Case Based Questions: Statistics | Mathematics (Maths) Class 10 PDF Download

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Q1: Read the source below and answer the questions that follow:
Vocational training complements traditional education by providing practical skills and

i. What is the lower limit of the modal class of the above data? (1 mark)
ii. Find the median class of the above data. (1 mark)
iii. Find the number of participants of age less than 50 years who undergo vocational training. (1 mark)
iv. Give the empirical relationship between mean, median and mode. (1 mark)

Ans:
First convert the given table in exclusive form subtract 0.5 from lower limit and add 0.5 to upper limit, so the new table will be:
i. Modal class is the class with highest frequency, so, it is 19.5 – 24.5. hence, lower limit will be ‘19.5’.
ii. (a)
N/2 = 365/2 = 182.5
Medium class will be 19.5 – 24.5.
iii. Approx 361 participants are there at Class Interval 44.5-49.5 showing 361 cummulative frequency.
Hence 361 participants are less than 50 year of age who undergo vocational training.
iv. Empirical relationship between mean, median and mode.
Mode = 3 Median – 2 Mean

Q2: Read the source below and answer the questions that follow:

i. Write the modal class. (1 mark)
ii. Find the median of the given data. (1 mark)
iii. Find the mean rainfall in this season. (1 mark)
iv. If a sub-division having at least 1000 mm rainfall during monsoon season, is considered a good rainfall sub-division, then how many sub-divisions had good rainfall?  (1 mark)

Ans:
i. Here, maximum class frequency is 7 and class corresponding to this frequency is 600-800, so the modal class is 600-800.
ii. Here n/2 = 24/2 = 12
Class whose cumulative frequency just greater than and nearest to n/2 is called median class.
Here, cf. = 13 (>12) and corresponding class 600 - 800 is : median class.
l = 600, c.f. = 6, f = 7, h = 200
Median = l + (n/2 - c.f.)/f × h
= 600 + (12 - 6)/7 × 200
= 600 + (6/7) × 200
= 771.429
So, the median of the given data is 771.429.
iii.
Assumed mean a = 1100 and class size, h = 400 - 200 = 200
Mean = a + (h / Σ fi) [Σ fi ui]
= 1100 + (200/24) × (-30)
= 1100 - (6000/24)
= 850
So, mean rainfall in the season is 850 mm.
iv. Number of sub-division having good rainfall
= 2 + 3 + 1 + 1 = 7

Q3: Read the source below and answer the questions that follow:

i. If true class limits have been decided by making the classes of interval 10, then find the first class interval. (1 mark)
ii. Find the cumulative frequency table. (1 mark)
iii. Find the frequency of class preceding the median class. (1 mark)
iv. If the price of a ticket for the age group 30-40 is Rs30, then find the total amount spent by this age group. (1 mark)

Ans:
i. The age of any person is a positive number, so the first class interval must be 0-10.
ii. Cumulative frequency table:
iii. From the table, N = 71, therefore N/2 = 35.5
Now, the class interval whose cumulative frequency is just greater than 35.5 is 30-40.
Median class = 30-40
So, the frequency of class preceding the median class is 2.
iv. Number of persons, whose age lying in 30-40 is 18.
Total amount spent by people of this age group
= Rs (30 × 18) = 540