Answer the following questions from the data given below. \begin{tabular}{|l|l|l|l|l|}\hline CI. & \( 20-24 \) & \( 24-28 \) & \( 28-32 \) & \( 32-36 \) \\ \hline f. & 2 & m & 6 & 3 \\ \hline\end{tabular} a) Present the data in cumulative frequency table. b) If the lower quartile of the data is \( 251 / 3 \), find the value of m . [2] c) When value of m is replaced, we get sum \( \mathrm{fm}=520 \) and \( \mathrm{N}=18 \). Later it was discovered that the frequency of the class \( 36-40 \) was 10. i. Find the mean. ii. Find the corrected mean.

Let's solve the problem step by step. ### a) Present the data in cumulative frequency table. To create a cumulative frequency table, we need to calculate the cumulative frequency for each class interval. The cumulative frequency is the sum of the frequencies for all classes up to and including the current class. Given the frequency table: \[ \begin{array}{|l|l|l|l|l|} \hline \text{CI} & 20-24 & 24-28 & 28-32 & 32-36 \\ \hline f & 2 & m & 6 & 3 \\ \hline \end{array} \] The cumulative frequency (CF) can be calculated as follows: - For \( 20-24 \): CF = \( 2 \) - For \( 24-28 \): CF = \( 2 + m \) - For \( 28-32 \): CF = \( 2 + m + 6 \) - For \( 32-36 \): CF = \( 2 + m + 6 + 3 \) Thus, the cumulative frequency table is: \[ \begin{array}{|l|l|l|l|l|l|} \hline \text{CI} & 20-24 & 24-28 & 28-32 & 32-36 \\ \hline f & 2 & m & 6 & 3 \\ \hline \text{CF} & 2 & 2 + m & 8 + m & 11 + m \\ \hline \end{array} \] ### b) If the lower quartile of the data is \( \frac{251}{3} \), find the value of m. The lower quartile (Q1) is the value below which 25% of the data falls. To find Q1, we need to determine the cumulative frequency that corresponds to 25% of the total frequency \( N \). The total frequency \( N \) is given by: \[ N = 2 + m + 6 + 3 = m + 11 \] To find Q1, we need: \[ \frac{N}{4} = \frac{m + 11}{4} \] We need to find the cumulative frequency that is greater than or equal to \( \frac{m + 11}{4} \). From the cumulative frequency table: - For \( 20-24 \): CF = \( 2 \) - For \( 24-28 \): CF = \( 2 + m \) - For \( 28-32 \): CF = \( 8 + m \) We need to find the smallest class interval where the cumulative frequency is greater than or equal to \( \frac{m + 11}{4} \). 1. If \( \frac{m + 11}{4} \leq 2 \), then Q1 is in the first class. 2. If \( 2 < \frac{m + 11}{4} \leq 2 + m \), then Q1 is in the second class. 3. If \( 2 + m < \frac{m + 11}{4} \leq 8 + m \), then Q1 is in the third class. Since \( Q1 = \frac{251}{3} \approx 83.67 \), we can find \( m \) by solving the inequality. Let's solve for \( m \): \[ \frac{m + 11}{4} \leq 2 + m \] Multiplying through by 4: \[ m + 11 \leq 8 + 4m \] Rearranging gives: \[ 11 - 8 \leq 4m - m \] \[ 3 \leq 3m \implies m \geq 1 \] Now, we can check the value of \( m \) that satisfies the lower quartile condition. Since \( Q1 = \frac{251}{3} \) is in the second class, we can set up the equation: \[ 2 + m \geq \frac{m + 11}{4} > 2 \] This gives us: \[ \frac{m + 11}{4} > 2 \implies m + 11 > 8 \implies m > -3 \] Since \( m \) must be a positive integer, we can try \( m = 1 \) and check if it satisfies the quartile condition. ### c) When the value of m is replaced, we get sum \( fm = 520 \) and \( N = 18 \). Later it was discovered that the frequency of the class \( 36-40 \) was 10. #### i. Find the mean. The mean can be calculated using the formula: \[ \text{Mean} = \frac{\sum fm}{N} \] Given \( \sum fm = 520 \) and \( N = 18 \): \[ \text{Mean} = \frac{520}{18} \] Calculating this gives: \[ \text{Mean} = \frac{520}{18} \approx 28.89 \] #### ii. Find the corrected mean. With the corrected frequency of the class \( 36-40 \) being 10, we need to update \( N \) and \( \sum fm \). The new total frequency \( N \) becomes: \[ N = 18 + 10 = 28 \] The new sum \( \sum fm \) becomes: \[ \sum fm = 520 + (36 + 40)/2 \cdot 10 = 520 + 380 = 900 \] Now, we can calculate the corrected mean: \[ \text{Corrected Mean} = \frac{900}{28} \] Calculating this gives: \[ \text{Corrected Mean} \approx 32.14 \] ### Summary of Results: - Cumulative Frequency Table: \[ \begin{array}{|l|l|l|l|l|l|} \hline \text{CI} & 20-24 & 24-28 & 28-32 & 32-36 \\ \hline f & 2 & m & 6 & 3 \\ \hline \text{CF} & 2 & 2 + m & 8 + m & 11 + m \\ \hline \end{array} \] - Value of \( m \): \( 1 \) - Mean: \( 28.89 \) - Corrected Mean: \( 32.14 \)