Question 1 - Case Based Questions (MCQ) - Chapter 13 Class 10 Statistics

Last updated at May 29, 2023 by Teachoo

The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans. The following tables shows the age distribution of case admitted during a day in two different hospitals

Refer to Table 1

Question 1

The average age for which maximum cases occurred is
(a) 32.24
(b) 34.36
(c) 36.82
(d) 42.24

Question 2

The upper limit of modal class is
(a) 15
(b) 25
(c) 35
(d) 45

Question 3

The mean of the given data is
(a) 26.2
(b) 32.4
(c) 33.5
(d) 35.4

Refer to table 2

Question 4

The mode of the given data is
(a) 41.4
(b) 48.2
(c) 55.3
(d) 64.6

Question 5

The median of the given data is
(a) 32.7
(b) 40.2
(c) 42.3
(d) 48.6

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Question The COVID-19 pandemic, also known as coronavirus pandemic, is an ongoing pandemic of coronavirus disease caused by the transmission of severe acute respiratory syndrome coronavirus 2 (SARS-CoV-2) among humans. The following tables shows the age distribution of case admitted during a day in two different hospitalsRefer to Table 1
Question 1 The average age for which maximum cases occurred is (a) 32.24 (b) 34.36 (c) 36.82 (d) 42.24(c) 36.82
Now,
When we want to find average age, we find mean
But, when we need to find average age of maximum cases, we need to find mode
Finding Mode
Mode = l + (šš ā šš)/(ššš ā šš ā šš) Ć h
Modal class = Interval with highest frequency
= 35 ā 45
where l = lower limit of modal class
h = class-interval
f1 = frequency of the modal class
f0 = frequency of class before modal class
f2 = frequency of class after modal class
Putting values in formula
Mode = l + (š1 āš0)/(2š1 āš0 āš2) Ć h
= 35 + (šš āšš)/(š(šš) ā šš ā šš) Ć 10
= 35 + 2/(46 ā 35) Ć 10
= 35 + 2/11 Ć 10
= 35 + 1.818
= 36.818
ā 36.82
So, the correct answer is (c)
Question 2 The upper limit of modal class is (a) 15 (b) 25 (c) 35 (d) 45(d) 45
Modal class is class with highest frequency
Since 35 ā 45 has highest frequency 23
ā“ It is the modal class
And Upper limit of Modal Class = 45
So, the correct answer is (d)
Question 3 The mean of the given data is (a) 26.2 (b) 32.4 (c) 33.5 (d) 35.4(d) 35.4
Mean(š„ Ģ ) = (āāšššš)/(āāšš)
= 2830/80
= 35.375
= 35.4
So, the correct answer is (d)
Refer to table 2 Question 4 The mode of the given data is (a) 41.4 (b) 48.2 (c) 55.3 (d) 64.6(a) 41.4
Mode = l + (šš ā šš)/(ššš ā šš ā šš) Ć h
Modal class = Interval with highest frequency
= 35 ā 45
where l = lower limit of modal class
h = class-interval
f1 = frequency of the modal class
f0 = frequency of class before modal class
f2 = frequency of class after modal class
Putting values in formula
Mode = l + (š1 āš0)/(2š1 āš0 āš2) Ć h
= 35 + (šš ā šš)/(š(šš) āšš āšš) Ć 10
= 35 + 32/(84 ā34) Ć 10
= 35 + 32/50 Ć 10
= 35 + 32/5
= 35 + 6.4
= 41.4
Question 5 The median of the given data is (a) 32.7 (b) 40.2 (c) 42.3 (d) 48.6(b) 40.2
Median = l + (šµ/š āšš)/š Ć h
Here,
šµ/š
ā“ 35 ā 45 is the median class
And,
l = lower limit of median class
h = class-interval
cf = cumulative frequency of the class before median class
f = frequency of the median class
Putting values in formula
Median = l + (š/2 āšš)/š Ć h
= 35 + (šš āšš)/šš Ć 10
= 35 + 22/42 Ć 10
= 35 + 110/21
= 35 + 5.23
= 40.23
So, the correct answer is (b)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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