Use the data given below, which shows the retirement ages for a sample of engineers, to construct a cumulative frequency distribution using SIX (6) classes. (8 marks) 60 65 68 63 66 67 69 67 58 65 67 61 63 65 62 64 73 50 61 71 62 69 72 63 B. Construct a cumulative frequency curve using the frequency distribution constructed in (A). (6 marks) C. Using your curve or otherwise determine: i. the median retirement age (2 marks) ii. the age that represents the 30th percentile, explain briefly what this value indicates

**A. Construct a cumulative frequency distribution using SIX (6) classes:**

To construct the cumulative frequency distribution, start by sorting the data in ascending order:

50 58 60 61 61 62 62 63 63 63 64 65 65 65 66 67 67 67 68 69 69 71 72 73

Next, determine the range of the data: 73 - 50 = 23 .

Calculate the class width: \frac{23}{6} \approx 3.83 , round up to 4.

Now, construct the cumulative frequency distribution using 6 classes:

\begin{array}{|c|c|c|}\hline\text{Class} & \text{Frequency} & \text{Cumulative Frequency} \\hline50-53 & 1 & 1 \\hline54-57 & 1 & 2 \\hline58-61 & 4 & 6 \\hline62-65 & 6 & 12 \\hline66-69 & 6 & 18 \\hline70-73 & 6 & 24 \\hline\end{array}

**B. Construct a cumulative frequency curve:**

To construct the cumulative frequency curve, plot the points (lower class boundary, cumulative frequency) on a graph and join them with a smooth curve.

**C. i. Calculate the median retirement age:**

The median is the middle value of a dataset when it is ordered. Since we have 24 data points, the median would be the average of the 12th and 13th values.

The 12th and 13th values are both in the class 62-65. So, the median would be 62 + \frac{64-62}{2} = 63

**C. ii. Calculate the age that represents the 30th percentile:**

The 30th percentile corresponds to the value below which 30% of the data falls.

From the cumulative frequency distribution, we can see that the 30th percentile lies in the class 58-61.

Since 30% of the data fall below this percentile, we can use the formula for percentile interpolation:

L + \left(\frac{N}{100} \times w\right)

where:
- L = Lower class boundary of the class interval (58)
- N = Desired percentile (30)
- w = Width of the class interval (4)
- f = Frequency of the class interval (4)
- C = Cumulative frequency of the class before the desired percentile (2)

Substitute the values into the formula:

58 + \left(\frac{30}{100} \times 4\right) = 58 + 1.2 = 59.2

Therefore, the age that represents the 30th percentile is 60. This value indicates that it is greater than 30% of the value

\boxed{\text{Answer: 59.2}}