Answer :
Sure, let’s solve each part of the question step by step.
### Part (a)
a) ‘i’ in the median formula [tex]$\left(M_d\right)=L+\cfrac{\cfrac{N}{2}-cf}{f} \times i$[/tex] represents the class interval width, which is the difference between the upper and lower boundaries of any class.
### Part (b)
b) To find the median from the given data:
1. Determine N (total number of students):
[tex]\[ N = 4 + 6 + 8 + 5 + 7 + 10 = 40. \][/tex]
2. Find [tex]$\cfrac{N}{2}$[/tex]:
[tex]\[ \cfrac{N}{2} = \cfrac{40}{2} = 20. \][/tex]
3. Locate the median class (the class where the cumulative frequency just exceeds [tex]$\cfrac{N}{2}$[/tex]):
- Cumulative frequencies:
[tex]\[ 4, \quad 4 + 6 = 10, \quad 10 + 8 = 18, \quad 18 + 5 = 23, \quad 23 + 7 = 30, \quad 30 + 10 = 40. \][/tex]
The median class is [tex]$30 - 40$[/tex] because cumulative frequency surpasses 20 in this class.
4. Identify the required values from the median class:
- Lower boundary of the median class, [tex]\(L = 30\)[/tex].
- Cumulative frequency before the median class ([tex]\(cf\)[/tex]): 18.
- Frequency of the median class ([tex]\(f\)[/tex]): 5.
5. Class interval width (‘i’):
[tex]\[ i = 40 - 30 = 10. \][/tex]
6. Calculate the median using the median formula:
[tex]\[ M_d = L + \cfrac{ \cfrac{N}{2} - cf }{ f } \times i \][/tex]
[tex]\[ M_d = 30 + \cfrac{ 20 - 18 }{ 5 } \times 10 \][/tex]
[tex]\[ M_d = 30 + \cfrac{2}{5} \times 10 \][/tex]
[tex]\[ M_d = 30 + 4 = 34. \][/tex]
### Part (c)
c) To calculate the mean score from the given data:
1. Find the midpoints (x) of each class interval:
- [tex]$10 - 20$[/tex]: Midpoint [tex]\( = \cfrac{10+20}{2} = 15 \)[/tex]
- [tex]$20 - 30$[/tex]: Midpoint [tex]\( = \cfrac{20+30}{2} = 25 \)[/tex]
- [tex]$30 - 40$[/tex]: Midpoint [tex]\( = \cfrac{30+40}{2} = 35 \)[/tex]
- [tex]$40 - 50$[/tex]: Midpoint [tex]\( = \cfrac{40+50}{2} = 45 \)[/tex]
- [tex]$50 - 60$[/tex]: Midpoint [tex]\( = \cfrac{50+60}{2} = 55 \)[/tex]
2. Multiply each midpoint by the frequency of its class and sum them up (Σ(fx)):
[tex]\[ Σ(fx) = 4 \times 15 + 6 \times 25 + 8 \times 35 + 5 \times 45 + 7 \times 55 + 10 \times 55 \][/tex]
[tex]\[ Σ(fx) = 60 + 150 + 280 + 225 + 385 + 550 = 1755. \][/tex]
3. Calculate the mean:
[tex]\[ \text{Mean} = \cfrac{ Σ(fx) }{ N } = \cfrac{ 1755 }{ 40 } = 43.75. \][/tex]
### Part (d)
d) To find the number of students in the [tex]$50-60$[/tex] class to make the average score 30:
1. Let the number of students needed in the 50-60 class be [tex]\( x \)[/tex].
2. Calculate the current total sum of scores excluding the 50-60 class:
[tex]\[ Σ(fx)_{\text{current}} = 1755 - (10 \times 55) = 1755 - 550 = 1205. \][/tex]
3. Calculate the new required target sum of scores:
[tex]\[ \text{New required total sum of scores} = 30 \times (N + x). \][/tex]
[tex]\[ 30 \times (40 - 10 + x) = 30 \times (30 + x). \][/tex]
4. Form an equation and solve for [tex]\( x \)[/tex]:
[tex]\[ \text{Target Total} = 30 \times (30 + x) \][/tex]
[tex]\[ 1205 + 55x = 30 \times (30 + x) \][/tex]
[tex]\[ 1205 + 55x = 900 + 30x \][/tex]
[tex]\[ 1205 - 900 = 30x - 55x \][/tex]
[tex]\[ 305 = 25x \][/tex]
[tex]\[ x = \cfrac{305}{25} = 12.2 \approx 13. \][/tex]
In summary:
- (a) The interval.
- (b) Median: [tex]\(34\)[/tex].
- (c) Mean: [tex]\(43.75\)[/tex].
- (d) The number of students needed in the 50-60 class to make the average score 30 is approximately 13 students.
### Part (a)
a) ‘i’ in the median formula [tex]$\left(M_d\right)=L+\cfrac{\cfrac{N}{2}-cf}{f} \times i$[/tex] represents the class interval width, which is the difference between the upper and lower boundaries of any class.
### Part (b)
b) To find the median from the given data:
1. Determine N (total number of students):
[tex]\[ N = 4 + 6 + 8 + 5 + 7 + 10 = 40. \][/tex]
2. Find [tex]$\cfrac{N}{2}$[/tex]:
[tex]\[ \cfrac{N}{2} = \cfrac{40}{2} = 20. \][/tex]
3. Locate the median class (the class where the cumulative frequency just exceeds [tex]$\cfrac{N}{2}$[/tex]):
- Cumulative frequencies:
[tex]\[ 4, \quad 4 + 6 = 10, \quad 10 + 8 = 18, \quad 18 + 5 = 23, \quad 23 + 7 = 30, \quad 30 + 10 = 40. \][/tex]
The median class is [tex]$30 - 40$[/tex] because cumulative frequency surpasses 20 in this class.
4. Identify the required values from the median class:
- Lower boundary of the median class, [tex]\(L = 30\)[/tex].
- Cumulative frequency before the median class ([tex]\(cf\)[/tex]): 18.
- Frequency of the median class ([tex]\(f\)[/tex]): 5.
5. Class interval width (‘i’):
[tex]\[ i = 40 - 30 = 10. \][/tex]
6. Calculate the median using the median formula:
[tex]\[ M_d = L + \cfrac{ \cfrac{N}{2} - cf }{ f } \times i \][/tex]
[tex]\[ M_d = 30 + \cfrac{ 20 - 18 }{ 5 } \times 10 \][/tex]
[tex]\[ M_d = 30 + \cfrac{2}{5} \times 10 \][/tex]
[tex]\[ M_d = 30 + 4 = 34. \][/tex]
### Part (c)
c) To calculate the mean score from the given data:
1. Find the midpoints (x) of each class interval:
- [tex]$10 - 20$[/tex]: Midpoint [tex]\( = \cfrac{10+20}{2} = 15 \)[/tex]
- [tex]$20 - 30$[/tex]: Midpoint [tex]\( = \cfrac{20+30}{2} = 25 \)[/tex]
- [tex]$30 - 40$[/tex]: Midpoint [tex]\( = \cfrac{30+40}{2} = 35 \)[/tex]
- [tex]$40 - 50$[/tex]: Midpoint [tex]\( = \cfrac{40+50}{2} = 45 \)[/tex]
- [tex]$50 - 60$[/tex]: Midpoint [tex]\( = \cfrac{50+60}{2} = 55 \)[/tex]
2. Multiply each midpoint by the frequency of its class and sum them up (Σ(fx)):
[tex]\[ Σ(fx) = 4 \times 15 + 6 \times 25 + 8 \times 35 + 5 \times 45 + 7 \times 55 + 10 \times 55 \][/tex]
[tex]\[ Σ(fx) = 60 + 150 + 280 + 225 + 385 + 550 = 1755. \][/tex]
3. Calculate the mean:
[tex]\[ \text{Mean} = \cfrac{ Σ(fx) }{ N } = \cfrac{ 1755 }{ 40 } = 43.75. \][/tex]
### Part (d)
d) To find the number of students in the [tex]$50-60$[/tex] class to make the average score 30:
1. Let the number of students needed in the 50-60 class be [tex]\( x \)[/tex].
2. Calculate the current total sum of scores excluding the 50-60 class:
[tex]\[ Σ(fx)_{\text{current}} = 1755 - (10 \times 55) = 1755 - 550 = 1205. \][/tex]
3. Calculate the new required target sum of scores:
[tex]\[ \text{New required total sum of scores} = 30 \times (N + x). \][/tex]
[tex]\[ 30 \times (40 - 10 + x) = 30 \times (30 + x). \][/tex]
4. Form an equation and solve for [tex]\( x \)[/tex]:
[tex]\[ \text{Target Total} = 30 \times (30 + x) \][/tex]
[tex]\[ 1205 + 55x = 30 \times (30 + x) \][/tex]
[tex]\[ 1205 + 55x = 900 + 30x \][/tex]
[tex]\[ 1205 - 900 = 30x - 55x \][/tex]
[tex]\[ 305 = 25x \][/tex]
[tex]\[ x = \cfrac{305}{25} = 12.2 \approx 13. \][/tex]
In summary:
- (a) The interval.
- (b) Median: [tex]\(34\)[/tex].
- (c) Mean: [tex]\(43.75\)[/tex].
- (d) The number of students needed in the 50-60 class to make the average score 30 is approximately 13 students.
