wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The following are the marks obtained in Science by students in class X.
The mean of the given data is 63.33.
Is this mean an appropriate representation of the average marks obtained?
If no, what would be the most appropriate measure of central tendency for the correct representation of the marks distribution?
What is the value of this measure of central tendency for the given data?

MarksFrequency10−20120−30030−40040−50050−60160−70570−801580−902090−1003


A

No, median, value of median = 80.12

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

No, mode, value of mode = 85

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Yes, mean is the best measure of an average

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

Yes, the value of mean represents this data the best

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A

No, median, value of median = 80.12


Most of the class got marks above 70.
However, since one student got marks below 20, the mean got lowered to 63.33, which is a poor representation of the performance by the rest of the class.
A better representation would be the median, which would be the typical value.
The median can be calculated as follows:
MarksFrequencyCumulative frequency102011203001304001405001506012607057708015228090204290100345
Here n = 45 and n2=22.5. Hence, the median class is 80-90.
Therefore,median=l+n2cff×h=80+22.52242×10=80.12


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Mode
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon