The problem provides a table showing the distribution of marks scored by students in a test. (a) Given the mean mark is $3\frac{6}{23}$, we need to find the value of $m$. (b) We need to find the interquartile range and the probability of selecting a student who scored at least 4 marks.

1. Problem Description

The problem provides a table showing the distribution of marks scored by students in a test.

(a) Given the mean mark is $3\frac{6}{23}$, we need to find the value of $m$.

(b) We need to find the interquartile range and the probability of selecting a student who scored at least 4 marks.

2. Solution Steps

(a) The mean mark is given by the sum of (marks * number of students) divided by the total number of students. The total number of students is $(m+2) + (m-1) + (2m-3) + (m+5) + (3m-4) = 8m - 1$.

The sum of (marks * number of students) is $1(m+2) + 2(m-1) + 3(2m-3) + 4(m+5) + 5(3m-4) = m+2 + 2m-2 + 6m-9 + 4m+20 + 15m-20 = 28m - 9$.

The mean is given as $3\frac{6}{23} = \frac{3*23 + 6}{23} = \frac{69+6}{23} = \frac{75}{23}$.

Therefore, $\frac{28m-9}{8m-1} = \frac{75}{23}$.

Cross-multiplying gives $23(28m-9) = 75(8m-1)$.

$644m - 207 = 600m - 75$

$644m - 600m = 207 - 75$

$44m = 132$

$m = \frac{132}{44} = 3$.

(b) Now that we know $m=3$, we can find the number of students for each mark:

Mark 1: $m+2 = 3+2 = 5$

Mark 2: $m-1 = 3-1 = 2$

Mark 3: $2m-3 = 2(3)-3 = 6-3 = 3$

Mark 4: $m+5 = 3+5 = 8$

Mark 5: $3m-4 = 3(3)-4 = 9-4 = 5$

The total number of students is $8m-1 = 8(3)-1 = 24-1 = 23$.

The marks are: 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5

The cumulative frequencies are:

Mark 1: 5

Mark 2: 5+2 = 7

Mark 3: 7+3 = 10

Mark 4: 10+8 = 18

Mark 5: 18+5 = 23

$Q_1$ is the value at the $\frac{1}{4}(n+1)$ position.

$Q_1$ position is $\frac{1}{4}(23+1) = \frac{24}{4} = 6$. The 6th student has a mark of 2, so $Q_1 = 2$.

$Q_3$ is the value at the $\frac{3}{4}(n+1)$ position.

$Q_3$ position is $\frac{3}{4}(23+1) = \frac{3}{4}(24) = 18$. The 18th student has a mark of 4, so $Q_3 = 4$.

Interquartile range (IQR) = $Q_3 - Q_1 = 4 - 2 = 2$.

(ii) Probability of selecting a student who scored at least 4 marks:

The number of students who scored at least 4 marks is the number of students who scored 4 plus the number of students who scored 5, which is $8+5 = 13$.

The total number of students is

2

3. The probability is $\frac{13}{23}$.

3. Final Answer

(a) $m = 3$

(b) (i) Interquartile range = 2

(ii) Probability of selecting a student who scored at least 4 marks = $\frac{13}{23}$