Answer :
Sure, let's work through the problem step-by-step:
### 2.1 Modal Class
The modal class is the class interval with the highest frequency.
Given frequencies for each interval:
- [tex]\( 90 \leq x < 100 \)[/tex]: Frequency = 4
- [tex]\( 100 \leq x < 110 \)[/tex]: Frequency = 8
- [tex]\( 110 \leq x < 120 \)[/tex]: Frequency = 7
- [tex]\( 120 \leq x < 130 \)[/tex]: Frequency = 5
- [tex]\( 130 \leq x < 140 \)[/tex]: Frequency = 4
- [tex]\( 140 \leq x < 150 \)[/tex]: Frequency = 2
From these, the interval [tex]\( 100 \leq x < 110 \)[/tex] has the highest frequency, which is 8.
Modal Class: [tex]\( 100 \leq x < 110 \)[/tex]
### 2.2 Median Class
To find the median class, we first need to determine the position of the median in the cumulative frequency distribution.
First, calculate the cumulative frequencies:
- Cumulative frequency up to [tex]\( 90 \leq x < 100 \)[/tex]: 4
- Cumulative frequency up to [tex]\( 100 \leq x < 110 \)[/tex]: 4 + 8 = 12
- Cumulative frequency up to [tex]\( 110 \leq x < 120 \)[/tex]: 12 + 7 = 19
- Cumulative frequency up to [tex]\( 120 \leq x < 130 \)[/tex]: 19 + 5 = 24
- Cumulative frequency up to [tex]\( 130 \leq x < 140 \)[/tex]: 24 + 4 = 28
- Cumulative frequency up to [tex]\( 140 \leq x < 150 \)[/tex]: 28 + 2 = 30
Total number of observations (N) = 30
The median position is given by [tex]\( \frac{N}{2} = \frac{30}{2} = 15 \)[/tex].
The interval where the 15th observation lies is found by looking at the cumulative frequencies:
- From the cumulative frequencies, we see that 15 falls in the cumulative frequency range of 12 to 19, which corresponds to the interval [tex]\( 110 \leq x < 120 \)[/tex].
Median Class: [tex]\( 110 \leq x < 120 \)[/tex]
### 2.3 Mean IQ Score
The mean IQ score is estimated using the midpoints of each class interval and the corresponding frequencies.
First, calculate the midpoints for each interval:
- Midpoint of [tex]\( 90 \leq x < 100 \)[/tex] = [tex]\( \frac{90 + 100}{2} = 95 \)[/tex]
- Midpoint of [tex]\( 100 \leq x < 110 \)[/tex] = [tex]\( \frac{100 + 110}{2} = 105 \)[/tex]
- Midpoint of [tex]\( 110 \leq x < 120 \)[/tex] = [tex]\( \frac{110 + 120}{2} = 115 \)[/tex]
- Midpoint of [tex]\( 120 \leq x < 130 \)[/tex] = [tex]\( \frac{120 + 130}{2} = 125 \)[/tex]
- Midpoint of [tex]\( 130 \leq x < 140 \)[/tex] = [tex]\( \frac{130 + 140}{2} = 135 \)[/tex]
- Midpoint of [tex]\( 140 \leq x < 150 \)[/tex] = [tex]\( \frac{140 + 150}{2} = 145 \)[/tex]
Next, multiply each midpoint by the corresponding frequency and sum these products:
- [tex]\( 95 \times 4 = 380 \)[/tex]
- [tex]\( 105 \times 8 = 840 \)[/tex]
- [tex]\( 115 \times 7 = 805 \)[/tex]
- [tex]\( 125 \times 5 = 625 \)[/tex]
- [tex]\( 135 \times 4 = 540 \)[/tex]
- [tex]\( 145 \times 2 = 290 \)[/tex]
Total sum of products = 380 + 840 + 805 + 625 + 540 + 290 = 3480
Finally, divide by the total number of observations:
- Mean IQ score = [tex]\( \frac{3480}{30} = 116.0 \)[/tex]
Mean IQ Score: 116.0
This concludes our step-by-step solution to the question.
### 2.1 Modal Class
The modal class is the class interval with the highest frequency.
Given frequencies for each interval:
- [tex]\( 90 \leq x < 100 \)[/tex]: Frequency = 4
- [tex]\( 100 \leq x < 110 \)[/tex]: Frequency = 8
- [tex]\( 110 \leq x < 120 \)[/tex]: Frequency = 7
- [tex]\( 120 \leq x < 130 \)[/tex]: Frequency = 5
- [tex]\( 130 \leq x < 140 \)[/tex]: Frequency = 4
- [tex]\( 140 \leq x < 150 \)[/tex]: Frequency = 2
From these, the interval [tex]\( 100 \leq x < 110 \)[/tex] has the highest frequency, which is 8.
Modal Class: [tex]\( 100 \leq x < 110 \)[/tex]
### 2.2 Median Class
To find the median class, we first need to determine the position of the median in the cumulative frequency distribution.
First, calculate the cumulative frequencies:
- Cumulative frequency up to [tex]\( 90 \leq x < 100 \)[/tex]: 4
- Cumulative frequency up to [tex]\( 100 \leq x < 110 \)[/tex]: 4 + 8 = 12
- Cumulative frequency up to [tex]\( 110 \leq x < 120 \)[/tex]: 12 + 7 = 19
- Cumulative frequency up to [tex]\( 120 \leq x < 130 \)[/tex]: 19 + 5 = 24
- Cumulative frequency up to [tex]\( 130 \leq x < 140 \)[/tex]: 24 + 4 = 28
- Cumulative frequency up to [tex]\( 140 \leq x < 150 \)[/tex]: 28 + 2 = 30
Total number of observations (N) = 30
The median position is given by [tex]\( \frac{N}{2} = \frac{30}{2} = 15 \)[/tex].
The interval where the 15th observation lies is found by looking at the cumulative frequencies:
- From the cumulative frequencies, we see that 15 falls in the cumulative frequency range of 12 to 19, which corresponds to the interval [tex]\( 110 \leq x < 120 \)[/tex].
Median Class: [tex]\( 110 \leq x < 120 \)[/tex]
### 2.3 Mean IQ Score
The mean IQ score is estimated using the midpoints of each class interval and the corresponding frequencies.
First, calculate the midpoints for each interval:
- Midpoint of [tex]\( 90 \leq x < 100 \)[/tex] = [tex]\( \frac{90 + 100}{2} = 95 \)[/tex]
- Midpoint of [tex]\( 100 \leq x < 110 \)[/tex] = [tex]\( \frac{100 + 110}{2} = 105 \)[/tex]
- Midpoint of [tex]\( 110 \leq x < 120 \)[/tex] = [tex]\( \frac{110 + 120}{2} = 115 \)[/tex]
- Midpoint of [tex]\( 120 \leq x < 130 \)[/tex] = [tex]\( \frac{120 + 130}{2} = 125 \)[/tex]
- Midpoint of [tex]\( 130 \leq x < 140 \)[/tex] = [tex]\( \frac{130 + 140}{2} = 135 \)[/tex]
- Midpoint of [tex]\( 140 \leq x < 150 \)[/tex] = [tex]\( \frac{140 + 150}{2} = 145 \)[/tex]
Next, multiply each midpoint by the corresponding frequency and sum these products:
- [tex]\( 95 \times 4 = 380 \)[/tex]
- [tex]\( 105 \times 8 = 840 \)[/tex]
- [tex]\( 115 \times 7 = 805 \)[/tex]
- [tex]\( 125 \times 5 = 625 \)[/tex]
- [tex]\( 135 \times 4 = 540 \)[/tex]
- [tex]\( 145 \times 2 = 290 \)[/tex]
Total sum of products = 380 + 840 + 805 + 625 + 540 + 290 = 3480
Finally, divide by the total number of observations:
- Mean IQ score = [tex]\( \frac{3480}{30} = 116.0 \)[/tex]
Mean IQ Score: 116.0
This concludes our step-by-step solution to the question.
