Mathematics Questions and Answers – Statistics – Median of Grouped Data

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Statistics – Median of Grouped Data”.

1. What is the median?
a) Difference between higher half and lower half of the data set
b) Mean of the highest and lowest number in a data sample
c) Value separating higher half from the lower half of a data sample
d) Difference between the highest and lowest number.
View Answer

Answer: c
Explanation: In statistics, the Median is also called the ‘Middle Value’ as it is the value that separates the Highest half from the lower half of a data sample.
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2. What is the Median of the following data sample?
2, 7, 4, 8, 9, 10, 6, 12, 13
a) 8
b) 11
c) 9
d) 10
View Answer

Answer: a
Explanation: Arranging the data sample in ascending order 2, 4, 6, 7, 8, 9, 10, 12, 13.
Median is the middle value which separates the higher half from the lower half in a data sample. After arranging them in ascending order, 8 is the Middle value.

3. What is the Median of the following data sample?
3, 7, 4, 8, 9, 6, 10, 12, 13, 15
a) 7.5
b) 9
c) 8.5
d) 10
View Answer

Answer: c
Explanation: Arranging the data set in ascending order 3, 4, 6, 7, 8, 9, 10, 12, 13, 15
8 and 9 are the two middle numbers.
Median is the mean of the middle two numbers.
Median is the mean of the middle two numbers = \(\frac {8+9}{2}\) = 8.5
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4. Some of the samples are given below. Find the median.
90, 45, 67, 34, 26, 76, 44, 55.
a) 55
b) 45
c) 40
d) 50
View Answer

Answer: d
Explanation: There is an even number of terms in this Data Set. So, the median will be the mean of the middle two numbers.
Arranging them in ascending order 26, 34, 44, 45, 55, 67, 76, 90.
45 and 55 are the middle two numbers. So, Median = \(\frac {45+55}{2}\) = 50

5. If the mean and the mode are given as 35 and 30. Find the Median.
a) 75
b) 33.33
c) 19
d) 32
View Answer

Answer: b
Explanation: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
Given Mean = 35, Mode = 30
35 – 30 = 3(35 – Median )
5 = 105 – 3 Median
Median = 33.33
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6. If the Mean and Mode are 25, then find the Median.
a) 13
b) 9
c) 25
d) 0
View Answer

Answer: c
Explanation: The empirical mean median mode relation is given as
Mean – Mode = 3(Mean – Median)
25 – 25 = 3(25 – Median )
Median = 25

7. What is the formula for the median of Grouped data?
a) Median = L + [(n / 2 – cf) / f] * h
b) Median = L + [(n / 2 + cf) / f] * h
c) Median = L + [(n / 2 – cf) / f] + h
d) Median = L * [(n / 2 – cf) / f] * h
View Answer

Answer: a
Explanation: Formula for the Median of Grouped Data
Median = L + [(n / 2 – c.f) / f] * h
L = Lower limit of Median Class
cf = Cumulative frequency of the class prior to median class
f = Frequency of Median Class
h = Class size
n = Total frequency
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8. Find the Median of the following grouped data.

Marks Frequency
0-10 9
10-20 10
20-30 24
30-40 16
40-50 11

a) 15
b) 20
c) 26.66
d) 35
View Answer

Answer: c
Explanation: Total frequency n = 9 + 10 + 24 + 16 + 11 = 70
\(\frac {n}{2} = \frac {70}{2}\) = 35

Marks Frequency Cumulative Frequency
0-10 9 9
10-20 10 9 + 10 = 19
20-30 24 19 + 24 = 43
30-40 16 43 + 16 = 59
40-50 11 59 + 11 = 70
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35 is less than 43 and greater than 19
So, 20 – 30 is the Median Class
Now L = 20, h = 10, cf = 19, f = 24
Median = L + [(n / 2 – c.f) / f] * h
= 20 + [(35 – 19) / 24] * 10
= 26.66

9. Find the Median of given Grouped data.

Rating Frequency
0-5 12
5-10 20
10-15 10
15-20 6

a) 10
b) 18
c) 8
d) 17.5
View Answer

Answer: c
Explanation: Total Frequency = 12 + 20 + 10 + 6 = 48 and \(\frac {n}{2} = \frac {48}{2}\) = 24

Rating Frequency Cumulative Frequency
0-5 12 12
5-10 20 20 + 12 = 32
10-15 10 32 + 10 = 42
15-20 6 42 + 6 = 48

24 is less than 32 and greater than 12. So, the Median Class is 5 – 10.
L = 5, h = 5, cf = 12, f = 20
Median = L + [(n / 2 – c.f) / f] * h
= 5 + [(24 – 12) / 20] * 5
= 8

10. Find the Median of the following grouped data.

Results Frequency
0-20 5
20-40 10
40-60 30
60-80 15

a) 52
b) 50
c) 34
d) 45
View Answer

Answer: b
Explanation: Total Frequency n = 5 + 10 + 30 + 15 = 60 and \(\frac {n}{2} = \frac {60}{2}\) = 30

Results Frequency Cumulative Frequency
0-20 5 5
20-40 10 5 + 10 = 15
40-60 30 15 + 30 = 45
60-80 15 45 + 15 = 60

30 is less than 45 and greater than 15. So, the median class is 40 – 60.
L = 40, h = 20, cf = 15, f = 30
Median = L + [(n / 2 – c.f) / f] * h
= 40 + [(30 – 15) / 30] * 20
= 50

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter