$x_i$ 5 7 9 10 12 15 $f_i$ 8 6 2 2 2 6
is equal to:
Calculating Mean Deviation About Mean for Frequency Data
To find the Mean Deviation (MD) about the mean for the given data, we follow these steps:
- Calculate the total frequency ($N$).
- Calculate the mean ($\bar{x}$) of the data.
- Calculate the absolute difference between each $x_i$ and the mean ($\bar{x}$), i.e., $|x_i - \bar{x}|$.
- Multiply each absolute difference by its corresponding frequency ($f_i$), i.e., $f_i |x_i - \bar{x}|$.
- Sum these products ($\sum f_i |x_i - \bar{x}|$).
- Divide the sum by the total frequency ($N$) to get the Mean Deviation about the mean.
Step-by-Step Calculation
1. Calculate Total Frequency ($N$)
The total frequency is the sum of all $f_i$ values:
$N = \sum f_i = 8 + 6 + 2 + 2 + 2 + 6 = 26$2. Calculate the Mean ($\bar{x}$)
First, calculate the sum of the products of $x_i$ and $f_i$ ($\sum f_i x_i$):
$\sum f_i x_i = (8 \times 5) + (6 \times 7) + (2 \times 9) + (2 \times 10) + (2 \times 12) + (6 \times 15)$ $\sum f_i x_i = 40 + 42 + 18 + 20 + 24 + 90 = 234$Now, calculate the mean:
$\bar{x} = \frac{\sum f_i x_i}{N} = \frac{234}{26} = 9$3. Calculate Deviations and Weighted Deviations
We create a table to find $|x_i - \bar{x}|$ and $f_i |x_i - \bar{x}|$. Here, $\bar{x} = 9$.
| $x_i$ | $f_i$ | $|x_i - \bar{x}|$ | $f_i |x_i - \bar{x}|$ |
|---|---|---|---|
| 5 | 8 | $|5 - 9| = 4$ | $8 \times 4 = 32$ |
| 7 | 6 | $|7 - 9| = 2$ | $6 \times 2 = 12$ |
| 9 | 2 | $|9 - 9| = 0$ | $2 \times 0 = 0$ |
| 10 | 2 | $|10 - 9| = 1$ | $2 \times 1 = 2$ |
| 12 | 2 | $|12 - 9| = 3$ | $2 \times 3 = 6$ |
| 15 | 6 | $|15 - 9| = 6$ | $6 \times 6 = 36$ |
| Sum | $N = 26$ | - | $\sum f_i |x_i - \bar{x}| = 88$ |
4. Calculate Mean Deviation About the Mean (MD)
Use the formula:
$MD = \frac{\sum f_i |x_i - \bar{x}|}{N}$ $MD = \frac{88}{26}$Simplify the fraction:
$MD = \frac{44}{13}$The Mean Deviation about the mean for the given data is $\frac{44}{13}$.
Similar Questions
- 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
- A random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$ respectively, where $a, b \in \mathbf{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\frac{a}{b}$ is equal to :
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 :
- 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:
- For 10 observations $x_1, x_2, \dots, x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:
- Let the mean and the standard deviation of the observation $2, 3, 3, 4, 5, 7, a, b$ be 4 and $\sqrt{2}$ respectively. Then the mean deviation about the mode of these observations is :
- The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2, 3, 5, 10, 11, 13, 15, 21, then the mean deviation about the median of all the 10 observations is
If the mean and median of the data
x 0-10 10-20 20-30 30-40 40-50 f 3 6 2 x y $\Sigma f = 20$ are equal, then $xy^2$ is equal to
If the mean of the data: 7,8,9,7,8,7,$\lambda$,8 is 8, then the variance of this data is :-
Four dice are thrown simultaneously and the numbers shown on these dice are recorded in $2\times2$ matrices. The probability that such formed matrices have all different entries and are non-singular, is :
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
- A random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a+1}{30}, \frac{8a-1}{30}, \frac{4a+1}{30}, b$ respectively, where $a, b \in \mathbf{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^2 + \mu^2 = 2$. Then $\frac{a}{b}$ is equal to :
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 :
- 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:
- For 10 observations $x_1, x_2, \dots, x_{10}$, if $\sum_{i=1}^{10} (x_i + 2)^2 = 180$ and $\sum_{i=1}^{10} (x_i - 1)^2 = 90$, then their standard deviation is:
