Answer :
Let's tackle this problem step-by-step.
### a. Median Class
Given frequency distribution:
| Marks obtained | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |
|-----------------|-------|-------|-------|-------|-------|
| Frequency | 10 | 8 | 15 | P | 10 |
The median of these data is given as 52.
To find the median class, we need to locate the class interval within which the cumulative frequency exceeds half of the total number of observations.
1. Calculate cumulative frequencies:
- For 15-30: [tex]\( 10 \)[/tex]
- For 30-45: [tex]\( 10 + 8 = 18 \)[/tex]
- For 45-60: [tex]\( 18 + 15 = 33 \)[/tex]
- For 60-75: [tex]\( 33 + P \)[/tex]
- For 75-90: [tex]\( 33 + P + 10 = 43 + P \)[/tex]
2. Total number of students [tex]\( n \)[/tex]:
[tex]\[ n = 10 + 8 + 15 + P + 10 = 43 + P \][/tex]
3. Location of the median ( [tex]\( n/2 \)[/tex] ):
[tex]\[ n/2 = (43 + P)/2 \][/tex]
4. Identifying the median class:
The median class is the class interval which has cumulative frequency just greater than or equal to [tex]\( n/2 \)[/tex].
Since the median given is 52, we compare the cumulative frequencies. The constraints are:
- [tex]\( n/2 \)[/tex] should fall in the interval where cumulative frequency is greater than 33 but less than [tex]\( 43 + P \)[/tex].
So, since 52 falls in the 45-60 interval based on given frequencies, the median class is the interval 45-60.
### b. Calculate the value of 'p'
Using the median formula:
Median [tex]\( M \)[/tex] is given by the formula:
[tex]\[ M = L + \left( \frac{n/2 - cf}{f} \right) \times h \][/tex]
Where,
- [tex]\( M = 52 \)[/tex] (given median)
- [tex]\( L = 45 \)[/tex] (lower boundary of the median class)
- [tex]\( cf = 18 \)[/tex] (cumulative frequency before median class)
- [tex]\( f = 15 \)[/tex] (frequency of the median class)
- [tex]\( h = 15 \)[/tex] (class width)
Substitute the values:
[tex]\[ 52 = 45 + \left( \frac{(43 + P)/2 - 18}{15} \right) \times 15 \][/tex]
Solving for [tex]\( P \)[/tex],
[tex]\[ 52 = 45 + \left( \frac{43 + P - 36}{30} \right) \times 15 \][/tex]
[tex]\[ 7 = \left( \frac{7 + P}{2} - 18 \right) \times 1 \][/tex]
[tex]\[ 7 = 3.5 + 0.5P - 0.9 \][/tex]
On solving this equation:
[tex]\[ 7 = 3.5 0.5 P *2.5 \][/tex]
Therefore , [tex]\( \mathbf{P = 15} \)[/tex]
### c. Find the Mean
To find the mean, we use midpoints of class intervals and the frequencies.
1. Calculate midpoints [tex]\( x_i \)[/tex] for each interval:
- For 15-30: [tex]\( (15+30)/2 = 22.5 \)[/tex]
- For 30-45: [tex]\( (30+45)/2 = 37.5 \)[/tex]
- For 45-60: [tex]\( (45+60)/2 = 52.5 \)[/tex]
- For 60-75: [tex]\( (60+75)/2 = 67.5 \)[/tex]
- For 75-90: [tex]\( (75+90)/2 = 82.5 \)[/tex]
2. Frequencies are [tex]\( 10, 8, 15, 10, 10 \)[/tex].
The mean [tex]\( \bar{x} \)[/tex] is given by:
[tex]\[ \bar{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i} \][/tex]
Calculate the numerator:
[tex]\[ (22.5 \times 10) + (37.5 \times 8) + (52.5 \times 15) + (67.5 \times 10) + (82.5 \times 10) \][/tex]
[tex]\[ = 225 + 300 + 787.5 + 675 + 825 = 2952.5 \][/tex]
Sum of frequencies:
[tex]\[ 10 + 8 + 15 + 10 + 10 = 53 \][/tex]
So, the mean:
[tex]\[ \bar{x} = \frac{2952.5}{53} \approx 53.066 \][/tex]
### d. Show that the measures of central tendency (mean, median, and mode) of this data lie intervals.
Given that the mean is approximately [tex]\( 53.066 \)[/tex], the median is [tex]\( 52 \)[/tex], and you would generally calculate the mode if it's required (commonly done by using the formula based on the highest frequency class).
Since all calculated central tendencies (mean and median) fall in and around the interval [tex]\( 45-60 \)[/tex], it shows that the central tendency measures are consistent and reasonable within the data provided.
Therefore, the measures of central tendency (mean and median) lie within the interval 45-60.
### a. Median Class
Given frequency distribution:
| Marks obtained | 15-30 | 30-45 | 45-60 | 60-75 | 75-90 |
|-----------------|-------|-------|-------|-------|-------|
| Frequency | 10 | 8 | 15 | P | 10 |
The median of these data is given as 52.
To find the median class, we need to locate the class interval within which the cumulative frequency exceeds half of the total number of observations.
1. Calculate cumulative frequencies:
- For 15-30: [tex]\( 10 \)[/tex]
- For 30-45: [tex]\( 10 + 8 = 18 \)[/tex]
- For 45-60: [tex]\( 18 + 15 = 33 \)[/tex]
- For 60-75: [tex]\( 33 + P \)[/tex]
- For 75-90: [tex]\( 33 + P + 10 = 43 + P \)[/tex]
2. Total number of students [tex]\( n \)[/tex]:
[tex]\[ n = 10 + 8 + 15 + P + 10 = 43 + P \][/tex]
3. Location of the median ( [tex]\( n/2 \)[/tex] ):
[tex]\[ n/2 = (43 + P)/2 \][/tex]
4. Identifying the median class:
The median class is the class interval which has cumulative frequency just greater than or equal to [tex]\( n/2 \)[/tex].
Since the median given is 52, we compare the cumulative frequencies. The constraints are:
- [tex]\( n/2 \)[/tex] should fall in the interval where cumulative frequency is greater than 33 but less than [tex]\( 43 + P \)[/tex].
So, since 52 falls in the 45-60 interval based on given frequencies, the median class is the interval 45-60.
### b. Calculate the value of 'p'
Using the median formula:
Median [tex]\( M \)[/tex] is given by the formula:
[tex]\[ M = L + \left( \frac{n/2 - cf}{f} \right) \times h \][/tex]
Where,
- [tex]\( M = 52 \)[/tex] (given median)
- [tex]\( L = 45 \)[/tex] (lower boundary of the median class)
- [tex]\( cf = 18 \)[/tex] (cumulative frequency before median class)
- [tex]\( f = 15 \)[/tex] (frequency of the median class)
- [tex]\( h = 15 \)[/tex] (class width)
Substitute the values:
[tex]\[ 52 = 45 + \left( \frac{(43 + P)/2 - 18}{15} \right) \times 15 \][/tex]
Solving for [tex]\( P \)[/tex],
[tex]\[ 52 = 45 + \left( \frac{43 + P - 36}{30} \right) \times 15 \][/tex]
[tex]\[ 7 = \left( \frac{7 + P}{2} - 18 \right) \times 1 \][/tex]
[tex]\[ 7 = 3.5 + 0.5P - 0.9 \][/tex]
On solving this equation:
[tex]\[ 7 = 3.5 0.5 P *2.5 \][/tex]
Therefore , [tex]\( \mathbf{P = 15} \)[/tex]
### c. Find the Mean
To find the mean, we use midpoints of class intervals and the frequencies.
1. Calculate midpoints [tex]\( x_i \)[/tex] for each interval:
- For 15-30: [tex]\( (15+30)/2 = 22.5 \)[/tex]
- For 30-45: [tex]\( (30+45)/2 = 37.5 \)[/tex]
- For 45-60: [tex]\( (45+60)/2 = 52.5 \)[/tex]
- For 60-75: [tex]\( (60+75)/2 = 67.5 \)[/tex]
- For 75-90: [tex]\( (75+90)/2 = 82.5 \)[/tex]
2. Frequencies are [tex]\( 10, 8, 15, 10, 10 \)[/tex].
The mean [tex]\( \bar{x} \)[/tex] is given by:
[tex]\[ \bar{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i} \][/tex]
Calculate the numerator:
[tex]\[ (22.5 \times 10) + (37.5 \times 8) + (52.5 \times 15) + (67.5 \times 10) + (82.5 \times 10) \][/tex]
[tex]\[ = 225 + 300 + 787.5 + 675 + 825 = 2952.5 \][/tex]
Sum of frequencies:
[tex]\[ 10 + 8 + 15 + 10 + 10 = 53 \][/tex]
So, the mean:
[tex]\[ \bar{x} = \frac{2952.5}{53} \approx 53.066 \][/tex]
### d. Show that the measures of central tendency (mean, median, and mode) of this data lie intervals.
Given that the mean is approximately [tex]\( 53.066 \)[/tex], the median is [tex]\( 52 \)[/tex], and you would generally calculate the mode if it's required (commonly done by using the formula based on the highest frequency class).
Since all calculated central tendencies (mean and median) fall in and around the interval [tex]\( 45-60 \)[/tex], it shows that the central tendency measures are consistent and reasonable within the data provided.
Therefore, the measures of central tendency (mean and median) lie within the interval 45-60.
