Quartile Deviation.pptx

Quartile Deviation.pptx

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  Quartile Deviation
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  Introduction One of themetrics of dispersion is quartile deviation. Before we go any further, let's review quartiles and how we might define them. The numbers that divide a list of numerical data into three-quarters, such as Q1, Q2, and Q3, are known as quartiles. The middle quarter of the three quarters measures the distribution's central point and displays data values near the midpoint or the central value.
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  Quartiles Quartiles divide acollection into four equal sections. So there are three quartiles, denoted by the letters Q1, Q2, and Q3, correspondingly. Q2 is nothing more than the median because it represents the item's location in the list and hence is a positional average. To find the quartiles of a set of data, organize data in ascending order We can measure the distribution in the median using the lower and upper quartiles. Aside from mean and median, there are other statistical measures that can divide data into particular equal portions. A median is used to divide a series into two equal sections. A data set's values can be divided into three categories:
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  Quartiles Formula Suppose, Q3is the upper quartile is the median of the upper half of the data set. Whereas, Q1 is the lower quartile and median of the lower half of the data set. Q2 is the median. Consider, we have n number of items in a data set. Then the quartiles are given by; Q1 = [(n+1)/4]th item Q2 = [(n+1)/2]th item Q3 = [3(n+1)/4]th item Hence, the formula for quartile can be given by;
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  Here, Qr isthe rth quartile l1 is the lower limit l2 is the upper limit f is the frequency c is the cumulative frequency of the class preceding the quartile class.
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  Use of Quartilesin Statistics Similarly to the median, which divides the data in half so that 50% of the estimation is below the median and 50% is above it, the quartile divides the data into quarters so that 25% of the estimation is less than the lower quartile, 50% is less than the mean, and 75% is less than the upper quartile. Typically, data is sorted from smallest to largest: ❏ First quartile: 25% from lowest to biggest number ❏ Between 25.1 and 50 percent in the second quartile (till median) ❏ The third quartile ranges from 51% to 75%. (above the median) ❏ Fourth quartile: 25% of the highest numbers
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  Quartile deviation =(Q3-Q1)/2 Question 1: Find the quartiles of the following data: 4, 6, 7, 8, 10, 23, 34. Solution: Here the numbers are arranged in the ascending order and number of items, n = 7 Lower quartile, Q1 = [(n+1)/4] th item Q1= 7+1/4 = 2nd item = 6 Median, Q2 = [(n+1)/2]th item Q2= 7+1/2 item = 4th item = 8 Upper Quartile, Q3 = [3(n+1)/4]th item Q3 = 3(7+1)/4 item = 6th item = 23
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  Question 2: Findthe Quartiles of the following age:- 23, 13, 37, 16, 26, 35, 26, 35 Solution: First, we need to arrange the numbers in increasing order. Therefore, 13, 16, 23, 26, 26, 35, 35, 37 Number of items, n = 8 Lower quartile, Q1 = [(n+1)/4] th item Q1 = 8+1/4 = 9/4 = 2.25th term P.T.O
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  From the quartileformula we can write; Q1 = 2nd term + 0.25(3rd term-2nd term) Q1= 16+0.25(23-26) = 15.25 Similarly, Median, Q2 = [(n+1)/2]th item Q2 = 8+1/2 = 9/2 = 4.5 Q2 = 4th term+0.5 (5th term-4th term) Q2= 26+0.5(26-26) = 26 And, Upper Quartile, Q3 = [3(n+1)/4]th item Q3 = 3(8+1)/4 = 6.75th term Q3 = 6th term + 0.75(7th term-6th term) Q3 = 35+0.75(35-35) = 35
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  Acknowledgement I'd want tothank my professor for providing me with this fantastic opportunity to work on this project. I'd also like to express my heartfelt gratitude to my Brainware university . I also want to thank my parents and friends for their unwavering support and assistance during this effort. It would have been extremely impossible to complete our assignment without their assistance.
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  THANK YOU