Questions: Ms. Alexander gave her students a biology test last week. Here are the test scores for each of the fifteen students. Test scores: 83, 92, 97, 81, 80, 80, 91, 80, 93, 85, 80, 95, 90, 89, 79.

Ms. Alexander gave her students a biology test last week. Here are the test scores for each of the fifteen students. Test scores: 83, 92, 97, 81, 80, 80, 91, 80, 93, 85, 80, 95, 90, 89, 79.
Transcript text: Ms. Alexander gave her students a biology test last week. Here are the test scores for each of the fifteen students. \begin{tabular}{|ccccc|} \hline \multicolumn{5}{|c|}{ Test scores } \\ 83 & 92 & 97 & 81 & 80 \\ 80 & 91 & 80 & 93 & 85 \\ 80 & 95 & 90 & 89 & 79 \\ \hline \end{tabular}
failed

Solution

failed
failed

Find the descriptive statistics for the biology test scores.

Organize the data

I'll organize the data from the given test scores: \begin{tabular}{|ccccc|} \hline \multicolumn{5}{|c|}{ Test scores } \\ 83 & 92 & 97 & 81 & 80 \\ 80 & 91 & 80 & 93 & 85 \\ 80 & 95 & 90 & 89 & 79 \\ \hline \end{tabular}

The complete list of scores is: 83, 92, 97, 81, 80, 80, 91, 80, 93, 85, 80, 95, 90, 89, 79

Calculate measures of central tendency

Mean: \(\bar{x} = \frac{83+92+97+81+80+80+91+80+93+85+80+95+90+89+79}{15} = \frac{1295}{15} = 86.33\)

Median: First I'll arrange the data in ascending order: 79, 80, 80, 80, 80, 81, 83, 85, 89, 90, 91, 92, 93, 95, 97

Since there are 15 values (odd number), the median is the 8th value: 85

Mode: The value 80 appears 4 times, which is more frequent than any other value, so the mode is 80.

Calculate measures of dispersion

Range: Maximum - Minimum = 97 - 79 = 18

Variance: \(s^2 = \frac{\sum(x_i - \bar{x})^2}{n-1}\)

I'll calculate the sum of squared deviations: \(\sum(x_i - \bar{x})^2 = (83-86.33)^2 + (92-86.33)^2 + ... + (79-86.33)^2\)

Computing this sum gives approximately 642.67

Therefore: \(s^2 = \frac{642.67}{14} = 45.91\)

Standard Deviation: \(s = \sqrt{45.91} = 6.77\)

Calculate quartiles

With the ordered data: 79, 80, 80, 80, 80, 81, 83, 85, 89, 90, 91, 92, 93, 95, 97

Q1 (First quartile): The median of the lower half of the data Lower half: 79, 80, 80, 80, 80, 81, 83 Q1 = 80

Q3 (Third quartile): The median of the upper half of the data Upper half: 89, 90, 91, 92, 93, 95, 97 Q3 = 92

Interquartile Range (IQR) = Q3 - Q1 = 92 - 80 = 12

\(\boxed{\begin{array}{ll} \text{Mean} = 86.33 \\ \text{Median} = 85 \\ \text{Mode} = 80 \\ \text{Range} = 18 \\ \text{Standard Deviation} = 6.77 \\ \text{Variance} = 45.91 \\ \text{Q1} = 80 \\ \text{Q3} = 92 \\ \text{IQR} = 12 \end{array}}\)

\(\boxed{\begin{array}{ll} \text{Mean} = 86.33 \\ \text{Median} = 85 \\ \text{Mode} = 80 \\ \text{Range} = 18 \\ \text{Standard Deviation} = 6.77 \\ \text{Variance} = 45.91 \\ \text{Q1} = 80 \\ \text{Q3} = 92 \\ \text{IQR} = 12 \end{array}}\)

Was this solution helpful?
failed
Unhelpful
failed
Helpful