Question

find the variance for the given data. round your answer to one more decimal place than the original data. 7) compute the variance. the owner of a small manufacturing plant employs six people. as part of their personnel file, she asked each one to record to the nearest one-tenth of a mile the distance they travel one way from home to work. the six distances are listed below: 17 18 35 31 21 50

Explanation:

Step1: Find the mean of the data

The data set is \(17, 18, 35, 31, 21, 50\). The number of data points \(n = 6\).
The mean \(\bar{x}=\frac{17 + 18+35 + 31+21 + 50}{6}\)
First, calculate the sum: \(17+18 = 35\), \(35+35=70\), \(70 + 31=101\), \(101+21 = 122\), \(122+50=172\)
So, \(\bar{x}=\frac{172}{6}\approx28.6667\)

Step2: Calculate the squared differences from the mean

For each data point \(x_i\), calculate \((x_i-\bar{x})^2\):
- For \(x_1 = 17\): \((17 - 28.6667)^2=(- 11.6667)^2\approx136.1113\)
- For \(x_2 = 18\): \((18 - 28.6667)^2=(-10.6667)^2\approx113.7779\)
- For \(x_3 = 35\): \((35 - 28.6667)^2=(6.3333)^2\approx40.1111\)
- For \(x_4 = 31\): \((31 - 28.6667)^2=(2.3333)^2\approx5.4443\)
- For \(x_5 = 21\): \((21 - 28.6667)^2=(-7.6667)^2\approx58.7779\)
- For \(x_6 = 50\): \((50 - 28.6667)^2=(21.3333)^2\approx455.1111\)

Step3: Find the sum of the squared differences

Sum of \((x_i - \bar{x})^2\) is \(136.1113+113.7779 + 40.1111+5.4443+58.7779+455.1111\)
\(136.1113+113.7779=249.8892\); \(249.8892 + 40.1111=290.0003\); \(290.0003+5.4443 = 295.4446\); \(295.4446+58.7779=354.2225\); \(354.2225 + 455.1111=809.3336\)

Step4: Calculate the variance

Since this is a sample (the six people are a sample of employees), the sample variance \(s^2=\frac{\sum(x_i - \bar{x})^2}{n - 1}\)
Here, \(n=6\), so \(n - 1=5\)
\(s^2=\frac{809.3336}{5}\approx161.8667\)
If we consider it as a population (though in this case, since it's all employees, population variance \(\sigma^2=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{809.3336}{6}\approx134.8889\), but usually, for a small group like this, if it's the entire population of employees, we use population variance. Wait, the problem says "the owner of a small manufacturing plant employs six people" so it's the entire population. Let's recalculate with \(n = 6\)
\(\sigma^2=\frac{809.3336}{6}\approx134.8889\approx134.89\) (but wait, let's check the steps again)

Wait, let's recalculate the mean correctly: \(17+18 = 35\), \(35+35 = 70\), \(70+31=101\), \(101 + 21=122\), \(122+50 = 172\). Mean \(\bar{x}=\frac{172}{6}=28.666\cdots\)

Squared differences:
- \(17-28.666\cdots=- 11.666\cdots\), square is \(136.111\cdots\)
- \(18 - 28.666\cdots=-10.666\cdots\), square is \(113.777\cdots\)
- \(35-28.666\cdots = 6.333\cdots\), square is \(40.111\cdots\)
- \(31-28.666\cdots=2.333\cdots\), square is \(5.444\cdots\)
- \(21-28.666\cdots=-7.666\cdots\), square is \(58.777\cdots\)
- \(50 - 28.666\cdots=21.333\cdots\), square is \(455.111\cdots\)

Sum of squares: \(136.111+113.777 = 249.888\); \(249.888+40.111=290\); \(290 + 5.444=295.444\); \(295.444+58.777=354.221\); \(354.221+455.111 = 809.332\)

Population variance (since it's all 6 employees) is \(\frac{809.332}{6}\approx134.8887\approx134.89\) (but let's check the problem statement: "the owner of a small manufacturing plant employs six people" so it's the entire population. So variance (population) is \(\sigma^2=\frac{\sum(x_i - \bar{x})^2}{N}\), where \(N = 6\)

Wait, but maybe the problem considers it as a sample? Wait, the problem says "compute the variance" without specifying, but in statistics, if it's the entire population, we use population variance, if it's a sample, sample variance. Since it's all employees (six people), it's population.

Wait, let's recalculate the mean: \(17+18+35+31+21+50 = 172\), mean \(=172/6\approx28.6667\)

Sum of squared deviations:
\((17 - 28.6667)^2=(-11.6667)^2 = 136.1113\)
\((18 - 28.6667)^2=(-10.6667)^2=113.7779\)
\((35 - 28.6667)^2=(6.3333)^2 = 40.1111\)
\((31 - 28.6667)^2=(2.3333)^2=5.4443\)
\((21 - 28.6667)^2=(-7.6667)^2=58.7779\)
\((50 - 28.6667)^2=(21.3333)^2=455.1111\)

Sum: \(136.1113+113.7779 = 249.8892\); \(249.8892+40.1111 = 290.0003\); \(290.0003+5.4443=295.4446\); \(295.4446+58.7779 = 354.2225\); \(354.2225+455.1111 = 809.3336\)

Population variance: \(\frac{809.3336}{6}\approx134.8889\approx134.89\) (if we round to one more decimal place than original data. Original data is to the nearest tenth (one decimal place), so we need to round to two decimal places? Wait, original data: 17, 18, 35, 31, 21, 50 – these are whole numbers, so one decimal place would be, for example, 17.0, 18.0 etc. So the original data has 0 decimal places (since they are whole numbers, but recorded to nearest tenth, so maybe they are considered as having one decimal place: 17.0, 18.0, 35.0, 31.0, 21.0, 50.0). So we need to round the variance to two decimal places (one more than one decimal place).

Wait, let's check again. The problem says "Round your answer to one more decimal place than the original data." The original data is "to the nearest one - tenth of a mile", so each data point has one decimal place (e.g., 17.0, 18.0). So we need to round the variance to two decimal places.

But let's confirm the calculation of variance.

Alternatively, maybe the problem expects sample variance. Let's check both:

Sample variance: \(s^2=\frac{809.3336}{5}\approx161.8667\approx161.87\)

But which is it? The problem says "the owner of a small manufacturing plant employs six people" – so it's the entire population of employees, so population variance.

Wait, let's check the sum of the data again: 17 + 18 = 35; 35+35 = 70; 70+31 = 101; 101+21 = 122; 122+50 = 172. Correct. Mean is 172/6 ≈28.6667. Correct.

Sum of squared deviations: as calculated, 809.3336.

Population variance: 809.3336 / 6 ≈134.8889 ≈134.89 (rounded to two decimal places, since original data has one decimal place (nearest tenth)).

Wait, but maybe I made a mistake in assuming population. Let's see, in statistics, when the data is the entire population, we use \(\sigma^2\), when it's a sample, \(s^2\). Since there are six employees, and we have data from all six, it's population. So variance is \(\frac{\sum(x_i - \bar{x})^2}{N}=\frac{809.3336}{6}\approx134.89\)

Answer:

\(134.89\) (if population) or \(161.87\) (if sample). But since it's all employees, population variance is appropriate. So the answer is approximately \(134.89\) (rounded to two decimal places, one more than the original data's one decimal place (nearest tenth)).