Question

the ages (in years) of a random sample of shoppers at a gaming store are shown. determine the range, mean, variance, and standard deviation of the sample data set. 12, 18, 23, 15, 14, 16, 22, 16, 13, 18 the range is 11. (simplify your answer.) the mean is 16.7. (simplify your answer. round to the nearest tenth as needed.) the variance is 3.7. (simplify your answer. round to the nearest tenth as needed.)

Explanation:

Step1: Find the Range

The range is calculated as the difference between the maximum and minimum values in the data set.
First, identify the maximum and minimum values. The data set is \(12, 18, 23, 15, 14, 16, 22, 16, 13, 18\).
The maximum value is \(23\) and the minimum value is \(12\).
So, the range \(= 23 - 12 = 11\).

Step2: Find the Mean

The mean (\(\bar{x}\)) of a sample data set is calculated by the formula \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(x_{i}\) are the data points and \(n\) is the number of data points.
First, sum up all the data points:
\(12 + 18+23 + 15+14 + 16+22 + 16+13 + 18\)
\(=(12 + 18)+(23 + 15)+(14 + 16)+(22 + 16)+(13 + 18)\)
\(= 30+38 + 30+38+31\)
\(=30\times2+38\times2 + 31\)
\(=60 + 76+31\)
\(=167\)
The number of data points \(n = 10\).
So, the mean \(\bar{x}=\frac{167}{10}=16.7\).

Step3: Find the Variance (Sample Variance)

The formula for sample variance (\(s^{2}\)) is \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}\)
We know that \(\bar{x}=16.7\) and \(n = 10\)
Calculate \((x_{i}-\bar{x})^{2}\) for each data point:
- For \(x = 12\): \((12 - 16.7)^{2}=(- 4.7)^{2}=22.09\)
- For \(x = 18\): \((18 - 16.7)^{2}=(1.3)^{2}=1.69\)
- For \(x = 23\): \((23 - 16.7)^{2}=(6.3)^{2}=39.69\)
- For \(x = 15\): \((15 - 16.7)^{2}=(-1.7)^{2}=2.89\)
- For \(x = 14\): \((14 - 16.7)^{2}=(-2.7)^{2}=7.29\)
- For \(x = 16\): \((16 - 16.7)^{2}=(-0.7)^{2}=0.49\)
- For \(x = 22\): \((22 - 16.7)^{2}=(5.3)^{2}=28.09\)
- For \(x = 16\): \((16 - 16.7)^{2}=(-0.7)^{2}=0.49\)
- For \(x = 13\): \((13 - 16.7)^{2}=(-3.7)^{2}=13.69\)
- For \(x = 18\): \((18 - 16.7)^{2}=(1.3)^{2}=1.69\)

Now, sum up these squared differences:
\(22.09+1.69 + 39.69+2.89+7.29+0.49+28.09+0.49+13.69+1.69\)
\(=(22.09+39.69)+(1.69 + 1.69)+(2.89+7.29)+(0.49+0.49)+(28.09+13.69)\)
\(=61.78+3.38+10.18+0.98+41.78\)
\(=61.78+3.38 = 65.16\); \(65.16+10.18=75.34\); \(75.34 + 0.98=76.32\); \(76.32+41.78 = 118.1\)

Now, divide by \(n - 1=9\):
\(s^{2}=\frac{118.1}{9}\approx13.1\) (Wait, there is a mistake in the original variance value. Let's recalculate the sum of squared differences correctly)

Let's recalculate the sum of \((x_{i}-\bar{x})^{2}\) again:

\(x_i\) values: \(12,18,23,15,14,16,22,16,13,18\)

\(\bar{x} = 16.7\)

1. \(12-16.7=-4.7\), \((-4.7)^2 = 22.09\)
2. \(18 - 16.7 = 1.3\), \((1.3)^2=1.69\)
3. \(23-16.7 = 6.3\), \((6.3)^2 = 39.69\)
4. \(15 - 16.7=-1.7\), \((-1.7)^2=2.89\)
5. \(14 - 16.7=-2.7\), \((-2.7)^2 = 7.29\)
6. \(16 - 16.7=-0.7\), \((-0.7)^2=0.49\)
7. \(22 - 16.7 = 5.3\), \((5.3)^2=28.09\)
8. \(16 - 16.7=-0.7\), \((-0.7)^2=0.49\)
9. \(13 - 16.7=-3.7\), \((-3.7)^2 = 13.69\)
10. \(18 - 16.7 = 1.3\), \((1.3)^2=1.69\)

Now sum these:

\(22.09+1.69=23.78\)

\(23.78 + 39.69=63.47\)

\(63.47+2.89 = 66.36\)

\(66.36+7.29=73.65\)

\(73.65+0.49 = 74.14\)

\(74.14+28.09=102.23\)

\(102.23+0.49 = 102.72\)

\(102.72+13.69=116.41\)

\(116.41+1.69 = 118.1\) (Wait, same as before. Then \(s^{2}=\frac{118.1}{9}\approx13.1\). The original variance value of \(3.7\) is incorrect. Maybe it is a population variance? Let's check population variance.

Population variance (\(\sigma^{2}\)) is \(\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}\)

\(\frac{118.1}{10}=11.81\approx11.8\). Still not \(3.7\). Maybe there is a miscalculation in the mean? Wait, the sum of the data points: \(12 + 18+23+15+14+16+22+16+13+18\)

Let's add again:

\(12+18 = 30\); \(30+23 = 53\); \(53+15 = 68\); \(68+14 = 82\); \(82+16 = 98\); \(98+22 = 120\); \(120+16 = 136\); \(136+13 = 149\); \(149+18 = 167\). So mean is \(167/10 = 16.7\) (correct).

Wait, maybe the data set is different? Let's check the data set again: \(12,18,23,15,14,16,22,16,13,18\). Number of data points is 10.

Wait, maybe the original problem has a typo, but let's proceed with the correct calculation for variance and standard deviation.

Step4: Find the Variance (Correct Sample Variance)

Using the formula \(s^{2}=\frac{\sum(x_{i}-\bar{x})^{2}}{n - 1}\)

We have \(\sum(x_{i}-\bar{x})^{2}=118.1\) and \(n - 1 = 9\)

So \(s^{2}=\frac{118.1}{9}\approx13.1\)

Step5: Find the Standard Deviation

The standard deviation (\(s\)) is the square root of the variance.

\(s=\sqrt{s^{2}}=\sqrt{\frac{118.1}{9}}\approx\sqrt{13.1}\approx3.6\) (Wait, but let's do it properly)

\(\frac{118.1}{9}\approx13.122\)

\(\sqrt{13.122}\approx3.62\approx3.6\) (rounded to nearest tenth)

But let's check the original wrong variance. If we assume that the variance is calculated as population variance (divided by \(n\)):

\(\sigma^{2}=\frac{118.1}{10}=11.81\approx11.8\), standard deviation \(\sigma=\sqrt{11.81}\approx3.4\)

But the original variance is given as \(3.7\), which is incorrect. Maybe the data set is different? Let's check the data set again. Wait, maybe the data points are \(12, 18, 23, 15, 14, 16, 22, 16, 13, 18\) – no, that's what is given.

Wait, maybe I made a mistake in calculating the sum of squared differences. Let's recalculate each \((x_i - \bar{x})^2\):

1. \(12 - 16.7=-4.7\), square is \(22.09\)
2. \(18 - 16.7 = 1.3\), square is \(1.69\)
3. \(23 - 16.7 = 6.3\), square is \(39.69\)
4. \(15 - 16.7=-1.7\), square is \(2.89\)
5. \(14 - 16.7=-2.7\), square is \(7.29\)
6. \(16 - 16.7=-0.7\), square is \(0.49\)
7. \(22 - 16.7 = 5.3\), square is \(28.09\)
8. \(16 - 16.7=-0.7\), square is \(0.49\)
9. \(13 - 16.7=-3.7\), square is \(13.69\)
10. \(18 - 16.7 = 1.3\), square is \(1.69\)

Now sum them:

\(22.09+1.69 = 23.78\)

\(23.78+39.69 = 63.47\)

\(63.47+2.89 = 66.36\)

\(66.36+7.29 = 73.65\)

\(73.65+0.49 = 74.14\)

\(74.14+28.09 = 102.23\)

\(102.23+0.49 = 102.72\)

\(102.72+13.69 = 116.41\)

\(116.41+1.69 = 118.1\) (correct). So sample variance is \(118.1/9\approx13.1\), population variance is \(118.1/10 = 11.81\)

Now, standard deviation:

- Sample standard deviation: \(\sqrt{13.1}\approx3.6\)
- Population standard deviation: \(\sqrt{11.81}\approx3.4\)

But the original variance is given as \(3.7\), which is wrong. Maybe the data set is different? Let's check if the data points are \(12, 18, 23, 15, 14, 16, 22, 16, 13, 18\) – yes.

Wait, maybe the problem is about population (not sample). Let's recalculate with population variance:

Population variance \(\sigma^{2}=\frac{\sum(x_{i}-\bar{x})^{2}}{n}=\frac{118.1}{10}=11.81\approx11.8\)

Standard deviation \(\sigma=\sqrt{11.81}\approx3.4\)

But the original variance is \(3.7\), which is incorrect. Maybe there is a mistake in the problem or in the original answer.

Assuming that we follow the correct method:

Range:

Maximum value \(= 23\), Minimum value \(= 12\)

Range \(= 23 - 12 = 11\) (correct as per original)

Mean:

Sum of data points \(= 167\), \(n = 10\)

Mean \(=\frac{167}{10}=16.7\) (correct as per original)

Variance (Sample):

\(s^{2}=\frac{\sum(x_{i}-\bar{x})^{2}}{n - 1}=\frac{118.1}{9}\approx13.1\)

Variance (Population):

\(\sigma^{2}=\frac{\sum(x_{i}-\bar{x})^{2}}{n}=\frac{118.1}{10}=11.8\)

Standard Deviation (Sample):

\(s=\sqrt{13.1}\approx3.6\)

Standard Deviation (Population):

\(\sigma=\sqrt{11.8}\approx3.4\)

Since the original variance is wrong, but let's proceed with the correct calculation.

Answer:

• Range: \(\boldsymbol{11}\)
• Mean: \(\boldsymbol{16.7}\)
• Variance (Sample): \(\boldsymbol{\approx13.1}\) (Variance (Population): \(\boldsymbol{\approx11.8}\))
• Standard Deviation (Sample): \(\boldsymbol{\approx3.6}\) (Standard Deviation (Population): \(\boldsymbol{\approx3.4}\))

(Note: The original variance value of \(3.7\) is incorrect. The correct variance (sample) is approximately \(13.1\) and (population) is approximately \(11.8\))