Exercises in Measure of Central Tendency-Grouped and Ungrouped ... - Free Printable

Problem Analysis and Solution

The provided document contains several statistical problems involving ungrouped data. We will solve each problem step by step, calculating the mean, median, and mode where applicable.

---

Problem 1: Number of Incorrect Answers

The number of incorrect answers on a true-false test for 15 students is given as:
\[ 2, 3, 0, 1, 3, 6, 0, 3, 3, 5, 2, 1, 4, 2 \]

#### a) Mean
The mean is calculated as:
\[
\text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}
\]

First, calculate the sum:
\[
2 + 3 + 0 + 1 + 3 + 6 + 0 + 3 + 3 + 5 + 2 + 1 + 4 + 2 = 37
\]

There are 15 values, so:
\[
\text{Mean} = \frac{37}{15} \approx 2.47
\]

#### b) Median
The median is the middle value when the data is arranged in ascending order. If there is an even number of values, the median is the average of the two middle values.

Arrange the data in ascending order:
\[
0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 5, 6
\]

Since there are 15 values (odd), the median is the 8th value:
\[
\text{Median} = 3
\]

#### c) Mode
The mode is the value that appears most frequently.

From the ordered data:
\[
0, 0, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 5, 6
\]

The value \(3\) appears most frequently (4 times).

\[
\text{Mode} = 3
\]

#### Final Answer for Problem 1:
\[
\boxed{\text{Mean} = 2.47, \text{Median} = 3, \text{Mode} = 3}
\]

---

Problem 2: Number of Building Permits

The number of building permits issued to 12 construction firms is:
\[ 4, 7, 0, 7, 11, 4, 1, 15, 3, 5, 8, 7 \]

#### a) Mean
Calculate the sum:
\[
4 + 7 + 0 + 7 + 11 + 4 + 1 + 15 + 3 + 5 + 8 + 7 = 72
\]

There are 12 values, so:
\[
\text{Mean} = \frac{72}{12} = 6
\]

#### b) Median
Arrange the data in ascending order:
\[
0, 1, 3, 4, 4, 5, 7, 7, 7, 8, 11, 15
\]

Since there are 12 values (even), the median is the average of the 6th and 7th values:
\[
\text{Median} = \frac{5 + 7}{2} = 6
\]

#### c) Mode
From the ordered data:
\[
0, 1, 3, 4, 4, 5, 7, 7, 7, 8, 11, 15
\]

The value \(7\) appears most frequently (3 times).

\[
\text{Mode} = 7
\]

#### Final Answer for Problem 2:
\[
\boxed{\text{Mean} = 6, \text{Median} = 6, \text{Mode} = 7}
\]

---

Problem 3: Reaction Times

The reaction times for 9 subjects are:
\[ 2.5, 3.6, 3.1, 4.3, 2.9, 2.3, 4.1, 3.4 \]

#### a) Mean
Calculate the sum:
\[
2.5 + 3.6 + 3.1 + 4.3 + 2.9 + 2.3 + 4.1 + 3.4 = 26.2
\]

There are 9 values, so:
\[
\text{Mean} = \frac{26.2}{9} \approx 2.91
\]

#### b) Median
Arrange the data in ascending order:
\[
2.3, 2.5, 2.9, 3.1, 3.4, 3.6, 4.1, 4.3
\]

Since there are 9 values (odd), the median is the 5th value:
\[
\text{Median} = 3.4
\]

#### Final Answer for Problem 3:
\[
\boxed{\text{Mean} = 2.91, \text{Median} = 3.4}
\]

---

Problem 4: Employee Donations

The donations pledged by employees are:
\[ 10, 40, 25, 5, 20, 10, 50, 30, 10, 5, 15, 25, 50, 10, 30, 5, 25, 45, 15 \]

#### a) Mean
Calculate the sum:
\[
10 + 40 + 25 + 5 + 20 + 10 + 50 + 30 + 10 + 5 + 15 + 25 + 50 + 10 + 30 + 5 + 25 + 45 + 15 = 450
\]

There are 19 values, so:
\[
\text{Mean} = \frac{450}{19} \approx 23.68
\]

#### b) Mode
Arrange the data in ascending order and count frequencies:
\[
5, 5, 5, 10, 10, 10, 10, 15, 15, 20, 25, 25, 25, 30, 30, 40, 45, 50, 50
\]

The value \(10\) appears most frequently (4 times).

\[
\text{Mode} = 10
\]

#### Final Answer for Problem 4:
\[
\boxed{\text{Mean} = 23.68, \text{Mode} = 10}
\]

---

Problem 5: Phosphates per Load of Laundry

The amount of phosphates per load of laundry for various detergents is given in the table:

| Laundry Detergent | Phosphates per Load (gm) |
|-------------------|--------------------------|
| A & P Blue Sail | 48 |
| Dash | 47 |
| Concentrated All | 42 |
| Cold Water All | 42 |
| Breeze | 41 |
| Oxydol | 34 |
| Ajax | 31 |
| Sears | 30 |
| Fab | 29 |
| Cold Power | 29 |
| Bold | 29 |
| Rinso | 26 |

#### a) Mean
Calculate the sum of phosphates:
\[
48 + 47 + 42 + 42 + 41 + 34 + 31 + 30 + 29 + 29 + 29 + 26 = 438
\]

There are 12 detergents, so:
\[
\text{Mean} = \frac{438}{12} = 36.5
\]

#### b) Median
Arrange the data in ascending order:
\[
26, 29, 29, 29, 30, 31, 34, 41, 42, 42, 47, 48
\]

Since there are 12 values (even), the median is the average of the 6th and 7th values:
\[
\text{Median} = \frac{31 + 34}{2} = 32.5
\]

#### Final Answer for Problem 5:
\[
\boxed{\text{Mean} = 36.5, \text{Median} = 32.5}
\]

---

Summary of Final Answers:

1. \(\boxed{\text{Mean} = 2.47, \text{Median} = 3, \text{Mode} = 3}\)
2. \(\boxed{\text{Mean} = 6, \text{Median} = 6, \text{Mode} = 7}\)
3. \(\boxed{\text{Mean} = 2.91, \text{Median} = 3.4}\)
4. \(\boxed{\text{Mean} = 23.68, \text{Mode} = 10}\)
5. \(\boxed{\text{Mean} = 36.5, \text{Median} = 32.5}\)