Answer :
- Calculate Q1 and Q3: $Q1 = 53.55$, $Q3 = 65.95$.
- Calculate IQR: $IQR = Q3 - Q1 = 12.4$.
- Determine mild outlier bounds: $(34.95, 84.55)$.
- Determine extreme outlier bounds: $(16.35, 103.15)$.
- Count mild and extreme outliers: 0 mild outliers, 0 extreme outliers. The final answer is $\boxed{0}$.
### Explanation
1. Understand the problem
We are given a data set of 48 values and asked to find the number of mild and extreme outliers. The data set is nearly normal.
2. Calculate Q1 and Q3
First, we need to calculate the first quartile (Q1) and the third quartile (Q3) of the data set. The data set is:
[43.2, 44.3, 45.3, 49, 50.2, 50.6, 51.1, 51.7, 51.9, 52.7, 53.5, 53.6, 54.2, 54.9, 54.9, 55.3, 55.4, 55.8, 56.4, 56.7, 56.8, 57.1, 57.7, 58.8, 59, 60.1, 60.3, 60.6, 61, 61.9, 62, 63, 63.8, 64.2, 64.7, 65.6, 66.3, 67.1, 67.2, 67.6, 67.6, 70.2, 70.7, 71, 71.2, 72.9, 73.3, 79.4]
Q1 is the 25th percentile, and Q3 is the 75th percentile. Since $n = 48$, Q1 is the value at position $(48+1)/4 = 12.25$, so we interpolate between the 12th and 13th values. Q3 is the value at position $3(48+1)/4 = 36.75$, so we interpolate between the 36th and 37th values.
3. Determine quartiles
The 12th value is 53.6, and the 13th value is 54.2. Therefore, $Q1 = 0.75 * 53.6 + 0.25 * 54.2 = 53.55 + 0.15 = 53.7$.
The 36th value is 65.6, and the 37th value is 66.3. Therefore, $Q3 = 0.25 * 65.6 + 0.75 * 66.3 = 66.125$.
4. Calculate IQR
Next, we calculate the interquartile range (IQR): $IQR = Q3 - Q1 = 65.95 - 53.55 = 12.4$.
5. Calculate mild outlier bounds
Now, we determine the bounds for mild outliers:
Lower bound: $Q1 - 1.5 * IQR = 53.55 - 1.5 * 12.4 = 53.55 - 18.6 = 34.95$
Upper bound: $Q3 + 1.5 * IQR = 65.95 + 1.5 * 12.4 = 65.95 + 18.6 = 84.55$
6. Calculate extreme outlier bounds
And the bounds for extreme outliers:
Lower bound: $Q1 - 3 * IQR = 53.55 - 3 * 12.4 = 53.55 - 37.2 = 16.35$
Upper bound: $Q3 + 3 * IQR = 65.95 + 3 * 12.4 = 65.95 + 37.2 = 103.15$
7. Count outliers
Now we count the number of mild and extreme outliers. Mild outliers are values outside the range (34.95, 84.55) but inside the range [16.35, 103.15]. Extreme outliers are values outside the range [16.35, 103.15].
Looking at the data, all values are within the range (34.95, 84.55), so there are no mild outliers. Also, all values are within the range [16.35, 103.15], so there are no extreme outliers.
8. State the final answer
Therefore, the number of mild outliers is 0, and the number of extreme outliers is 0.
### Examples
Outlier detection is used in fraud detection to identify unusual transactions that may be fraudulent. It is also used in medical diagnosis to identify unusual symptoms that may indicate a disease. In environmental science, it can help detect anomalies in pollution levels.
- Calculate IQR: $IQR = Q3 - Q1 = 12.4$.
- Determine mild outlier bounds: $(34.95, 84.55)$.
- Determine extreme outlier bounds: $(16.35, 103.15)$.
- Count mild and extreme outliers: 0 mild outliers, 0 extreme outliers. The final answer is $\boxed{0}$.
### Explanation
1. Understand the problem
We are given a data set of 48 values and asked to find the number of mild and extreme outliers. The data set is nearly normal.
2. Calculate Q1 and Q3
First, we need to calculate the first quartile (Q1) and the third quartile (Q3) of the data set. The data set is:
[43.2, 44.3, 45.3, 49, 50.2, 50.6, 51.1, 51.7, 51.9, 52.7, 53.5, 53.6, 54.2, 54.9, 54.9, 55.3, 55.4, 55.8, 56.4, 56.7, 56.8, 57.1, 57.7, 58.8, 59, 60.1, 60.3, 60.6, 61, 61.9, 62, 63, 63.8, 64.2, 64.7, 65.6, 66.3, 67.1, 67.2, 67.6, 67.6, 70.2, 70.7, 71, 71.2, 72.9, 73.3, 79.4]
Q1 is the 25th percentile, and Q3 is the 75th percentile. Since $n = 48$, Q1 is the value at position $(48+1)/4 = 12.25$, so we interpolate between the 12th and 13th values. Q3 is the value at position $3(48+1)/4 = 36.75$, so we interpolate between the 36th and 37th values.
3. Determine quartiles
The 12th value is 53.6, and the 13th value is 54.2. Therefore, $Q1 = 0.75 * 53.6 + 0.25 * 54.2 = 53.55 + 0.15 = 53.7$.
The 36th value is 65.6, and the 37th value is 66.3. Therefore, $Q3 = 0.25 * 65.6 + 0.75 * 66.3 = 66.125$.
4. Calculate IQR
Next, we calculate the interquartile range (IQR): $IQR = Q3 - Q1 = 65.95 - 53.55 = 12.4$.
5. Calculate mild outlier bounds
Now, we determine the bounds for mild outliers:
Lower bound: $Q1 - 1.5 * IQR = 53.55 - 1.5 * 12.4 = 53.55 - 18.6 = 34.95$
Upper bound: $Q3 + 1.5 * IQR = 65.95 + 1.5 * 12.4 = 65.95 + 18.6 = 84.55$
6. Calculate extreme outlier bounds
And the bounds for extreme outliers:
Lower bound: $Q1 - 3 * IQR = 53.55 - 3 * 12.4 = 53.55 - 37.2 = 16.35$
Upper bound: $Q3 + 3 * IQR = 65.95 + 3 * 12.4 = 65.95 + 37.2 = 103.15$
7. Count outliers
Now we count the number of mild and extreme outliers. Mild outliers are values outside the range (34.95, 84.55) but inside the range [16.35, 103.15]. Extreme outliers are values outside the range [16.35, 103.15].
Looking at the data, all values are within the range (34.95, 84.55), so there are no mild outliers. Also, all values are within the range [16.35, 103.15], so there are no extreme outliers.
8. State the final answer
Therefore, the number of mild outliers is 0, and the number of extreme outliers is 0.
### Examples
Outlier detection is used in fraud detection to identify unusual transactions that may be fraudulent. It is also used in medical diagnosis to identify unusual symptoms that may indicate a disease. In environmental science, it can help detect anomalies in pollution levels.
