Age (in years) | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 |
No. of participants | 62 | 132 | 96 | 37 | 13 | 11 | 10 | 4 |
From the above answer the following questions :
(i) What is the lower limit of the modal class of the above data? [1 Mark]
(ii) (a) Find the median class of the above data. [2 Marks]
OR
(b) Find the number of participants of age less than 50 years who undergo vocational training.
(iii) Give the empirical relationship between the mean, median and mode. [1 Mark]
CBSE 2024 (30/3/1)
Solution :
This is a case study based question form Class 10th Statistics.
Here the class intervals are not continuous so first of all we have to make them continuous.
As we can see that here the gap is 1 (15-19, 20-24)
So we will subtract 0.5 (1/2=0.5) from the lower limit and add 0.5 to the upper limit.
Modified continuous frequency distribution table is given below
Age (in years) | 14.5-19.5 | 19.5-24.5 | 24.5-29.5 | 29.5-34.5 | 34.5-40.5 | 39.5-44.5 | 44.5-49.5 | 49.5-54.5 |
No. of participants | 62 | 132 | 96 | 37 | 13 | 11 | 10 | 4 |
i) Maximum frequency = 132
Modal Class = 19.5-24.5
Lower Limit of the modal class = 19.5
Modal Class = 19.5-24.5
Lower Limit of the modal class = 19.5
ii)
Age (in years) | 14.5-19.5 | 19.5-24.5 | 24.5-29.5 | 29.5-34.5 | 34.5-40.5 | 39.5-44.5 | 44.5-49.5 | 49.5-54.5 |
No. of participants | 62 | 132 | 96 | 37 | 13 | 11 | 10 | 4 |
Cumulative Frequency (CF) | 62 | 194 | 290 | 327 | 340 | 351 | 361 | 365 |
(a) Total number of observations (N) = 365
N/2 = 365/2 = 182.5
N/2 = 365/2 = 182.5
Hence, the required Median Class is 19.5-24.5
OR
(b) Number of participants of age less than 50 years = 365-4 = 361
(b) Number of participants of age less than 50 years = 365-4 = 361
(iii) 3Media = 2Mean+Mode
which is the required empirical relation between mean, median and mode.
which is the required empirical relation between mean, median and mode.