Education with vocational training is helpful in making a student self-reliant and to help and
serve the society. Keeping this in view, a teacher made the following table giving the
frequency distribution of a student undergoing vocational training from the training
institute.
Age (in years). Frequency(no.
of participants)
15-19. 62
20-24. 132
25-29. 96
30-34. 37
35-39. 13
40-44. 8
45-49. 6
50-54. 4
55-59. 4
60-above. 3
Median class of above data:
a. 20 - 24
b. 20.5 - 24.5
c. 19.5 - 24.5
d. 24.5 - 29.5
ii. Calculate the median.
a. 24.06
b. 30.07
c. 24.77
d. 42.07
iii. The empirical relationship between mean, median, mode:
a. Mode = 3 Median + 2 Mean
b. Mode = 3 Median - 2 Mean
c. Mode = 3 Mean + 2 Median
d. 3 Mode = Median - 2 Mean
iv. If mode = 80 and mean = 110, then find the median.
a. 200
b. 500
c. 190
d. 100
v. The mode is the:
a. middlemost frequent value
b. least frequent value
c. maximum frequent value
d. none of thes
Answers
Answer:
misdiagnosis
Answer:
i. c. 19.5-24.5
ii. a. 24.06
iii. b. mode= 3median-2mode
iv. d.100
v. c. maximum frequent value
Step-by-step explanation:
arrange the data with equal class sizes and calculate the cumulative frequency
i.e,
age frequency cumulative frequency
14.5-19.5 62 62
19.5-24.5 132 194
24.5-29.5 96 290
29.5-34.5 37 327
34.5-39.5 13 340
39.5-44.5 8 348
44.5-49.5 6 354
49.5-54.5 4 358
54.5-59.5 4 362
59.5-above 3 365
i. here, n=365
similarly, n/2=365/2=182.5
therefore, 194 is the nearest value for 182.5...hence, "19.5-24.5" is the median class
option c
ii. substitute the values from median class,
median= l+(n/2-cf/f)h
= 19.5+(182.5-62/132)5
= 19.5+(120.5*5/132)
= 19.5+4.56
=24.06
option a
iii. 3median= mode + 2mean
on transposing mode to the LHS
mode= 3median-2mean
iv. 3median=mode+2mean
median=80+(2*110)/3=80+220/3=300/3=100
option d
v. mode is the most frequent value...
option c