The numbers of students in 30 schools in Kikuyu sub-county were recorded as follows. (a) Using a class width of 50 and ending with the class 395-444, make a frequency distribution table for the data. (b) Calculate the mean. (c) On the grid provided, draw a histogram and use it to find the median.

The data points are: 200, 350, 258, 400, 198, 240, 202, 338, 372, 392, 303, 250, 250, 300, 400, 405, 338, 302, 330, 370, 345, 209, 220, 230, 275, 280, 390, 403, 322, 340.
The minimum value is 198 and the maximum value is 405.

The class width is 50, and the last class is 395-444. Working backward with a class width of 50:

• 395 - 444
• 345 - 394
• 295 - 344
• 245 - 294
• 195 - 244 (This class includes the minimum value of 198).

• 195 - 244: 198, 200, 202, 209, 220, 230, 240 (Frequency = 7)
• 245 - 294: 250, 250, 258, 275, 280 (Frequency = 5)
• 295 - 344: 300, 302, 303, 322, 330, 338, 338, 340 (Frequency = 8)
• 345 - 394: 345, 350, 370, 372, 390, 392 (Frequency = 6)
• 395 - 444: 400, 400, 403, 405 (Frequency = 4) The total frequency is $7+5+8+6+4 = 30$, which matches the number of schools.

The frequency distribution table is:

The number of data points is $N = 30$.

The mean is $\boxed{310}$.

We need class boundaries for the histogram. The class width is 50.

• Class 195 - 244 has boundaries 194.5 - 244.5 (Frequency = 7)
• Class 245 - 294 has boundaries 244.5 - 294.5 (Frequency = 5)
• Class 295 - 344 has boundaries 294.5 - 344.5 (Frequency = 8)
• Class 345 - 394 has boundaries 344.5 - 394.5 (Frequency = 6)
• Class 395 - 444 has boundaries 394.5 - 444.5 (Frequency = 4)

• Draw a horizontal axis (x-axis) for the number of students, labeled with the class boundaries: 194.5, 244.5, 294.5, 344.5, 394.5, 444.5.
• Draw a vertical axis (y-axis) for the frequency, starting from 0 and going up to at least 8 (the maximum frequency).
• Draw rectangular bars for each class interval. The width of each bar should correspond to the class width (50 units on the x-axis), and the height of each bar should correspond to its frequency on the y-axis.
  • Bar 1: From 194.5 to 244.5, height 7.
  • Bar 2: From 244.5 to 294.5, height 5.
  • Bar 3: From 294.5 to 344.5, height 8.
  • Bar 4: From 344.5 to 394.5, height 6.
  • Bar 5: From 394.5 to 444.5, height 4.

The total number of data points is $N = 30$. The median position is $\frac{N}{2} = \frac{30}{2} = 15^\text{th}$ value.
To find the median class, we calculate the cumulative frequencies:

• Class 194.5 - 244.5: Cumulative frequency = 7
• Class 244.5 - 294.5: Cumulative frequency = $7 + 5 = 12$
• Class 294.5 - 344.5: Cumulative frequency = $12 + 8 = 20$ The 15th value falls within the class 294.5 - 344.5. This is the median class.

Using the formula for the median of grouped data:

Where:

• $L$ = lower class boundary of the median class = 294.5
• $N$ = total frequency = 30
• $cf$ = cumulative frequency of the class before the median class = 12
• $f_m$ = frequency of the median class = 8
• $c$ = class width = 50

Substitute the values into the formula:

The median is $\boxed{313.25}$.