Consider the following frequency distribution : If mean = $\frac{309}{22}$ and median=$14$, then the value $(a - b)^2$ is equal to ________.Class 0 - 6 6 - 12 12 - 18 18 - 24 24 - 30 Frequency a b 12 9 5
Solving Frequency Distribution Problems with Mean and Median
We are given a frequency distribution and need to find the value of $(a - b)^2$. The mean and median are provided.
Given Data:
| Class | Frequency |
|---|---|
| 0 - 6 | $a$ |
| 6 - 12 | $b$ |
| 12 - 18 | 12 |
| 18 - 24 | 9 |
| 24 - 30 | 5 |
Mean = $\frac{309}{22}$
Median = 14
Median Calculation
First, let's find the total number of observations, $N$.
$N = a + b + 12 + 9 + 5 = a + b + 26$
The median value is 14. Since the class width is 6, the median (14) falls within the 12 - 18 class.
The formula for the median is:
Median = $L + \frac{\frac{N}{2} - CF_{prev}}{f} \times h$
- $L$ (Lower limit of the median class) = 12
- $h$ (Class width) = 6
- $f$ (Frequency of the median class) = 12
- $CF_{prev}$ (Cumulative frequency of the class preceding the median class) = $a + b$
- $N = a + b + 26$
Substituting the values into the formula:
$14 = 12 + \frac{\frac{a + b + 26}{2} - (a + b)}{12} \times 6$
$2 = \frac{\frac{a + b + 26 - 2(a + b)}{2}}{2}$
$4 = \frac{26 - a - b}{2}$
$8 = 26 - a - b$
Rearranging gives our first equation:
$a + b = 18$ (Equation 1)
Mean Calculation
Now, let's use the mean formula. First, we find the class marks ($x_i$):
- Class 0-6: $x_i = \frac{0+6}{2} = 3$
- Class 6-12: $x_i = \frac{6+12}{2} = 9$
- Class 12-18: $x_i = \frac{12+18}{2} = 15$
- Class 18-24: $x_i = \frac{18+24}{2} = 21$
- Class 24-30: $x_i = \frac{24+30}{2} = 27$
The mean formula is:
Mean = $\frac{\sum f_i x_i}{N}$
Calculate $\sum f_i x_i$:
$\sum f_i x_i = (a \times 3) + (b \times 9) + (12 \times 15) + (9 \times 21) + (5 \times 27)$
$\sum f_i x_i = 3a + 9b + 180 + 189 + 135 = 3a + 9b + 504$
We know $N = a + b + 26$. From Equation 1, $a + b = 18$. So, $N = 18 + 26 = 44$.
Substitute the values into the mean formula:
$\frac{309}{22} = \frac{3a + 9b + 504}{44}$
Multiply both sides by 44:
$(\frac{309}{22}) \times 44 = 3a + 9b + 504$
$309 \times 2 = 3a + 9b + 504$
$618 = 3a + 9b + 504$
$3a + 9b = 618 - 504$
$3a + 9b = 114$
Divide by 3:
$a + 3b = 38$ (Equation 2)
Solving for 'a' and 'b'
We have a system of two linear equations:
- $a + b = 18$
- $a + 3b = 38$
Subtract Equation 1 from Equation 2:
$(a + 3b) - (a + b) = 38 - 18$
$2b = 20$
$b = 10$
Substitute $b=10$ into Equation 1:
$a + 10 = 18$
$a = 8$
Final Calculation
We need to find the value of $(a - b)^2$.
$(a - b)^2 = (8 - 10)^2$
$(a - b)^2 = (-2)^2$
$(a - b)^2 = 4$
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