Question

the distribution of test scores for two random samples of students is shown in the stem - and - leaf plots.
group a:
5 | 8
6 | 3 8
7 | 2 3 5 5 6
8 | 0 2 4 5 5 5 7 7
9 | 1 3 4 5 6 7 7 8 8 8 8 9
10 | 0 0 0 0
key: 5|8 represents 58
group b:
5 | 4 7
6 | 2 3 7
7 | 1 1 2 3 5 6 7 8 8 8 8
8 | 2 2 3 3 3 3 4 4 5 7 9
9 | 1 2 8
10 | 0 0
key: 5|4 represents 54
part a: calculate the mean, median, range, and interquartile range for each data set. (2 points)
part b: using the appropriate measures, compare the center and variability of the two groups of students. explain your reasoning based on the shapes of the distributions. (2 points)

Answer:

Part A: Group A

Step 1: Extract Data from Stem - Leaf Plot

The stem - leaf plot for Group A:

• Stem 5: Leaf 8 → 58
• Stem 6: Leaf 3, 8 → 63, 68
• Stem 7: Leaf 2, 3, 5, 5, 6 → 72, 73, 75, 75, 76
• Stem 8: Leaf 0, 2, 4, 5, 5, 5, 7, 7 → 80, 82, 84, 85, 85, 85, 87, 87
• Stem 9: Leaf 1, 3, 4, 5, 6, 7, 7, 8, 8, 8, 8, 9 → 91, 93, 94, 95, 96, 97, 97, 98, 98, 98, 98, 99
• Stem 10: Leaf 0, 0, 0, 0 → 100, 100, 100, 100

Now, we count the number of data points. Let's sum up the number of leaves: 1 (stem 5) + 2 (stem 6)+5 (stem 7)+8 (stem 8)+12 (stem 9)+4 (stem 10) = 32 data points.

Step 2: Calculate the Mean

The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)

First, we calculate the sum of all data points:

• Stem 5: \(58\times1 = 58\)
• Stem 6: \(63 + 68=131\)
• Stem 7: \(72+73 + 75+75+76=371\)
• Stem 8: \(80 + 82+84+85+85+85+87+87=675\)
• Stem 9: \(91+93+94+95+96+97+97+98+98+98+98+99\)

\(=91+(93 + 94)+(95+96)+(97+97)+(98\times4)+99\)
\(=91 + 187+191+194+392+99\)
\(=91+187 = 278; 278+191 = 469; 469+194 = 663; 663+392 = 1055; 1055 + 99=1154\)

• Stem 10: \(100\times4 = 400\)

Total sum \(S=58 + 131+371+675+1154+400\)
\(58+131 = 189; 189+371 = 560; 560+675 = 1235; 1235+1154 = 2389; 2389+400 = 2789\)

Mean \(\bar{x}_{A}=\frac{2789}{32}\approx87.16\)

Step 3: Calculate the Median

Since \(n = 32\) (even), the median is the average of the \(\frac{n}{2}=16\)th and \((\frac{n}{2}+1)=17\)th values when the data is ordered.

We order the data:

First, list all the data points in order:

58, 63, 68, 72, 73, 75, 75, 76, 80, 82, 84, 85, 85, 85, 87, 87, 91, 93, 94, 95, 96, 97, 97, 98, 98, 98, 98, 99, 100, 100, 100, 100

The 16th value is 87 and the 17th value is 91.

Median \(M_{A}=\frac{87 + 91}{2}=89\)

Step 4: Calculate the Range

Range \(= \text{Maximum}-\text{Minimum}\)

Maximum value in Group A is 100, Minimum value is 58.

Range \(R_{A}=100 - 58 = 42\)

Step 5: Calculate the Inter - Quartile Range (IQR)

First, find the first quartile (\(Q_1\)) and the third quartile (\(Q_3\)).

For \(n = 32\), the position of \(Q_1\) is \(\frac{n}{4}=\frac{32}{4}=8\)th value (when data is ordered), and the position of \(Q_3\) is \(\frac{3n}{4}=\frac{3\times32}{4}=24\)th value.

From the ordered data:

• 8th value: 76 (so \(Q_1 = 76\))
• 24th value: 98 (so \(Q_3 = 98\))

IQR \(=Q_3 - Q_1=98 - 76 = 22\)

Group B

Step 1: Extract Data from Stem - Leaf Plot

The stem - leaf plot for Group B:

• Stem 5: Leaf 4, 7 → 54, 57
• Stem 6: Leaf 2, 3, 7 → 62, 63, 67
• Stem 7: Leaf 1, 1, 2, 3, 5, 6, 7, 8, 8, 8, 8 → 71, 71, 72, 73, 75, 76, 77, 78, 78, 78, 78
• Stem 8: Leaf 2, 2, 3, 3, 3, 3, 4, 4, 5, 7, 9 → 82, 82, 83, 83, 83, 83, 84, 84, 85, 87, 89
• Stem 9: Leaf 1, 2, 8 → 91, 92, 98
• Stem 10: Leaf 0, 0 → 100, 100

Number of data points: 2 (stem 5)+3 (stem 6)+11 (stem 7)+11 (stem 8)+3 (stem 9)+2 (stem 10)=32 data points.

Step 2: Calculate the Mean

Sum of data points:

• Stem 5: \(54 + 57=111\)
• Stem 6: \(62+63 + 67=192\)
• Stem 7: \(71+71+72+73+75+76+77+78+78+78+78\)

\(=(71\times2)+72 + 73+75+76+77+(78\times4)\)
\(=142+72+73+75+76+77 + 312\)
\(=142+72 = 214; 214+73 = 287; 287+75 = 362; 362+76 = 438; 438+77 = 515; 515+312 = 827\)

• Stem 8: \(82+82+83+83+83+83+84+84+85+87+89\)

\(=(82\times2)+(83\times4)+(84\times2)+85+87+89\)
\(=164+332+168+85+87+89\)
\(=164+332 = 496; 496+168 = 664; 664+85 = 749; 749+87 = 836; 836+89 = 925\)

• Stem 9: \(91+92+98 = 281\)
• Stem 10: \(100\times2 = 200\)

Total sum \(S = 111+192+827+925+281+200\)
\(111+192 = 303; 303+827 = 1130; 1130+925 = 2055; 2055+281 = 2336; 2336+200 = 2536\)

Mean \(\bar{x}_{B}=\frac{2536}{32}=79.25\)

Step 3: Calculate the Median

Since \(n = 32\) (even), the median is the average of the 16th and 17th values.

Ordered data:

54, 57, 62, 63, 67, 71, 71, 72, 73, 75, 76, 77, 78, 78, 78, 78, 82, 82, 83, 83, 83, 83, 84, 84, 85, 87, 89, 91, 92, 98, 100, 100

16th value: 78, 17th value: 82

Median \(M_{B}=\frac{78 + 82}{2}=80\)

Step 4: Calculate the Range

Maximum value is 100, Minimum value is 54.

Range \(R_{B}=100 - 54 = 46\)

Step 5: Calculate the Inter - Quartile Range (IQR)

Position of \(Q_1=\frac{n}{4}=8\)th value, position of \(Q_3=\frac{3n}{4}=24\)th value.

From ordered data:

• 8th value: 72 (so \(Q_1 = 72\))
• 24th value: 84 (so \(Q_3 = 84\))

IQR \(=Q_3 - Q_1=84 - 72 = 12\)

Part B: Comparison of Center and Variability

Center (Median as a Measure, since distributions may be skewed)

• Group A has a median of 89, Group B has a median of 80. So, the center (median) of Group A is higher than that of Group B, indicating that on average (in terms of median) Group A students have higher test scores.

Variability (IQR and Range as Measures)

• Range: Group A has a range of 42, Group B has a range of 46. Group B has a slightly larger range.
• IQR: Group A has an IQR of 22, Group B has an IQR of 12. Group A has a larger IQR, indicating that the middle 50% of the data in Group A is more spread out than in Group B.

Shape of Distributions (to support the choice of median and IQR)

• For Group A: Looking at the stem - leaf plot, the data is more spread out in the upper half (stem 9 and 10 have more data points). The distribution may be slightly skewed left or approximately symmetric.
• For Group B: The data is more concentrated in the middle (stem 7 and 8) and the lower and upper tails are less heavy. The distribution may be approximately symmetric or slightly skewed right.

Since the distributions are not strongly skewed in a way that would make the mean a bad measure, but the median is also a good measure of center here. The IQR is a good measure of variability as it is less affected by outliers.

Final Answers for Part A

Group A

• Mean: \(\approx87.16\)
• Median: \(89\)
• Range: \(42\)
• IQR: \(22\)

Group B

• Mean: \(79.25\)
• Median: \(80\)
• Range: \(46\)
• IQR: \(12\)

Final Answer for Part B

The center (median) of Group A (\(89\)) is higher than that of Group B (\(80\)). The variability (IQR) of Group A (\(22\)) is greater than that of Group B (\(12\)), while the range of Group B (\(46\)) is slightly greater than that of Group A (\(42\)). The shape of Group A's distribution is more spread out in the upper half, and Group B's distribution is more concentrated in the middle, which supports using the median for center and IQR for variability.