OTHER PARTITION OR POSITIONAL MEASURES

Median of a distribution divides it into two equal parts. It is also possible to divide it into more than two equal parts. The values that divide a distribution into more than two equal parts are commonly known as partition values or fractiles. Some important partition values are discussed in the following sections.
Quartiles
The values of a variable that divide a distribution into four equal parts are called quartiles. Since three values are needed to divide a distribution into four parts, there are three quartiles, viz. Q1, Q2 and Q3, known as the first, second and the third quartile respectively. For a discrete distribution, the first quartile (Q1) is defined as that value of the variate such that at least 25% of the observations are less than or equal to it and at least 75% of the observations are greater than or equal to it.
For a continuous or grouped frequency distribution, Q1 is that value of the variate such that the area under the histogram to the left of the ordinate at Q1 is 25% and the area to its right is 75%. The formula for the computation of Q1 can be written by making suitable changes in the formula of median. After locating the first quartile class, the formula for Q1 can be written as follows:
Here, LQ1 is lower limit of the first quartile class, h is its width, fQ1 is its frequency and C is cumulative frequency of classes preceding the first quartile class.
By definition, the second quartile is median of the distribution. The third quartile (Q3) of a distribution can also be defined in a similar manner.
For a discrete distribution, Q3 is that value of the variate such that at least 75% of the observations are less than or equal to it and at least 25% of the observations are greater than or equal to it.
For a grouped frequency distribution, Q3 is that value of the variate such that area under the histogram to the left of the ordinate at Q3 is 75% and the area to its right is 25%. The formula for computation of Q3 can be written as
Deciles Deciles divide a distribution into 10 equal parts and there are, in all, 9 deciles denoted as D1, D2, ...... D9 respectively.
For a discrete distribution, the i th decile Di is that value of the variate such that at least (10i)% of the observation are less than or equal to it and at least (100 - 10i)% of the observations are greater than or equal to it (i = 1, 2, ...... 9).
For a continuous or grouped frequency distribution, Di is that value of the variate such that the area under the histogram to the left of the ordinate at Di is (10i)% and the area to its right is (100 - 10i)%. The formula for the ith decile can be written as
Percentiles Percentiles divide a distribution into 100 equal parts and there are, in all, 99 percentiles denoted as P1, P2, ...... P25, ...... P40, ...... P60, ...... P99 respectively.
For a discrete distribution, the kth percentile Pk is that value of the variate such that at least k% of the observations are less than or equal to it and at least (100 - k)% of the observations are greater than or equal to it.
For a grouped frequency distribution, Pk is that value of the variate such that the area under the histogram to the left of the ordinate at Pk is k% and the area to its right is (100 - k)% . The formula for the kth percentile can be written as
Remarks :
(i) We may note here that P25 = Q1, P50 = D5 = Q2 = Md, P75 = Q3, P10 = D1, P20 = D2, etc. (ii) In continuation of the above, the partition values are known as Quintiles (Octiles) if a distribution is divided in to 5 (8) equal parts. (iii) The formulae for various partition values of a grouped frequency distribution, given so far, are based on 'less than' type cumulative frequencies. The corresponding formulae based on 'greater than' type cumulative frequencies can be written in a similar manner, as given below:
Here UQ1 ,UQ3 ,UDi ,UPK are the upper limits of the corresponding classes and C denotes the greater than type cumulative frequencies.
Example: Locate Median, Q1, Q3, D4, D7, P15, P60 and P90 from the following data:
Solution: First we calculate the cumulative frequencies, as in the following table:
Determination of Median: Here N/2= 100. From the cumulative frequency column, we note that there are 102 (greater than 50% of the total) observations that are less than or equal to 78 and there are 133 observations that are greater than or equal to 78. Therefore, Md = Rs 78.Determination of Qand Q3First we determine N/4 which is equal to 50. From the cumulative frequency column, we note that there are 67 (which is greater than 25% of the total) observations that are less than or equal to 77 and there are 165 (which is greater than 75% of the total) observations that are greater than or equal to 77. Therefore, Q1 = Rs 77. Similarly, Q3 = Rs 80.Determination of Dand D7From the cumulative frequency column, we note that there are 102 (greater than 40% of the total) observations that are less than or equal to 78 and there are 133 (greater than 60% of the total) observations that are greater than or equal to 78. Therefore, D4 = Rs 78. Similarly, D7 = Rs 80.Determination of P15, P60 and P90From the cumulative frequency column, we note that there are 35 (greater than 15% of the total) observations that are less than or equal to 76 and there are 185 (greater than 85% of the total) observations that are greater than or equal to 76. Therefore, P15 = Rs 76. Similarly, P60 = Rs 79 and P90 = Rs 82.
Example : Calculate median, quartiles, 3rd and 6th deciles and 40th and 70th percentiles, from the following data
Also determine (i) The percentage of workers getting weekly wages between Rs 125 and Rs 260 and (ii) percentage of worker getting wages greater than Rs 340. Solution: First we make a cumulative frequency distribution table :
(i) Calculation of median: Here N = 500 so that
N/2= 250. Thus, median class is 250 - 300 and hence Lm = 250, fm = 125, h = 50 and C = 150. Substituting these values in the formula for median, we get Md = 250 + 250-150/125x50 = Rs 290
Hint: The given percentage of walkers and cyclists can be taken as frequencies. For calculation of mean, the necessary assumption is that the width of the first class is equal to the width of the following class, i.e., 1/4. On this assumption, the lower limit of the first class can be taken as 0. Similarly, on the assumption that width of the last class is equal to the width of last but one class, the upper limit of last class can be taken as 6. No assumption is needed for the calculation of median.
In a factory employing 3,000 persons, 5 percent earn less than Rs 3 per hour, 580 earn Rs 3.01 to 4.50 per hour, 30 percent earn from Rs 4.51 to Rs 6.00 per hour, 500 earn from 6.01 to Rs 7.5 per hour, 20 percent earn from Rs 7.51 to Rs 9.00 per hour and the rest earn Rs 9.01 or more per hour. What is the median wage?Hint: Write down the above information in the form of a frequency distribution. The class intervals given above are inclusive type. These should be converted into exclusive type for the calculation of median.The distribution of 2,000 houses of a new locality according to their distance from a milk booth is given in the following table :
Calculate the median distance of a house from milk booth.In the second phase of the construction of the locality, 500 additional houses were constructed out of which the distances of 200, 150 and 150 houses from the milk booth were in the intervals 450-500, 550-600 and 650-700 meters respectively. Calculate the median distance, taking all the 2500into account.
Hint: Add 200, 150 and 150 to the respective frequencies of the class intervals 450 - 500, 550 - 600 and 650 - 700.
The monthly salary distribution of 250 families in a certain locality of Agra is given below.
Draw a ‘less than’ ogive for the data given above and hence find out: (i) The limits of the income of the middle 50% of the families. (ii) If income tax is to be levied on families whose income exceeds Rs 1800 p.m., calculate the percentage of families which will be paying income tax.Hint: See example above.The following table gives the frequency distribution of marks of 800 candidates in an examination :
Draw 'less than' and 'more than' type ogives for the above data and answer the following from the graph: (i) If the minimum marks required for passing are 35, what percentage of candidates pass the examination? (ii) It is decided to allow 80% of the candidate to pass, what should be the minimum marks for passing? (iii) Find the median of the distribution.Hint: See example above.
Following are the marks obtained by a batch of 10 students in a certain class test in statistics (X) and accountancy (Y).
In which subject the level of knowledge of student is higher?Hint: Compare median of the two series.
The mean and median marks of the students of a class are 50% and 60% respectively. Is it correct to say that majority of the students have secured more than 50% marks? Explain.Hint: It is given that at least 50% of the students have got 60% or more marks.The monthly wages of 7 workers of a factory are : Rs 1,000, Rs 1,500, Rs 1,700, Rs 1,800, Rs 1,900, Rs 2,000 and Rs 3,000. Compute mean and median. Which measure is more appropriate? Which measure would you use to describe the situation if you were (i) a trade union leader, (ii) an employer?Hint: (i) median, (ii) mean.A boy saves Re. 1 on the first day, Rs 2 on the second day, ...... Rs 31 on the 31st day of a particular month. Compute the mean and median of his savings per day. If his father contributes Rs 10 and Rs 100 on the 32nd and 33rd day respectively, compute mean and median of his savings per day. Comment upon the results.Hint: Mean is too much affected by extreme observations.

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