statistics all formula class 11
Answers
Step-by-step explanation:
Statistics Formulas
Population mean = μ = ( Σ Xi ) / N.
Population standard deviation = σ = sqrt [ Σ ( Xi - μ )2 / N ]
Population variance = σ2 = Σ ( Xi - μ )2 / N.
Variance of population proportion = σP2 = PQ / n.
Standardized score = Z = (X - μ) / σ
hope it helps you dude ☺
Answer:
Measures of Dispersion
The dispersion or scatter in the data is measured based on the observations and the types of the measure of central tendency. The different types of measures of dispersion are:
Range
Quartile deviation
Mean deviation
Standard deviation.
In Class 11 statistics, you get a clear knowledge of all the measures of dispersion except quartile deviation, which will be studied in your higher classes.
Range
The range is the difference between the maximum value and the minimum value
Range = Maximum Value – Minimum Value
Mean Deviation
Mean deviation is the basic measure of deviations from value, and the value is generally a mean value or a median value. In order to find out the mean deviation, first take the mean of deviation for the observations from value is d = x – a Here x is the observation, and a is the fixed value.
The basic formula to find out the mean deviation is
Mean Deviation = Sum of absolute values of deviations from ‘a’ / Number of observations
Mean Deviation for Ungrouped Data
Calculation of mean deviation for ungrouped data involves the following steps :
Let us assume the observations x1, x2, x3, …..xn
Step 1: Calculate the central tendency measures about to find the mean deviation and let it be ‘a’.
Step 2: Find the deviation of xi from a, i.e., x1 – a, x2– a, x3 – a,. . . , xn– a
Step 3: Find the absolute values of the deviations, i.e., | x1 – a |, | x2– a |, |x3 – a|,. . . ,|xn– a| and the drop the minus sign (–), if it is there,
Step 4: calculate the mean of the absolute values of the deviations . This mean obtained is the mean deviation about a, i.e.,
M.D(a)=∑ni=1|xi−a|n
M.D(x¯)=∑ni=1∣∣xi−x¯∣∣n, where x¯ = Mean
M.D(M)=∑ni=1|xi−M|n, Where M = Median
Mean Deviation for Grouped Data
The data can be grouped into two ways namely,
Discrete frequency distribution
Continuous frequency distribution
The methods of finding the mean deviation for both the types are given below
Discrete Frequency Distribution
Here, the given data consist of n distinct values x1, x2, x3,….xn has frequencies f1, f2, f3,….fn respectively. This data is represented in the tabular form as and is called discrete frequency distribution, and the data are given below
x x1 x2 x3 …… xn
f f1 f2 f3 …… fn
Mean Deviation About Mean
First find the mean x¯, using the given formula :
x¯=∑ni=1xifi∑ni=1fi=1N∑ni=1xifi
Where the numerator denotes the sum of products of observations xi with the respective frequencies fi and the denominator denotes the sum of frequencies.
Now take the absolute values, |xi−x¯| , for all i = 1, 2, 3, ..n
Therefore, the required mean deviation about the mean is given by
M.D(x¯)=∑ni=1fi∣∣xi−x¯∣∣∑ni=1fi=1N∑ni=1fi|xi−x¯|
Mean Deviation About Median
To find the mean deviation about median for the given discrete frequency distribution. First arrange the observation in ascending order to get cumulative frequency which is equal or greater than N/2, where N is the sum of the frequencies.
Therefore, the mean deviation for median is given by,
M.D(M)=1N∑ni=1fi|xi−M|
Continuous Frequency Distribution
To find the mean deviation for the continuous frequency distribution, assume that the frequency in each class is centred at its midpoint. After finding the midpoint, proceed further to find the mean deviation similar to the discrete frequency distribution.
Standard Deviation
The positive square root of variance is called standard deviation (S.D) and it is denoted by the symbol, and the formula to find S.D is given by
σ=1n∑ni=1(xi−x¯)2−−−−−−−−−−−−−√
Where the variance is denoted as 2 and it is given by
σ2=1n∑ni=1(xi−x¯)2
Class 11 Statistics Examples
Here, statistics solved problem is given below. Go through once to get a clear idea to solve the problem.