Independent practice worksheet for calculating mean and standard deviation with data sets and frequency tables.

Here are the step-by-step solutions to the problems on your worksheet.

Part 1: Data Sets A, B, and C

1. Calculate the mean of each data set.
The mean is the average. You add all the numbers together and divide by how many numbers there are.

* Set A: $\{1, 2, 3, 4, 5\}$
* Sum: $1 + 2 + 3 + 4 + 5 = 15$
* Count: 5 numbers
* Mean: $15 \div 5 = \mathbf{3}$

* Set B: $\{2, 2, 2, 2, 2\}$
* Sum: $2 + 2 + 2 + 2 + 2 = 10$
* Count: 5 numbers
* Mean: $10 \div 5 = \mathbf{2}$

* Set C: $\{5, 7, 3, 11, 14\}$
* Sum: $5 + 7 + 3 + 11 + 14 = 40$
* Count: 5 numbers
* Mean: $40 \div 5 = \mathbf{8}$

2. Calculate the standard deviation of each data set.
Standard deviation measures how spread out the numbers are from the mean.
Formula steps:
1. Find the difference between each number and the mean.
2. Square that difference.
3. Add those squared differences together.
4. Divide by the count ($n$).
5. Take the square root.

* Set A (Mean = 3):
* Differences: $(1-3)=-2$, $(2-3)=-1$, $(3-3)=0$, $(4-3)=1$, $(5-3)=2$
* Squared: $4, 1, 0, 1, 4$
* Sum of squares: $4+1+0+1+4 = 10$
* Divide by 5: $10 \div 5 = 2$
* Square root: $\sqrt{2} \approx \mathbf{1.41}$

* Set B (Mean = 2):
* Differences: All numbers are 2, so $(2-2)=0$.
* Squared: $0, 0, 0, 0, 0$
* Sum of squares: $0$
* Divide by 5: $0$
* Square root: $\sqrt{0} = \mathbf{0}$

* Set C (Mean = 8):
* Differences: $(5-8)=-3$, $(7-8)=-1$, $(3-8)=-5$, $(11-8)=3$, $(14-8)=6$
* Squared: $9, 1, 25, 9, 36$
* Sum of squares: $9+1+25+9+36 = 80$
* Divide by 5: $80 \div 5 = 16$
* Square root: $\sqrt{16} = \mathbf{4}$

3. Which set has the largest standard deviation?
Comparing the results from Question 2:
* Set A: 1.41
* Set B: 0
* Set C: 4

Set C has the largest standard deviation.

4. Is it possible to answer question "c" without calculations of the standard deviation?
Yes. Standard deviation measures "spread" or variation.
* In Set B, all numbers are identical, so there is zero spread.
* In Set A, the numbers are close together (consecutive integers).
* In Set C, the numbers are much further apart (ranging from 3 to 14). Because the values in Set C are more spread out from their average than the values in Set A, Set C must have the larger standard deviation.

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Part 2: Frequency Table

First, let's organize the data to make calculation easier. We multiply the "Number of Children" by the "Frequency" to get the total for that row.

| Number of Children ($x$) | Frequency ($f$) | $x \cdot f$ |
| :--- | :--- | :--- |
| 20 | 2 | 40 |
| 40 | 2 | 80 |
| 60 | 6 | 360 |
| 80 | 5 | 400 |
| 90 | 3 | 270 |

* Total Frequency ($N$): $2 + 2 + 6 + 5 + 3 = 18$
* Sum of ($x \cdot f$): $40 + 80 + 360 + 400 + 270 = 1150$

5. Calculate the mean of the number of children of frequency.
$$ \text{Mean} = \frac{\text{Sum of } (x \cdot f)}{\text{Total Frequency}} $$
$$ \text{Mean} = \frac{1150}{18} \approx \mathbf{63.89} $$

6. Calculate the standard deviation of the number of children of frequency.
We use the formula for grouped data: $\sigma = \sqrt{\frac{\sum f(x - \mu)^2}{N}}$
*(Using $\mu \approx 63.89$)*

1. Find $(x - \mu)^2$ for each row:
* $x=20$: $(20 - 63.89)^2 = (-43.89)^2 \approx 1926.33$
* $x=40$: $(40 - 63.89)^2 = (-23.89)^2 \approx 570.73$
* $x=60$: $(60 - 63.89)^2 = (-3.89)^2 \approx 15.13$
* $x=80$: $(80 - 63.89)^2 = (16.11)^2 \approx 259.53$
* $x=90$: $(90 - 63.89)^2 = (26.11)^2 \approx 681.73$

2. Multiply by Frequency ($f$):
* $2 \times 1926.33 = 3852.66$
* $2 \times 570.73 = 1141.46$
* $6 \times 15.13 = 90.78$
* $5 \times 259.53 = 1297.65$
* $3 \times 681.73 = 2045.19$

3. Sum these values:
$3852.66 + 1141.46 + 90.78 + 1297.65 + 2045.19 = 8427.74$

4. Divide by Total Frequency ($N=18$):
$8427.74 \div 18 \approx 468.21$

5. Take the square root:
$\sqrt{468.21} \approx \mathbf{21.64}$

Final Answer:
1. Means:
* Set A: 3
* Set B: 2
* Set C: 8

2. Standard Deviations:
* Set A: $\approx 1.41$
* Set B: 0
* Set C: 4

3. Largest Standard Deviation: Set C

4. Possible without calculation? Yes, because Set C has the most spread-out data values.

5. Mean of Frequency Table: $\approx 63.89$

6. Standard Deviation of Frequency Table: $\approx 21.64$