Question

question 5 of 5 (20 points) | question attempt... places. 20.6 33 21.1 58.5 23.7 110.7 30.5 24.1 74.9 60.5 send data to excel part 1 of 3 find the range the range is 90.10 part 1 / 3 part 2 of 3 find the variance. the variance is blank

Explanation:

Step1: Calculate the mean

First, find the sum of the data set: \(20.6 + 33 + 21.1 + 58.5 + 23.7 + 110.7 + 30.5 + 24.1 + 74.9 + 60.5\)
\(= 20.6+33 = 53.6\); \(53.6+21.1 = 74.7\); \(74.7+58.5 = 133.2\); \(133.2+23.7 = 156.9\); \(156.9+110.7 = 267.6\); \(267.6+30.5 = 298.1\); \(298.1+24.1 = 322.2\); \(322.2+74.9 = 397.1\); \(397.1+60.5 = 457.6\)
The number of data points \(n = 10\), so the mean \(\bar{x}=\frac{457.6}{10}=45.76\)

Step2: Calculate the squared differences

For each data point \(x_i\), calculate \((x_i - \bar{x})^2\):
- \((20.6 - 45.76)^2 = (-25.16)^2 = 633.0256\)
- \((33 - 45.76)^2 = (-12.76)^2 = 162.8176\)
- \((21.1 - 45.76)^2 = (-24.66)^2 = 608.1156\)
- \((58.5 - 45.76)^2 = (12.74)^2 = 162.3076\)
- \((23.7 - 45.76)^2 = (-22.06)^2 = 486.6436\)
- \((110.7 - 45.76)^2 = (64.94)^2 = 4217.2036\)
- \((30.5 - 45.76)^2 = (-15.26)^2 = 232.8676\)
- \((24.1 - 45.76)^2 = (-21.66)^2 = 469.1556\)
- \((74.9 - 45.76)^2 = (29.14)^2 = 849.1396\)
- \((60.5 - 45.76)^2 = (14.74)^2 = 217.2676\)

Step3: Sum the squared differences

Sum these squared differences:
\(633.0256 + 162.8176 + 608.1156 + 162.3076 + 486.6436 + 4217.2036 + 232.8676 + 469.1556 + 849.1396 + 217.2676\)
\(= 633.0256+162.8176 = 795.8432\); \(795.8432+608.1156 = 1403.9588\); \(1403.9588+162.3076 = 1566.2664\); \(1566.2664+486.6436 = 2052.91\); \(2052.91+4217.2036 = 6270.1136\); \(6270.1136+232.8676 = 6502.9812\); \(6502.9812+469.1556 = 6972.1368\); \(6972.1368+849.1396 = 7821.2764\); \(7821.2764+217.2676 = 8038.544\)

Step4: Calculate the variance

Since this is a sample (assuming, as it's a sample of data), the sample variance \(s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1}\)
\(n - 1 = 9\), so \(s^2=\frac{8038.544}{9}\approx 893.17\) (if it's a population, we would divide by \(n\), but typically for such data sets, sample variance is used. Let's check both:
Population variance \(\sigma^2=\frac{8038.544}{10}=803.85\). But since the problem says "the data show the traveler spending... for a sample", we use sample variance. Wait, but maybe it's a population? Wait, the problem doesn't specify, but looking at the range calculation (110.7 - 20.6 = 90.1, which matches the given 90.10), so let's confirm the data points: 20.6, 33, 21.1, 58.5, 23.7, 110.7, 30.5, 24.1, 74.9, 60.5. That's 10 data points. If it's a sample, \(n - 1 = 9\), if population, \(n = 10\). Let's recalculate the sum of squared differences again to be precise:

Wait, let's recalculate the sum of \((x_i - \bar{x})^2\) with \(\bar{x}=45.76\):

- \(20.6 - 45.76 = -25.16\), squared: \(25.16^2 = 633.0256\)
- \(33 - 45.76 = -12.76\), squared: \(12.76^2 = 162.8176\)
- \(21.1 - 45.76 = -24.66\), squared: \(24.66^2 = 608.1156\)
- \(58.5 - 45.76 = 12.74\), squared: \(12.74^2 = 162.3076\)
- \(23.7 - 45.76 = -22.06\), squared: \(22.06^2 = 486.6436\)
- \(110.7 - 45.76 = 64.94\), squared: \(64.94^2 = 4217.2036\)
- \(30.5 - 45.76 = -15.26\), squared: \(15.26^2 = 232.8676\)
- \(24.1 - 45.76 = -21.66\), squared: \(21.66^2 = 469.1556\)
- \(74.9 - 45.76 = 29.14\), squared: \(29.14^2 = 849.1396\)
- \(60.5 - 45.76 = 14.74\), squared: \(14.74^2 = 217.2676\)

Now sum these:

\(633.0256 + 162.8176 = 795.8432\)

\(795.8432 + 608.1156 = 1403.9588\)

\(1403.9588 + 162.3076 = 1566.2664\)

\(1566.2664 + 486.6436 = 2052.91\)

\(2052.91 + 4217.2036 = 6270.1136\)

\(6270.1136 + 232.8676 = 6502.9812\)

\(6502.9812 + 469.1556 = 6972.1368\)

\(6972.1368 + 849.1396 = 7821.2764\)

\(7821.2764 + 217.2676 = 8038.544\)

Now, if it's a sample (since it's a "sample of the data"), we use \(n - 1 = 9\):

\(s^2=\frac{8038.544}{9}\approx 893.17\)

If it's a population, \(\sigma^2=\frac{8038.544}{10}=803.85\)

Wait, but let's check the range: max is 110.7, min is 20.6, 110.7 - 20.6 = 90.1, which matches the given 90.10 (rounded to two decimals). Now, let's see the problem statement: "the data show the traveler spending in billions of dollars for a recent year for a sample of..." So it's a sample, so we use sample variance.

Wait, but maybe I made a mistake in the mean? Let's recalculate the mean:

Sum: 20.6 + 33 = 53.6; +21.1 = 74.7; +58.5 = 133.2; +23.7 = 156.9; +110.7 = 267.6; +30.5 = 298.1; +24.1 = 322.2; +74.9 = 397.1; +60.5 = 457.6. Yes, sum is 457.6, n=10, mean=45.76. Correct.

So sample variance is \(\frac{8038.544}{9}\approx 893.17\) (rounded to two decimal places). If it's population variance, it's 803.85. But since it's a sample (as per "sample of the data"), we use sample variance.

Wait, but let's check with another approach. Maybe the problem is using population variance? Let's see:

If we consider it a population, variance is \(\frac{\sum (x_i - \bar{x})^2}{n}=\frac{8038.544}{10}=803.8544\approx 803.85\)

But which is it? Let's check the data points: 10 data points. If it's a sample, n-1=9; if population, n=10.

Wait, the problem says "the data show the traveler spending... for a recent year for a sample of..." So it's a sample, so sample variance. But maybe the problem is considering it a population? Let's see the range was calculated as max - min, which is correct for both. Let's see the variance calculation.

Wait, maybe I made a mistake in the sum of squared differences. Let's recalculate one of them: 110.7 - 45.76 = 64.94, squared: 64.94*64.94. Let's calculate 65^2 = 4225, minus 0.06*2*65 + 0.06^2 = 4225 - 7.8 + 0.0036 = 4217.2036. Correct.

20.6 - 45.76 = -25.16, squared: 25.16*25.16. 25^2=625, 0.16^2=0.0256, 2*25*0.16=8, so (25 + 0.16)^2=25^2 + 2*25*0.16 + 0.16^2=625 + 8 + 0.0256=633.0256. Correct.

So sum is correct. Then sample variance is 8038.544 / 9 ≈ 893.17, population variance is 803.85.

Wait, maybe the problem is using population variance. Let's check with the given range: 90.10, which is 110.7 - 20.6 = 90.1, so two decimal places. Let's see the variance:

If population variance: 803.85 (rounded to two decimals: 803.85)

If sample variance: 893.17 (rounded to two decimals: 893.17)

Which one is correct? Let's see the data: 10 values. If it's a sample, we use n-1. But maybe the problem considers it a population. Let's check with the formula.

Wait, let's recalculate the mean again: 457.6 / 10 = 45.76. Correct.

Sum of squared deviations: 8038.544. Correct.

So if it's population variance: 8038.544 / 10 = 803.8544 ≈ 803.85

If sample variance: 8038.544 / 9 ≈ 893.1716 ≈ 893.17

Now, let's see the problem statement: "Find the variance." It doesn't specify sample or population, but in many introductory statistics courses, if it's a sample of data (as it says "a sample of the data"), we use sample variance. But sometimes, if it's the entire population, we use population. Let's check the number of data points: 10. Maybe it's a population. Let's see, if we take population variance, the answer is 803.85, if sample, 893.17.

Wait, let's check with an example. Let's take a small data set. Suppose data is [1,2,3], mean is 2. Sum of squared deviations: (1-2)^2 + (2-2)^2 + (3-2)^2 = 1 + 0 + 1 = 2. Population variance: 2/3 ≈ 0.67, sample variance: 2/2 = 1. So depending on sample or population.

In the problem, since it's a "sample of the data", we should use sample variance. So 8038.544 / 9 ≈ 893.17.

Answer:

\(893.17\) (or \(803.85\) if population variance is considered, but likely sample variance is intended here)