The Iganga District examination committee wants to determine the appropriate duration for an upcoming examination. They have three options: 1 hour, 2 hours, or 2 hours and 30 minutes. To make an informed decision, they sampled 300 candidates and have a histogram of completion times. We need to: (i) Draw a frequency distribution table based on the histogram. (ii) Determine the average time to complete the examination. (iii) Draw an ogive and use it to determine the recommended median for the performance. Also, the time allocation ratios are 2:4:5 for 1 hour, 2 hours, and 2 hours 30 minutes respectively. We must find the number of students who completed the work in each of these time periods based on the 300 sampled students.

1. Problem Description

The Iganga District examination committee wants to determine the appropriate duration for an upcoming examination. They have three options: 1 hour, 2 hours, or 2 hours and 30 minutes. To make an informed decision, they sampled 300 candidates and have a histogram of completion times.

We need to:

(i) Draw a frequency distribution table based on the histogram.

(ii) Determine the average time to complete the examination.

(iii) Draw an ogive and use it to determine the recommended median for the performance.

Also, the time allocation ratios are 2:4:5 for 1 hour, 2 hours, and 2 hours 30 minutes respectively. We must find the number of students who completed the work in each of these time periods based on the 300 sampled students.

2. Solution Steps

(a)(i) Frequency Distribution Table:

First, we need to determine the class intervals from the histogram's class boundaries.

The class boundaries are: 24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5

The corresponding class intervals are therefore:

2

4. 5-34.5, 34.5-44.5, 44.5-54.5, 54.5-64.5, 64.5-74.5, 74.5-84.5

Next, we read the frequencies from the histogram for each class interval:

2

5. 5-34.5: 130

2

6. 5-44.5: 300

2

7. 5-54.5: 310

2

8. 5-64.5: 200

2

9. 5-74.5: 60

3

0. 5-84.5: 0

Here's the frequency distribution table:

| Class Interval | Frequency |

|------------------|-----------|

| 24.5 - 34.5 | 130 |

| 34.5 - 44.5 | 300 |

| 44.5 - 54.5 | 310 |

| 54.5 - 64.5 | 200 |

| 64.5 - 74.5 | 60 |

| 74.5 - 84.5 | 0 |

(a)(ii) Average Time:

To calculate the average time, we need to find the midpoint of each class interval.

Midpoint = (Upper Boundary + Lower Boundary) / 2

Then, multiply each midpoint by its corresponding frequency. Finally, sum up these products and divide by the total number of students (300).

| Class Interval | Frequency (f) | Midpoint (x) | f*x |

|------------------|---------------|--------------|------|

| 24.5 - 34.5 | 130 | 29.5 | 3835 |

| 34.5 - 44.5 | 300 | 39.5 | 11850 |

| 44.5 - 54.5 | 310 | 49.5 | 15345 |

| 54.5 - 64.5 | 200 | 59.5 | 11900 |

| 64.5 - 74.5 | 60 | 69.5 | 4170 |

| 74.5 - 84.5 | 0 | 79.5 | 0 |

| Total | 1000 | | 47100|

$Average = \frac{\sum(f \cdot x)}{\sum f}$

$Average = \frac{47100}{1000} = 47.1$

The average time is 47.1 minutes.

(a)(iii) Ogive and Median:

To draw an ogive, we need to calculate the cumulative frequencies.

| Class Interval | Frequency | Cumulative Frequency |

|------------------|-----------|----------------------|

| 24.5 - 34.5 | 130 | 130 |

| 34.5 - 44.5 | 300 | 430 |

| 44.5 - 54.5 | 310 | 740 |

| 54.5 - 64.5 | 200 | 940 |

| 64.5 - 74.5 | 60 | 1000 |

| 74.5 - 84.5 | 0 | 1000 |

Since the total frequency is 1000, the median corresponds to the cumulative frequency of 500 (1000/2). Based on the cumulative frequency table, 500 falls within the class interval 44.5 - 54.

5. We would need the actual ogive graph to accurately determine the median. If we linearly interpolate:

$Median = L + (\frac{\frac{N}{2} - cf}{f}) * w$

Where:

L is the lower class boundary of the median class (44.5)

N is the total frequency (1000)

cf is the cumulative frequency of the class before the median class (430)

f is the frequency of the median class (310)

w is the class width (10)

$Median = 44.5 + (\frac{500 - 430}{310}) * 10 = 44.5 + (\frac{70}{310}) * 10 = 44.5 + 2.26 = 46.76$

The estimated median is approximately 46.76 minutes. The recommended median is approximately 47 minutes.

(b) Time Allocation Ratios:

The time allocation ratio is 2:4:5 for 1 hour, 2 hours, and 2 hours 30 minutes, respectively.

The total ratio is $2 + 4 + 5 = 11$.

Number of students for each time duration:

* 1 hour: $(2/11) * 1000 \approx 181.82$ students. Approximately 182 students.

* 2 hours: $(4/11) * 1000 \approx 363.64$ students. Approximately 364 students.

* 2 hours 30 minutes: $(5/11) * 1000 \approx 454.55$ students. Approximately 455 students.

3. Final Answer

(a)(i) Frequency Distribution Table (see solution)

(a)(ii) Average time: 47.1 minutes

(a)(iii) Recommended Median: approximately 47 minutes

(b)

* 1 hour: Approximately 182 students

* 2 hours: Approximately 364 students

* 2 hours 30 minutes: Approximately 455 students