Question

For the following data set, calculate the percentage of data points that fall within one, two, and three standard deviation(s) of the mean. Does the data appear to be normally distributed? Show your work or explain how you got your answer. 20 12 19 40 22 25 10 24 36 23 15 22 23 13 18 19 17 32 24 21

Answer 1

To analyze the data set, we'll first calculate the mean and standard deviation, and then determine the percentage of data points that fall within one, two, and three standard deviations of the mean. Afterward, we'll assess whether the data appears to be normally distributed.

### Step 1: Calculate the Mean
The mean is calculated as the sum of all data points divided by the number of data points.

Data Set:
20, 12, 19, 40, 22, 25, 10, 24, 36, 23, 15, 22, 23, 13, 18, 19, 17, 32, 24, 21

1. **Sum of data points**:
\[
20 + 12 + 19 + 40 + 22 + 25 + 10 + 24 + 36 + 23 + 15 + 22 + 23 + 13 + 18 + 19 + 17 + 32 + 24 + 21 = 428
\]

2. **Number of data points (n)**:
\[
n = 20
\]

3. **Mean (\(\mu\))**:
\[
\mu = \frac{Sum}{n} = \frac{428}{20} = 21.4
\]

### Step 2: Calculate the Standard Deviation
1. **Variance (\(\sigma^2\)) calculation**:
\[
\sigma^2 = \frac{\sum (x_i - \mu)^2}{n}
\]
where \(x_i\) represents each data point.

2. **Calculate the deviations from the mean and their squares**:
- For each data point, compute \((x_i - 21.4)^2\):
\[
\begin{align*}
&(20 - 21.4)^2 = 1.96\\
&(12 - 21.4)^2 = 88.36\\
&(19 - 21.4)^2 = 5.76\\
&(40 - 21.4)^2 = 346.96\\
&(22 - 21.4)^2 = 0.36\\
&(25 - 21.4)^2 = 12.96\\
&(10 - 21.4)^2 = 128.16\\
&(24 - 21.4)^2 = 6.76\\
&(36 - 21.4)^2 = 216.36\\
&(23 - 21.4)^2 = 2.56\\
&(15 - 21.4)^2 = 41.16\\
&(22 - 21.4)^2 = 0.36\\
&(23 - 21.4)^2 = 2.56\\
&(13 - 21.4)^2 = 70.56\\
&(18 - 21.4)^2 = 11.56\\
&(19 - 21.4)^2 = 5.76\\
&(17 - 21.4)^2 = 19.36\\
&(32 - 21.4)^2 = 110.76\\
&(24 - 21.4)^2 = 6.76\\
&(21 - 21.4)^2 = 0.16\\
\end{align*}
\]

3. **Sum of the squared deviations**:
\[
1.96 + 88.36 + 5.76 + 346.96 + 0.36 + 12.96 + 128.16 + 6.76 + 216.36 + 2.56 + 41.16 + 0.36 + 2.56 + 70.56 + 11.56 + 5.76 + 19.36 + 110.76 + 6.76 + 0.16 = 515.2
\]

4. **Variance**:
\[
\sigma^2 = \frac{515.2}{20} = 25.76
\]

5. **Standard deviation (\(\sigma\))**:
\[
\sigma = \sqrt{25.76} \approx 5.08
\]

### Step 3: Determine Data Distribution Within Standard Deviations
1. **Calculate ranges for one, two, and three standard deviations**:
- **1 Standard Deviation**: \([ \mu - \sigma, \mu + \sigma ] = [21.4 - 5.08, 21.4 + 5.08] \approx [16.32, 26.48]\)
- **2 Standard Deviations**: \([ \mu - 2\sigma, \mu + 2\sigma ] = [21.4 - 2*5.08, 21.4 + 2*5.08] \approx [11.24, 31.56]\)
- **3 Standard Deviations**: \([ \mu - 3\sigma, \mu + 3\sigma ] = [21.4 - 3*5.08, 21.4 + 3*5.08] \approx [6.16, 36.64]\)

2. **Count data points within these ranges**:
- Between 1 SD (16.32 to 26.48): Data points = 10, count = (20, 22, 25, 24, 23, 22, 23, 24, 21) = 9 points
- Between 2 SD (11.24 to 31.56): All data points = 15, (10, 12, 13, 15, 17, 18, 19, 19, 20, 21, 22, 22, 23, 24, 24, 25, 32, 36, 40) = 18 points
- Between 3 SD (6.16 to 36.64): All data points = 20 points

### Summary of Percentages
- Within 1 standard deviation:
\[
\frac{9}{20} \times 100\% = 45\%
\]

- Within 2 standard deviations:
\[
\frac{18}{20} \times 100\% = 90\%
\]

- Within 3 standard deviations:
\[
\frac{20}{20} \times 100\% = 100\%
\]

### Conclusion on Normality
- **Observed distribution percentages**:
- For a normal distribution, we expect approximately 68% of data within 1 SD, 95% within 2 SD, and 99.7% within 3 SD.

- Our data shows 45% within 1 SD and 90% within 2 SD, indicating a deviation from normality.

Thus, we can conclude that the data does not appear to be normally distributed based on this analysis.