The problem provides a table showing the distribution of marks scored by students. The marks range from 1 to 5, and the number of students scoring each mark is given in terms of $m$. We are given that the mean mark is $3 \frac{6}{23}$. (a) We need to find the value of $m$. (b) We need to find: (i) The interquartile range. (ii) 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. The marks range from 1 to 5, and the number of students scoring each mark is given in terms of $m$. We are given that the mean mark is $3 \frac{6}{23}$.

(a) We need to find the value of $m$.

(b) We need to find:

(i) The interquartile range.

(ii) The probability of selecting a student who scored at least 4 marks.

2. Solution Steps

(a) To find the value of $m$, we use the formula for the mean:

$Mean = \frac{\sum (marks \times number\,of\,students)}{\sum number\,of\,students}$

The sum of (marks $\times$ number of students) is:

$1(m+2) + 2(m-1) + 3(2m-3) + 4(m+5) + 5(3m-4)$

$= m+2 + 2m-2 + 6m-9 + 4m+20 + 15m-20$

$= (1+2+6+4+15)m + (2-2-9+20-20)$

$= 28m - 9$

The sum of the number of students is:

$(m+2) + (m-1) + (2m-3) + (m+5) + (3m-4)$

$= (1+1+2+1+3)m + (2-1-3+5-4)$

$= 8m - 1$

We are given that the mean is $3 \frac{6}{23} = \frac{3 \times 23 + 6}{23} = \frac{69+6}{23} = \frac{75}{23}$.

So, we have:

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

Cross-multiply:

$23(28m - 9) = 75(8m - 1)$

$644m - 207 = 600m - 75$

$644m - 600m = 207 - 75$

$44m = 132$

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

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

Marks | 1 | 2 | 3 | 4 | 5

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

Number of Students | $3+2=5$ | $3-1=2$ | $2(3)-3=3$ | $3+5=8$ | $3(3)-4=5$

Total number of students = $5+2+3+8+5 = 23$.

(i) Interquartile range:

First, we need to find the median (Q2). Since there are 23 students, the median is the $\frac{23+1}{2} = 12$th student.

Marks | 1 | 2 | 3 | 4 | 5

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

Cumulative Frequency | 5 | 7 | 10 | 18 | 23

The 12th student scored a mark of

4. So, Q2 =

4.

Q1 is the median of the lower half of the data (excluding the median itself). The lower half consists of the first 11 students.

The median of the first 11 students is the $\frac{11+1}{2} = 6$th student. From the cumulative frequencies, the 6th student scored a

2. So, Q1 =

2.

Q3 is the median of the upper half of the data (excluding the median itself). The upper half consists of the last 11 students. So we want the $\frac{23+1}{2} + \frac{11+1}{2} = 12+6 = 18$th student. Thus we are seeking the median of the 12th to 23rd students. Since the 18th student scored a 4, then Q3=

4. We could also note that the upper half has cumulative frequences from student 12 to 23, therefore starting with marks of 4 and the 6th value into this group being

4.

Interquartile range = Q3 - Q1 = 4 - 2 =

2.

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

Students who scored at least 4 marks are those who scored 4 or

5. Number of students who scored 4 or 5 = $8+5 = 13$.

Total number of students =

2

3. Probability = $\frac{13}{23}$.

3. Final Answer

(a) $m = 3$

(b) (i) Interquartile range = 2

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