To solve this problem, let's calculate each of the required values step by step using the provided scores:

Given Scores:

43, 67, 67, 68, 70, 71, 71, 75, 79, 86, 86, 86, 91

1. Range:

The range is the difference between the maximum and minimum scores.

• Maximum Score: 91
• Minimum Score: 43

$\text{Range} = 91 - 43 = 48$

2. Interquartile Range (IQR):

The IQR is the difference between the third quartile (Q3) and the first quartile (Q1).

• Q1 (First Quartile): The median of the lower half of the data (excluding the median of the entire dataset).

  • Lower half: 43, 67, 67, 68, 70, 71
  • Median of this lower half: (67 + 68) / 2 = 67.5

• Q3 (Third Quartile): The median of the upper half of the data (excluding the median of the entire dataset).

  • Upper half: 71, 75, 79, 86, 86, 91
  • Median of this upper half: (79 + 86) / 2 = 82.5

$\text{IQR} = 82.5 - 67.5 = 15$

3. Standard Deviation:

The standard deviation measures the amount of variation or dispersion of the scores.

• Mean (already calculated): 73.85

We can use the formula for the sample standard deviation:

$s = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \bar{x})^2}{n-1}}$

Where $\bar{x}$ is the mean, $x_i$ are the individual data points, and $n$ is the number of data points.

Let's calculate it:

1. Subtract the mean from each score and square the result:

$(43 - 73.85)^2 = 950.8225$ $(67 - 73.85)^2 = 46.8225$ $(67 - 73.85)^2 = 46.8225$ $(68 - 73.85)^2 = 34.2225$ $(70 - 73.85)^2 = 14.8225$ $(71 - 73.85)^2 = 8.1225$ $(71 - 73.85)^2 = 8.1225$ $(75 - 73.85)^2 = 1.3225$ $(79 - 73.85)^2 = 26.6225$ $(86 - 73.85)^2 = 148.9225$ $(86 - 73.85)^2 = 148.9225$ $(86 - 73.85)^2 = 148.9225$ $(91 - 73.85)^2 = 297.0225$

2. Sum these squared differences:

$\text{Sum} = 950.8225 + 46.8225 + 46.8225 + 34.2225 + 14.8225 + 8.1225 + 8.1225 + 1.3225 + 26.6225 + 148.9225 + 148.9225 + 148.9225 + 297.0225 = 1881.485$

3. Divide by $n-1$:

$\frac{1881.485}{12} = 156.7904$

4. Take the square root to find the standard deviation:

$s = \sqrt{156.7904} \approx 12.52$

4. Outlier:

An outlier is typically any value more than 1.5 times the IQR above the third quartile or below the first quartile.

• Lower Bound: $Q1 - 1.5 \times IQR = 67.5 - 1.5 \times 15 = 45$
• Upper Bound: $Q3 + 1.5 \times IQR = 82.5 + 1.5 \times 15 = 105$

Since all data points are within this range (45 to 105), there is no outlier.

Final Answers:

• Range: 48
• Interquartile Range (IQR): 15
• Sample Standard Deviation: 12.52
• Outlier: None

Would you like more details or have any questions?

Here are 5 related questions:

1. How is the interquartile range used in detecting outliers?
2. What is the difference between sample and population standard deviation?
3. How would the range change if an outlier was added?
4. Can the mean, median, and mode be the same? Under what conditions?
5. Why is the standard deviation an important measure in statistics?

Tip: Knowing how to calculate and interpret standard deviation helps in understanding data variability, which is crucial for making informed decisions based on data analysis.