Question

The Mathematics test done shows a result presented as follows: 11-20(8), 21-30(12), 31-40(15), 41-50(21), 51-60(20), 61-70(8), 71-80(16), which shows a range of test scores and it's frequency in brackets. Find the following, 1. Q1 2. Median 3. Q3 4. 7th Deciles 5. 62nd percentile. 6. Mean.

Answer 1

(a) $Q_1$ = 33.83

(b) Median = 47.64

(c) $Q_3$ = 60

(d) 7th Deciles = 57.5

(e) 62nd percentile = 53.5

(f) Mean = 47.6

Step-by-step explanation:

We are given that the Mathematics test done shows a result presented as follows:

Scores        Revised C.I.     Frequency (f)       X         $X \times f$         cf

 11 - 20         10.5 - 20.5              8                 15.5         124           8

 21 - 30        20.5 - 30.5             12                25.5        306         20

 31 - 40        30.5 - 40.5             15                35.5        532.5      35

 41 - 50        40.5 - 50.5             21                45.5        955.5      56

 51 - 60        50.5 - 60.5             20               55.5         1110         76

 61 - 70         60.5 - 70.5              8                65.5         524        84

 71 - 80         70.5 - 80.5            16               75.5        1208     100

 Total                                         100                             4760  

(a) The formula for calculating first quartile ($Q_1$) is given by;

    $Q_1 = x_L +\frac{\frac{N}{4} -cf}{f}\times h$

where, $x_L$ = lower limit of the quartile class = 30.5

            N = Total of frequency = 100

            cf = cumulative frequency just above the quartile class = 20

             f = frequency of quartile class = 15

             h = width of class interval = 10

Here, $\frac{N}{4} =\frac{100}{4}$ = 25, and the cumulative frequency just greater or equal to 25 is 35. So, the quartile class is 30.5 - 40.5.

Now,  $Q_1 = 30.5 +\frac{\frac{100}{4} -20}{15}\times 10$

          $Q_1$ = 30.5 + 3.33 = 33.83

(b) The formula for calculating Median is given by;

                     $\text{Median} = x_L +\frac{\frac{N}{2} -cf}{f}\times h$

where, $x_L$ = lower limit of the median class = 40.5

            N = Total of frequency = 100

            cf = cumulative frequency just above the median class = 35

             f = frequency of median class = 21

             h = width of class interval = 10

Here, $\frac{N}{2} =\frac{100}{2}$ = 50, and the cumulative frequency just greater or equal to 50 is 56. So, the median class is 40.5 - 50.5.

Now,  $\text{Median} = 40.5 +\frac{\frac{100}{2} -35}{21}\times 10$

          Median = 40.5 + 7.14 = 47.64

(c) The formula for calculating third quartile ($Q_3$) is given by;

                        $Q_3 = x_L +\frac{\frac{3N}{4} -cf}{f}\times h$

where, $x_L$ = lower limit of the quartile class = 50.5

            N = Total of frequency = 100

            cf = cumulative frequency just above the quartile class = 56

             f = frequency of quartile class = 20

             h = width of class interval = 10

Here, $\frac{3N}{4} =\frac{3 \times 100}{4}$ = 75, and the cumulative frequency just greater or equal to 75 is 76. So, the quartile class is 50.5 - 60.5.

Now,  $Q_3 = 50.5 +\frac{\frac{3 \times 100}{4} -56}{20}\times 10$

          $Q_3$ = 50.5 + 9.5 = 60

(d) The formula for calculating seventh decile ($D_7$) is given by;

               $D_7 = x_L +\frac{\frac{7N}{10} -cf}{f}\times h$

where, $x_L$ = lower limit of the decile class = 50.5

            N = Total of frequency = 100

            cf = cumulative frequency just above the decile class = 56

             f = frequency of decile class = 20

             h = width of class interval = 10

Here, $\frac{7N}{10} =\frac{7 \times 100}{10}$ = 70, and the cumulative frequency just greater or equal to 70 is 76. So, the decile class is 50.5 - 60.5.

Now,  $D_7 = 50.5 +\frac{\frac{7 \times 100}{10} -56}{20}\times 10$

          $D_7$ = 50.5 + 7 = 57.5

(e) The formula for calculating 62nd percentile ($P_6_2$) is given by;

             $P_6_2= x_L +\frac{\frac{62N}{100} -cf}{f}\times h$

where, $x_L$ = lower limit of the percentile class = 50.5

            N = Total of frequency = 100

            cf = cumulative frequency just above the percentile class = 56

             f = frequency of percentile class = 20

             h = width of class interval = 10

Here, $\frac{6N}{100} =\frac{62 \times 100}{100}$ = 62, and the cumulative frequency just greater or equal to 62 is 76. So, the percentile class is 50.5 - 60.5.

Now,  $P_6_2 = 50.5 +\frac{\frac{62 \times 100}{100} -56}{20}\times 10$

          $P_6_2$ = 50.5 + 3 = 53.5

(f) The mean of the above data is given by;

              Mean, $\bar X$  =  $\frac{\sum X\times f}{n}$

                              =  $\frac{4760}{100}$  =  47.6