Question

transforming data
statistics and probability
name: tennarda
mr. reimer is considering two options for bumping up the test scores for unit 1.
but how would this affect the distribution of scores?
below is a list of the test scores from unit 1:
60 65 65 75 75 77 77 78 79 80 80 80 80 81 82 87 90 93 97 99
find the following values for this data set:

shape	center		variability
	mean	med	range	standard deviation

option 1: the 5 point gift. mr. reimer adds 5 to each of the scores.
fill in the table and the dotplot below.

test score	60	65	65	75	75	77	77	78	79	80	80	80	80	81	82	87	90	93	97	99
test score + 5

dotplot for test score
dotplot with 60, 65, 70, 75, 80, 85, 90, 95 as test score
dotplot for test score + 5
dotplot axis with 60, 65, 70, 75, 80, 85, 90, 95 as test score + 5
recalculate:

shape	center		variability
	mean	med	range	standard deviation

1. what happens to the shape, center, and variability when adding 5?

a. shape:
b. center:
c. variability:

Answer:

Part 1: Original Data Calculations

Step 1: Count the number of data points

The data set is: \( 60, 65, 65, 75, 75, 77, 77, 78, 79, 80, 80, 80, 80, 81, 82, 87, 90, 93, 97, 99 \).
Number of data points \( n = 20 \).

Step 2: Calculate the Mean

Mean \( \bar{x} = \frac{\sum x}{n} \)
Sum of data:
\( 60 + 65 + 65 + 75 + 75 + 77 + 77 + 78 + 79 + 80 + 80 + 80 + 80 + 81 + 82 + 87 + 90 + 93 + 97 + 99 \)
Let's compute step-by-step:
\( 60 + 65 = 125 \); \( 125 + 65 = 190 \); \( 190 + 75 = 265 \); \( 265 + 75 = 340 \); \( 340 + 77 = 417 \); \( 417 + 77 = 494 \); \( 494 + 78 = 572 \); \( 572 + 79 = 651 \); \( 651 + 80 = 731 \); \( 731 + 80 = 811 \); \( 811 + 80 = 891 \); \( 891 + 80 = 971 \); \( 971 + 81 = 1052 \); \( 1052 + 82 = 1134 \); \( 1134 + 87 = 1221 \); \( 1221 + 90 = 1311 \); \( 1311 + 93 = 1404 \); \( 1404 + 97 = 1501 \); \( 1501 + 99 = 1600 \).
Mean \( \bar{x} = \frac{1600}{20} = 80 \).

Step 3: Calculate the Median

Since \( n = 20 \) (even), median is the average of the \( 10^{\text{th}} \) and \( 11^{\text{th}} \) values.
Order of data (already ordered): \( 60, 65, 65, 75, 75, 77, 77, 78, 79, 80, 80, 80, 80, 81, 82, 87, 90, 93, 97, 99 \).
\( 10^{\text{th}} \) value: \( 80 \); \( 11^{\text{th}} \) value: \( 80 \).
Median \( = \frac{80 + 80}{2} = 80 \).

Step 4: Calculate the Range

Range \( = \text{Max} - \text{Min} \)
Max \( = 99 \); Min \( = 60 \).
Range \( = 99 - 60 = 39 \).

Step 5: Calculate Standard Deviation (brief)

First, find the squared deviations from the mean, sum them, divide by \( n \), then take the square root.
Deviations: \( (60-80)^2, (65-80)^2, \dots, (99-80)^2 \).
Sum of squared deviations: Let's compute a few:
\( (60-80)^2 = 400 \); \( (65-80)^2 = 225 \) (two times); \( (75-80)^2 = 25 \) (two times); \( (77-80)^2 = 9 \) (two times); \( (78-80)^2 = 4 \); \( (79-80)^2 = 1 \); \( (80-80)^2 = 0 \) (four times); \( (81-80)^2 = 1 \); \( (82-80)^2 = 4 \); \( (87-80)^2 = 49 \); \( (90-80)^2 = 100 \); \( (93-80)^2 = 169 \); \( (97-80)^2 = 289 \); \( (99-80)^2 = 361 \).
Summing these:
\( 400 + 225(2) + 25(2) + 9(2) + 4 + 1 + 0(4) + 1 + 4 + 49 + 100 + 169 + 289 + 361 \)
\( = 400 + 450 + 50 + 18 + 4 + 1 + 0 + 1 + 4 + 49 + 100 + 169 + 289 + 361 \)
\( = 400 + 450 = 850 \); \( 850 + 50 = 900 \); \( 900 + 18 = 918 \); \( 918 + 4 = 922 \); \( 922 + 1 = 923 \); \( 923 + 1 = 924 \); \( 924 + 4 = 928 \); \( 928 + 49 = 977 \); \( 977 + 100 = 1077 \); \( 1077 + 169 = 1246 \); \( 1246 + 289 = 1535 \); \( 1535 + 361 = 1896 \).
Variance \( = \frac{1896}{20} = 94.8 \).
Standard Deviation \( = \sqrt{94.8} \approx 9.74 \).

Step 6: Shape

The data has a low outlier (60) and more values on the left (lower end) relative to the peak at 80? Wait, no—wait, the data: 60 (1), 65 (2), 75 (2), 77 (2), 78 (1), 79 (1), 80 (4), 81 (1), 82 (1), 87 (1), 90 (1), 93 (1), 97 (1), 99 (1). Wait, actually, the peak is at 80, and the left tail (lower scores) is shorter? Wait, no—60 is a single point, then 65 (two), 75 (two), etc. Wait, maybe it's slightly left-skewed? Or maybe symmetric? Wait, the mean and median are both 80, which suggests symmetry. Wait, but 60 is a low value. Wait, maybe the shape is approximately symmetric (since mean = median).

Part 2: Option 1 (Adding 5 to each score)

Step 1: New Data Set

Each score \( x \) becomes \( x + 5 \). So the new data is:
\( 65, 70, 70, 80, 80, 82, 82, 83, 84, 85, 85, 85, 85, 86, 87, 92, 95, 98, 102, 104 \).

Step 2: Mean of New Data

Mean of new data \( = \frac{\sum (x + 5)}{n} = \frac{\sum x + 5n}{n} = \bar{x} + 5 = 80 + 5 = 85 \).

Step 3: Median of New Data

The \( 10^{\text{th}} \) and \( 11^{\text{th}} \) values of the original data were 80 and 80. Adding 5, they become 85 and 85. So median \( = \frac{85 + 85}{2} = 85 \).

Step 4: Range of New Data

New Max \( = 99 + 5 = 104 \); New Min \( = 60 + 5 = 65 \).
Range \( = 104 - 65 = 39 \) (same as original).

Step 5: Standard Deviation of New Data

Adding a constant to each data point does not change the spread (variance/standard deviation). So standard deviation remains \( \approx 9.74 \).

Step 6: Shape of New Data

Since we added 5 to each point, the distribution shifts right but the shape (relative frequencies, skewness) remains the same. So if original was symmetric, new is also symmetric.

Part 3: Answering the Questions (1a, 1b, 1c)

1a. Shape:

Adding a constant to all data points does not change the shape of the distribution (e.g., skewness, number of peaks, relative spread of data). So the shape remains the same (e.g., if original was symmetric, new is symmetric; if skewed, same skewness).

1b. Center:

Measures of center (mean, median) increase by the constant added (5). So the center (mean, median) increases by 5 (or shifts right by 5 units).

1c. Variability:

Measures of variability (range, standard deviation, variance) do not change when a constant is added to all data points, because the spread between data points remains the same. So variability (range, standard deviation) remains the same.

Final Answers (for Original Data Table):

Shape	Center	Variability

For Option 1 (Test Score + 5) Table:

Shape	Center	Variability

For Question 1:
a. Shape: Remains the same (no change in distribution shape).
b. Center: Increases by 5 (mean and median each increase by 5).
c. Variability: Remains the same (range and standard deviation do not change).