Question

The mean and mode of frequency distribution are 28 and 16 respectively . The median is ?

Answer 1

Let the missing frequency be f 1for class-interval 30 — 40 and f 2forclass-interval 50 — 60.Computation of MedianVariable Frequency (f) CumulativeFrequency (c.f.)10 — 20 12 1220 — 30 30 4230 — 40 f 142+f 1 40 — 50 65 107+f 1 50 — 60 f 2107+f 1+f 2 60 — 70 25 132+f 1+f 2 70 — 80 18 150+f 1+f 2 ∑f or N= 150 +f 1+f 2= 229 (given)Middle item is N/2 =229/2 =114.5Median value is 46 (given) and it lies in the class-interval 40 — 50. Using the Medianformula .M = l1+(l2-l1)/f 1(m-c) we get46 = 40+(50-40)/65 [114.5-(42+f1)]46 =40+(10/65×72.5-f 1)46 = 40+(725-10f_1)/65 or 6 = (725-10f_1)/65390 = 725-10f1 or -335 = -10f1f1 = 33.5 or 34 since the frequency cannot be in fractionNow ∑f = 150+f1+f2 (Given)or 229 = 150+34+f2 or f2 = 45(b) Now mean can be calculated by completing the series by putting the value of f1and f2. The calculation of mean would be done by the method explained earlier.The value of the mean in the question would come to 45.83.

Answer 2

Answer:

the median is 24

Step-by-step explanation:

mode=3median-2mean

16=3median-2[28]

16=3median-56

3median=16+56

3median=72

median=72/3

median=24