If the mean and median of the data are equal, then $xy^2$ is equal tox 0-10 10-20 20-30 30-40 40-50 f 3 6 2 x y $\Sigma f = 20$
Frequency Distribution and Data Analysis
The problem provides a frequency distribution with unknown frequencies $x$ and $y$. We are given that the mean and median of the data are equal, and the total frequency $\Sigma f = 20$. We need to find the value of $xy^2$.
Calculating Unknown Frequencies
First, use the total frequency information to establish a relationship between $x$ and $y$.
- Total frequency: $3 + 6 + 2 + x + y = 20$
- Simplifying, we get: $11 + x + y = 20$
- This gives us our first equation: $x + y = 9$ (Equation 1)
Finding the Median Class
To find the median, we need the cumulative frequencies (CF). The total number of observations $N = 20$, so $N/2 = 10$. The median class is the class where the cumulative frequency first equals or exceeds $N/2$.
| Class Interval | Frequency ($f$) | Midpoint ($x_i$) | Cumulative Frequency (CF) |
|---|---|---|---|
| 0-10 | 3 | 5 | 3 |
| 10-20 | 6 | 15 | $3 + 6 = 9$ |
| 20-30 | 2 | 25 | $9 + 2 = 11$ |
| 30-40 | $x$ | 35 | $11 + x$ |
| 40-50 | $y$ | 45 | $11 + x + y = 20$ |
- The cumulative frequency reaches 11 in the 20-30 class. Since $10$ falls within this cumulative frequency range, the median class is 20-30.
- Median Formula: Median $= L + \frac{\frac{N}{2} - CF}{f} \times h$
- Here, $L = 20$ (lower limit), $N/2 = 10$, $CF = 9$ (CF of preceding class), $f = 2$ (frequency of median class), $h = 10$ (class width).
- Median $= 20 + \frac{10 - 9}{2} \times 10 = 20 + \frac{1}{2} \times 10 = 20 + 5 = 25$.
Calculating the Mean
The mean is calculated using the midpoints of the classes.
- Mean Formula: Mean $= \frac{\Sigma (f \cdot x_i)}{\Sigma f}$
- $\Sigma (f \cdot x_i) = (3 \times 5) + (6 \times 15) + (2 \times 25) + (x \times 35) + (y \times 45)$
- $\Sigma (f \cdot x_i) = 15 + 90 + 50 + 35x + 45y = 155 + 35x + 45y$
- Mean $= \frac{155 + 35x + 45y}{20}$
Equating Mean and Median
We are given that Mean = Median.
- $\frac{155 + 35x + 45y}{20} = 25$
- $155 + 35x + 45y = 20 \times 25 = 500$
- $35x + 45y = 500 - 155 = 345$
- Divide by 5: $7x + 9y = 69$ (Equation 2)
Solving for x and y
Now, solve the system of two linear equations (Equation 1 and Equation 2).
- From Equation 1: $x = 9 - y$.
- Substitute this into Equation 2: $7(9 - y) + 9y = 69$
- $63 - 7y + 9y = 69$
- $63 + 2y = 69$
- $2y = 69 - 63 = 6$
- $y = 3$.
- Substitute $y=3$ back into $x = 9 - y$: $x = 9 - 3 = 6$.
- So, $x=6$ and $y=3$.
Calculating xy²
Finally, calculate the required value $xy^2$.
- $xy^2 = (6) \times (3)^2$
- $xy^2 = 6 \times 9$
- $xy^2 = 54$.
Similar Questions
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$x_i$ 5 7 9 10 12 15 $f_i$ 8 6 2 2 2 6
is equal to: - Suppose that the mean and median of the non-negative numbers 21, 8, 17, $a$, 51, 103, $b$, 13, 67, ($a > b$), are 40 and 21, respectively. If the mean deviation about the median is 26, then $2a$ is equal to:
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Important Questions from Measures of Dispersion and Probability
- Let $X = \{x \in \mathbb{N} : 1 \leq x \leq 19\}$ and for some $a, b \in \mathbb{R}, Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are 30 and 750, respectively, then the sum of all possible values of $b$ is
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Let the mean and variance of 8 numbers $-10, -7, -1, x, y, 9, 2, 16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.
Then the mean of 4 numbers $x, y, x + y + 1, |x - y|$ is :
- The mean deviation about the mean for the data
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is equal to: - Suppose that the mean and median of the non-negative numbers 21, 8, 17, $a$, 51, 103, $b$, 13, 67, ($a > b$), are 40 and 21, respectively. If the mean deviation about the median is 26, then $2a$ is equal to:
