Question #a7009

Question

1996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-15T08:14:41

The correct mean is #bar(x_c)=44.75#

From the data given we know, that:

#n=80# - the number of students

#bar(x)=45# - mean score before the correction

#x_m=92# - wrong value

#x_c=72# - correct value

The sum of all scores #s# can be calculated as:

#s=bar(x)xxn=45*80=3600#

This sum includes misread value. To calculate the correct sum we have to substract the miserad value and add the correct number:

#s_c=3600-92+72=3580#

Now we can calculate the correct mean as:

#bar(x_c)=s_c/n=3580/80=44.75#

Finally we can write the answer:

The correct mean score is #bar(x_c)=44.75#