Question

using other data representations
algebra 1 unit 1 supplemental lesson 1
task 1: measures of central tendency
find the mean and median of each data set below.
a. 23, 18, 28, 16, 19, 14, 22, 23, 24
184/9=20.8 median=22. (handwritten)
a1. add 53 to the set and recalculate the mean and median
a2. the ____ changed the most because:

b. 103, 112, 97, 68, 103, 112, 88, 89
name: emily mcnea
sec: 9th

b1 add 119 to the set and recalculate mean and median
b2. the mean changed by +/- ____
b3. by adding one more number, the median will change to __ but the mean __

Explanation:

Part a: Original Data Set (23, 18, 28, 16, 19, 14, 22, 23, 24)

Step 1: Find the Mean

First, sum all the numbers:
\(23 + 18 + 28 + 16 + 19 + 14 + 22 + 23 + 24\)
\(= (23+18)+(28+16)+(19+14)+(22+23)+24\)
\(= 41 + 44 + 33 + 45 + 24\)
\(= 41+44=85\); \(85+33=118\); \(118+45=163\); \(163+24=187\)

There are 9 numbers, so the mean is \(\frac{187}{9} \approx 20.78\) (wait, the handwritten note has 184/9, maybe a typo? Let's recalculate:
23+18=41; 41+28=69; 69+16=85; 85+19=104; 104+14=118; 118+22=140; 140+23=163; 163+24=187. Yes, 187/9 ≈20.78. Maybe the handwritten 184 was a mistake.)

Step 2: Find the Median

Sort the data: \(14, 16, 18, 19, 22, 23, 23, 24, 28\)
There are 9 (odd) numbers, so the median is the 5th term: \(22\) (matches the handwritten note).

Part a1: Add 53 to the Set (New Set: 14, 16, 18, 19, 22, 23, 23, 24, 28, 53)

Step 1: New Mean

Sum = \(187 + 53 = 240\)
Number of terms = \(10\)
Mean = \(\frac{240}{10} = 24\)

Step 2: New Median

Sort the new set: \(14, 16, 18, 19, 22, 23, 23, 24, 28, 53\)
There are 10 (even) numbers, so median is the average of the 5th and 6th terms: \(\frac{22 + 23}{2} = 22.5\)

Part a2: Which Changed More?

• Original mean: \(\approx 20.78\); New mean: \(24\); Change: \(24 - 20.78 \approx 3.22\)
• Original median: \(22\); New median: \(22.5\); Change: \(0.5\)

The mean changed the most because the mean is sensitive to extreme values (53 is much larger than other values), while the median is resistant to outliers.

Part b: Original Data Set (103, 112, 97, 68, 103, 112, 88, 89)

Step 1: Find the Mean

Sum: \(103 + 112 + 97 + 68 + 103 + 112 + 88 + 89\)
\(= (103+103)+(112+112)+(97+68)+(88+89)\)
\(= 206 + 224 + 165 + 177\)
\(= 206+224=430\); \(430+165=595\); \(595+177=772\)

Number of terms: \(8\)
Mean: \(\frac{772}{8} = 96.5\)

Step 2: Find the Median

Sort the data: \(68, 88, 89, 97, 103, 103, 112, 112\)
There are 8 (even) numbers, so median is the average of the 4th and 5th terms: \(\frac{97 + 103}{2} = 100\)

Part b1: Add 119 to the Set (New Set: 68, 88, 89, 97, 103, 103, 112, 112, 119)

Step 1: New Mean

Sum = \(772 + 119 = 891\)
Number of terms = \(9\)
Mean = \(\frac{891}{9} = 99\)

Step 2: New Median

Sort the new set: \(68, 88, 89, 97, 103, 103, 112, 112, 119\)
There are 9 (odd) numbers, so median is the 5th term: \(103\)

Part b2: Change in Mean

Original mean: \(96.5\); New mean: \(99\); Change: \(99 - 96.5 = 2.5\) (or \(+2.5\))

Part b2: Change in Median

Original median: \(100\); New median: \(103\); Change: \(103 - 100 = 3\) (or \(+3\))

Part b3: If We Add One More Number...

• Median: With 10 terms (even), the median would be the average of the 5th and 6th terms. The sorted set (after adding a number) would have the middle two terms around the current median range. For example, if we add a number ≤103, the 5th term is 103, 6th term is 103 (if added ≤103) or 103 (if added between 103-112), so median would be \(\frac{103 + 103}{2} = 103\) (if added ≤103) or similar. If added >112, 5th term 103, 6th term 103, still median 103.
• Mean: Depends on the number added. If we add \(x\), new mean = \(\frac{891 + x}{10}\). For example, if we add 103, mean = \(\frac{891 + 103}{10} = \frac{994}{10} = 99.4\), which changes by \(0.4\).

Summary (Key Answers):

• a: Mean ≈20.78, Median=22
• a1: Mean=24, Median=22.5
• a2: Mean (sensitive to outliers)
• b: Mean=96.5, Median=100
• b1: Mean=99, Median=103
• b2: Mean change=+2.5, Median change=+3
• b3: Median likely stays 103 (or similar), Mean changes based on the added number.

Answer:

Part a: Original Data Set (23, 18, 28, 16, 19, 14, 22, 23, 24)

Step 1: Find the Mean

First, sum all the numbers:
\(23 + 18 + 28 + 16 + 19 + 14 + 22 + 23 + 24\)
\(= (23+18)+(28+16)+(19+14)+(22+23)+24\)
\(= 41 + 44 + 33 + 45 + 24\)
\(= 41+44=85\); \(85+33=118\); \(118+45=163\); \(163+24=187\)

There are 9 numbers, so the mean is \(\frac{187}{9} \approx 20.78\) (wait, the handwritten note has 184/9, maybe a typo? Let's recalculate:
23+18=41; 41+28=69; 69+16=85; 85+19=104; 104+14=118; 118+22=140; 140+23=163; 163+24=187. Yes, 187/9 ≈20.78. Maybe the handwritten 184 was a mistake.)

Step 2: Find the Median

Sort the data: \(14, 16, 18, 19, 22, 23, 23, 24, 28\)
There are 9 (odd) numbers, so the median is the 5th term: \(22\) (matches the handwritten note).

Part a1: Add 53 to the Set (New Set: 14, 16, 18, 19, 22, 23, 23, 24, 28, 53)

Step 1: New Mean

Sum = \(187 + 53 = 240\)
Number of terms = \(10\)
Mean = \(\frac{240}{10} = 24\)

Step 2: New Median

Sort the new set: \(14, 16, 18, 19, 22, 23, 23, 24, 28, 53\)
There are 10 (even) numbers, so median is the average of the 5th and 6th terms: \(\frac{22 + 23}{2} = 22.5\)

Part a2: Which Changed More?

• Original mean: \(\approx 20.78\); New mean: \(24\); Change: \(24 - 20.78 \approx 3.22\)
• Original median: \(22\); New median: \(22.5\); Change: \(0.5\)

The mean changed the most because the mean is sensitive to extreme values (53 is much larger than other values), while the median is resistant to outliers.

Part b: Original Data Set (103, 112, 97, 68, 103, 112, 88, 89)

Step 1: Find the Mean

Sum: \(103 + 112 + 97 + 68 + 103 + 112 + 88 + 89\)
\(= (103+103)+(112+112)+(97+68)+(88+89)\)
\(= 206 + 224 + 165 + 177\)
\(= 206+224=430\); \(430+165=595\); \(595+177=772\)

Number of terms: \(8\)
Mean: \(\frac{772}{8} = 96.5\)

Step 2: Find the Median

Sort the data: \(68, 88, 89, 97, 103, 103, 112, 112\)
There are 8 (even) numbers, so median is the average of the 4th and 5th terms: \(\frac{97 + 103}{2} = 100\)

Part b1: Add 119 to the Set (New Set: 68, 88, 89, 97, 103, 103, 112, 112, 119)

Step 1: New Mean

Sum = \(772 + 119 = 891\)
Number of terms = \(9\)
Mean = \(\frac{891}{9} = 99\)

Step 2: New Median

Sort the new set: \(68, 88, 89, 97, 103, 103, 112, 112, 119\)
There are 9 (odd) numbers, so median is the 5th term: \(103\)

Part b2: Change in Mean

Original mean: \(96.5\); New mean: \(99\); Change: \(99 - 96.5 = 2.5\) (or \(+2.5\))

Part b2: Change in Median

Original median: \(100\); New median: \(103\); Change: \(103 - 100 = 3\) (or \(+3\))

Part b3: If We Add One More Number...

• Median: With 10 terms (even), the median would be the average of the 5th and 6th terms. The sorted set (after adding a number) would have the middle two terms around the current median range. For example, if we add a number ≤103, the 5th term is 103, 6th term is 103 (if added ≤103) or 103 (if added between 103-112), so median would be \(\frac{103 + 103}{2} = 103\) (if added ≤103) or similar. If added >112, 5th term 103, 6th term 103, still median 103.
• Mean: Depends on the number added. If we add \(x\), new mean = \(\frac{891 + x}{10}\). For example, if we add 103, mean = \(\frac{891 + 103}{10} = \frac{994}{10} = 99.4\), which changes by \(0.4\).

Summary (Key Answers):

• a: Mean ≈20.78, Median=22
• a1: Mean=24, Median=22.5
• a2: Mean (sensitive to outliers)
• b: Mean=96.5, Median=100
• b1: Mean=99, Median=103
• b2: Mean change=+2.5, Median change=+3
• b3: Median likely stays 103 (or similar), Mean changes based on the added number.