Question

The length of leaves of a plant are represented in the form of an data as shown, if

median of the given distribution is 46, find (4)

Class

interval

10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total

Frequency 12 30 p 65 q 25 18 230

(a) p=

(i) 75 (ii) 65 (iii) 34 (iv) 46

(b) q=

(i) 65 (ii) 46 (iii) 70 (iv) 34

(c) P+q =

(i) 70 (ii) 46 (iii) 70 (iv) 80

(d) Frequency of median class

(i) 56 (ii) 65 (iii) 70 (iv) 34

(e) Median class

(i) 50-60 (ii) 30-40(iii) 40-50 (iv) 20-30​

Answer 1

Given: The length of leaves of a plant are represented in the form of an data,

Median = 46

To Find: 1. 'p'

             2. 'q'

              3. 'p+q'

              4. Frequency of median class

              5. Median Class

Step-by-step explanation:

Class Interval         Frequency                  Cumulative Frequency

10-20                            12                                       12

20-30                           30                                      42

30-40                           p                                         42+p

40-50                           65                                       107+p

50-60                           q                                         107+p+q

60-70                            25                                      132+p+q

70-80                            18                                       150+p+q

                            N = 150+p+q  

                             N = 230

                      $\frac{N}{2} = \frac {230}{2} = 115$

Median Class is 40-50

Median = $l + (\frac{\frac{N}{2} - cf}{f})\times {h}$

where l = Lower limit of the  median class

          cf = cumulative frequency of the class preceding the median class

          f = frequency of the median class

          h = class size

i.e., l = 40, cf = 42 + p, f = 65, h = 10

$46 = 40 + (\frac{{115} - (42+p)}{65}) \times 10\\ 46 - 40 = (\frac{115-42-p}{65})\times 10\\6\times 65 = (73-p)\times 10\\390 = 730 - 10p\\10p = 730-390\\p = \frac{340}{10}\\p=34$

1. 'p' = 34

2. N = 230

   150 + p + q = 230 (putting the value of 'p' from 1.)

   150 + 34 + q = 230

   184 + q = 230     (taking 184 to the other side )

   q = 230 - 184

    q = 46

3.  p + q = 34 + 46

    p + q = 80

4. Frequency of the median class = 65

5. Median Class = 40-50

Answer 2

Answer:

answer

Step-by-step explanation:

all the best for ur exam