. - Multiply each midpoint by its frequency and add the products. - Divide by the total frequency; the result is an estimate because the exact values inside the classes are unknown.

1. The class midpoints are \(5\), \(15\), \(25\), and \(35\) minutes. 2. The weighted sum is \(5\cdot4+15\cdot9+25\cdot5+35\cdot2=350\). 3. The total frequency is \(4+9+5+2=20\). 4. The estimated mean is \(\frac{350}{20}=17.5\) minutes.

The estimated mean commute time is \(17.5\,\text{minutes}\).

- Separate the effects of shifting and scaling. - A vertical shift changes location but not distances between observations. - Multiplication by a constant scales standard deviation by the constant’s absolute value.

1. Subtracting \(7\) does not affect spread. 2. Multiplying by \(4\) multiplies standard deviation by \(4\). 3. The new standard deviation is \(4\cdot3.2=12.8\).

The sample standard deviation of \(Y\) is \(12.8\).

- Compare which observations influence each statistic. - Hold the central ordered positions fixed while imagining different values in the tails. - Ask which measure uses every distance from the mean.

1. The median and IQR depend only on central ordered positions. 2. Standard deviation uses every value and its distance from the mean. 3. Values outside the middle \(50\%\) can differ while the median and quartiles remain fixed, producing different standard deviations.

Yes. The same middle \(50\%\) can coexist with different tail values, which changes standard deviation but not the median or IQR.

- Use frequencies as weights when finding the total. - Locate the two middle positions from cumulative frequencies. - The mode is tied to the greatest frequency, not the greatest value.

1. The sample size is \(2+5+4+1=12\), and the total is \(1\cdot2+2\cdot5+3\cdot4+4\cdot1=28\). 2. The mean is \(\frac{28}{12}=\frac{7}{3}\approx2.33\). 3. The sixth and seventh ordered observations are both \(2\), so the median is \(2\). 4. The greatest frequency is \(5\) at value \(2\), so the mode is \(2\).

Mean: \(\frac{7}{3}\approx2.33\); median: \(2\); mode: \(2\).

- Convert each group mean into the corresponding group total. - Combine totals and sample sizes separately. - Equal weighting of group means is valid only for equal group sizes.

1. The first group total is \(12\cdot18=216\), and the second group total is \(8\cdot25=200\). 2. The combined total is \(216+200=416\). 3. The combined mean is \(\frac{416}{20}=20.8\). 4. The two group means cannot be averaged directly because the groups have different sizes.

The combined mean is \(20.8\).

- Relate each deviation to the reference value of \(30\) minutes. - Use the number of observations to reconstruct the original total. - After removing one observation, update both the total and the sample size.

1. The total of the eight times is \(8\cdot30+(-12)=228\) minutes. 2. The mean is \(\frac{228}{8}=28.5\) minutes. 3. Removing the \(38\)-minute time leaves a total of \(228-38=190\) minutes. 4. The remaining mean is \(\frac{190}{7}\approx27.14\) minutes.

a) \(28.5\) minutes b) \(\frac{190}{7}\approx27.14\) minutes

- Recover the original total from the reported mean. - Adjust the total by the difference between the incorrect and correct entries. - Keep the sample size unchanged.

1. The reported total is \(10\cdot50.4=504\). 2. Correcting the entry reduces the total by \(83-38=45\), giving \(504-45=459\). 3. The corrected mean is \(\frac{459}{10}=45.9\).

The corrected mean is \(45.9\).

- Apply the entire linear rule to the measure of center. - Standard deviation uses distances, so the sign of a scale factor does not make it negative. - A constant shift changes location but not spread.

1. The transformed mean is \(-3\cdot7+10=-11\). 2. Multiplying by \(-3\) multiplies all distances from the mean by \(3\); adding \(10\) does not change distances. 3. The transformed standard deviation is \(3\cdot4=12\).

The transformed mean is \(-11\), and the transformed standard deviation is \(12\).

- Consider how a value equal to the current mean changes the total. - Determine its contribution to squared deviation. - Update the sample degrees of freedom after increasing the sample size.

1. Adding a value equal to the original mean leaves the mean at \(10\). 2. The added observation has deviation \(0\), so the sum of squared deviations remains \(100\). 3. With \(5\) observations, the new sample variance is \(\frac{100}{5-1}=25\). 4. The new sample standard deviation is \(\sqrt{25}=5\).

The new mean is \(10\), and the new sample standard deviation is \(5\).

- The squared deviations are the same in both calculations. - The distinction is the divisor used for a population versus a sample. - Take the square root only after finding each variance.

1. The mean is \(\frac{2+4+6+8}{4}=5\). 2. The squared deviations sum to \((2-5)^2+(4-5)^2+(6-5)^2+(8-5)^2=20\). 3. The population variance is \(\frac{20}{4}=5\), and the population standard deviation is \(\sqrt{5}\approx2.236\). 4. The sample variance is \(\frac{20}{4-1}=\frac{20}{3}\approx6.667\), and the sample standard deviation is \(\sqrt{\frac{20}{3}}\approx2.582\).

Population: variance \(5\), standard deviation \(\sqrt{5}\approx2.236\). Sample: variance \(\frac{20}{3}\approx6.667\), standard deviation \(\sqrt{\frac{20}{3}}\approx2.582\).

- Use a defining property of deviations from the mean. - Find the missing deviation before calculating spread. - Apply the sample divisor to the squared deviations.

1. Deviations from the mean sum to \(0\), so \(-4-1+2+3+x=0\), giving \(x=0\). 2. The sum of squared deviations is \(16+1+4+9+0=30\). 3. The sample variance is \(\frac{30}{5-1}=7.5\). 4. The sample standard deviation is \(\sqrt{7.5}\approx2.739\).

\(x=0\), and the sample standard deviation is \(\sqrt{7.5}\approx2.739\).

- Locate the overall median before forming the halves. - Follow each stated convention literally. - Compare the resulting lower- and upper-half medians.

1. The overall median is \(5\). 2. Excluding the median gives lower half \(1,2,3,4\) and upper half \(6,7,8,9\), so \(Q_1=2.5\) and \(Q_3=7.5\). 3. Including the median gives lower half \(1,2,3,4,5\) and upper half \(5,6,7,8,9\), so \(Q_1=3\) and \(Q_3=7\). 4. Different accepted conventions can produce different quartiles for the same small data set, so the method should be stated.

a) \(Q_1=2.5\), \(Q_3=7.5\). b) \(Q_1=3\), \(Q_3=7\).

- Translate the known observation and its deviation into an equation. - Once the center is known, each deviation locates one observation relative to it. - Verify that the recovered observations average to the stated center.

1. A deviation equals observation minus mean, so \(12-\bar{x}=-3\). 2. Therefore \(\bar{x}=15\). 3. Add each deviation to \(15\) to obtain \(12,14,15,16,18\).

The mean is \(15\), and the observations are \(12,14,15,16,18\).

- Square the standard deviation to obtain the variance. - Relate the variance to the given squared-deviation total. - Remember the sample degrees of freedom.

1. The sample variance is \(s^2=3^2=9\). 2. Using \(s^2=\frac{72}{n-1}\), solve \(9=\frac{72}{n-1}\). 3. Then \(n-1=8\), so \(n=9\).

The sample size is \(9\).

- Distinguish the mean of the squares from the square of the mean. - Population variance is the difference between those two quantities. - The root mean square is the square root of the stated mean square.

1. The mean square is \(31\), and the square of the mean is \(5^2=25\). 2. The population variance is \(31-25=6\). 3. The population standard deviation is \(\sqrt{6}\approx2.449\). 4. The root mean square is \(\sqrt{31}\approx5.568\). 5. The relationship is \(\text{RMS}^2=\mu^2+\sigma^2\), since \(31=25+6\).

Population variance: \(6\) Population standard deviation: \(\sqrt{6}\approx2.449\) Root mean square: \(\sqrt{31}\approx5.568\)

- Track both the sum of squared deviations and the number of observations. - Compare the denominators used for population and sample variability. - Repeating every observation does not change the relative frequency distribution.

1. Both samples have mean \(5\). 2. For \(2,5,8\), the sum of squared deviations is \(18\). The population standard deviation is \(\sqrt{\frac{18}{3}}=\sqrt{6}\), and the sample standard deviation is \(\sqrt{\frac{18}{2}}=3\). 3. Duplicating every value doubles the sum of squared deviations to \(36\) and doubles the sample size to \(6\). 4. The population standard deviation remains \(\sqrt{\frac{36}{6}}=\sqrt{6}\). The sample standard deviation becomes \(\sqrt{\frac{36}{5}}=\sqrt{7.2}\approx2.683\). 5. Population variance divides by \(n\), so equal duplication leaves it unchanged. Sample variance divides by \(n-1\), so duplication changes the correction factor.

Both means are \(5\), and both population standard deviations are \(\sqrt{6}\approx2.449\). The sample standard deviation changes from \(3\) to \(\sqrt{7.2}\approx2.683\) because the denominator changes from \(n-1=2\) to \(n-1=5\).

- Compare deviations from the stated reference value with deviations from the mean. - Account for the distance between the reference value and the mean across all observations. - Use the sample-size correction only after finding the squared deviations about the mean.

1. The mean is \(24-20=4\) units above the reference value \(20\). 2. Use \(\sum (x_i-20)^2=\sum (x_i-\bar{x})^2+n(\bar{x}-20)^2\). 3. The sum of squared deviations from the sample mean is \(900-15\cdot4^2=660\). 4. The sample variance is \(s^2=\frac{660}{14}=\frac{330}{7}\approx47.143\). 5. The sample standard deviation is \(s=\sqrt{\frac{330}{7}}\approx6.866\).

Sample variance: \(\frac{330}{7}\approx47.143\) Sample standard deviation: \(\sqrt{\frac{330}{7}}\approx6.866\)

- List pair choices systematically so none are repeated or omitted. - Count how often each original observation appears among all pairs. - Compare the effective weight of each original value in the two means.

1. The six pairwise averages are \(5.5,7,8.5,8.5,10,11.5\). 2. Their mean is \(\frac{5.5+7+8.5+8.5+10+11.5}{6}=\frac{51}{6}=8.5\). 3. The original mean is \(\frac{4+7+10+13}{4}=8.5\). 4. Each original observation appears in exactly three pairs. In the sum of all pairwise averages, each value therefore contributes \(\frac{3}{2}\) times. Dividing by the six pairs gives the same weight, \(\frac{1}{4}\), for each original observation.

The mean of the six pairwise averages is \(8.5\), equal to the original mean. Each original observation receives equal total weight across all pairs.

- Write both the total frequency and the weighted sum in terms of the unknown. - Use the definition of the mean to form an equation. - Recheck the largest frequency after solving.

1. The total frequency is \(2+4+x+1=7+x\). 2. The weighted total is \(0\cdot2+1\cdot4+2x+3\cdot1=7+2x\). 3. Set \(\frac{7+2x}{7+x}=1.5\). Then \(7+2x=10.5+1.5x\), so \(x=7\). 4. Value \(2\) has frequency \(7\), the greatest frequency, so the mode is \(2\).

\(x=7\), and the mode is \(2\).

- To minimize the total, place every observation as low as its group permits. - To maximize the total, place every observation as high as its group permits. - Keep the frequency of each group fixed in both extreme cases.

1. The smallest total occurs when every observation is at the lower endpoint of its group: \(4\cdot0+6\cdot10+2\cdot20=100\). 2. The smallest possible mean is \(\frac{100}{12}=\frac{25}{3}\approx8.33\). 3. The largest total occurs when every observation is at the upper endpoint of its group: \(4\cdot9+6\cdot19+2\cdot29=208\). 4. The largest possible mean is \(\frac{208}{12}=\frac{52}{3}\approx17.33\).

Smallest possible mean: \(\frac{25}{3}\approx8.33\) Largest possible mean: \(\frac{52}{3}\approx17.33\)

- Compare each proposed reference value with the mean. - Shifting the reference away from the mean adds the same squared offset for every observation. - Use the given squared-deviation total as the baseline.

1. For any reference value \(c\), \(\sum (x_i-c)^2=\sum (x_i-\bar{x})^2+n(\bar{x}-c)^2\). 2. For \(c=10\), the total is \(80+8(12-10)^2=80+32=112\). 3. For \(c=15\), the total is \(80+8(12-15)^2=80+72=152\). 4. The total at \(c=12\) is the given \(80\), which is smaller than \(112\) and \(152\). 5. The mean minimizes the sum of squared distances.

\(\sum (x_i-10)^2=112\), and \(\sum (x_i-15)^2=152\). The reference value \(12\), the mean, gives the smallest total squared distance.

- Express each leave-one-out mean using the full total and one omitted value. - Add all six equations and count how often each original observation appears. - Recover each omitted value from its corresponding reduced mean.

1. Let the full total be \(S\). Each leave-one-out mean equals \(\frac{S-x_i}{5}\). 2. Adding the six leave-one-out means counts the full total exactly once, so \(12+13+14+15+16+17=S=87\). 3. The full mean is \(\frac{87}{6}=14.5\). 4. For each leave-one-out mean \(m_i\), the omitted observation is \(x_i=87-5m_i\). 5. The omitted values are \(27,22,17,12,7,2\), so the ordered data are \(2,7,12,17,22,27\).

The complete mean is \(14.5\). The six observations are \(2,7,12,17,22,27\).

- Recover both the original sum and the original sum of squares from the given summaries. - A corrected observation changes its contribution to each total differently. - Recompute variability around the corrected mean rather than the original mean.

1. The original sum is \(6\cdot20=120\). 2. From \(s^2=\frac{\sum x_i^2-n\bar{x}^2}{n-1}\), the original sum of squares is \(5\cdot4^2+6\cdot20^2=2480\). 3. Correcting the observation gives a new sum of \(120-18+24=126\), so the corrected mean is \(\frac{126}{6}=21\). 4. The corrected sum of squares is \(2480-18^2+24^2=2732\). 5. The corrected sum of squared deviations is \(2732-6\cdot21^2=86\). 6. The corrected sample standard deviation is \(\sqrt{\frac{86}{5}}=\sqrt{17.2}\approx4.147\).

Corrected mean: \(21\) Corrected sample standard deviation: \(\sqrt{17.2}\approx4.147\)

- Use the mean to obtain the sum of the two unknown observations. - Convert sample variance into a sum of squared deviations. - Relate the sum and sum of squares of two numbers to their product.

1. The mean condition gives \(2+4+x+y=4\cdot5\), so \(x+y=14\). 2. A sample variance of \(\frac{20}{3}\) gives a sum of squared deviations of \(3\cdot\frac{20}{3}=20\). 3. Using \(\sum x_i^2=\sum (x_i-\bar{x})^2+n\bar{x}^2\), the sum of squares is \(20+4\cdot5^2=120\). 4. Therefore, \(2^2+4^2+x^2+y^2=120\), so \(x^2+y^2=100\). 5. Since \((x+y)^2=x^2+y^2+2xy\), \(14^2=100+2xy\), giving \(xy=48\). 6. The numbers with sum \(14\) and product \(48\) are \(6\) and \(8\). Thus \(x=6\) and \(y=8\).

\(x=6\) and \(y=8\).

- Combine group totals to find the overall center. - Account for variation within each group and separation between group means. - Use the total sample degrees of freedom only after combining squared deviations.

1. The combined mean is \(\bar{x}=\frac{5\cdot10+7\cdot14}{12}=\frac{37}{3}\approx12.333\). 2. The within-group squared-deviation total is \((5-1)2^2+(7-1)3^2=70\). 3. The between-group contribution is \(5\left(10-\frac{37}{3}\right)^2+7\left(14-\frac{37}{3}\right)^2=\frac{140}{3}\). 4. The combined squared-deviation total is \(70+\frac{140}{3}=\frac{350}{3}\). 5. The combined sample variance is \(\frac{350/3}{12-1}=\frac{350}{33}\approx10.606\), so the sample standard deviation is \(\sqrt{\frac{350}{33}}\approx3.257\).

The combined mean is \(\frac{37}{3}\approx12.333\), and the combined sample standard deviation is \(\sqrt{\frac{350}{33}}\approx3.257\).