# Question #a7009

Jul 15, 2016

The correct mean is $\overline{{x}_{c}} = 44.75$

#### Explanation:

From the data given we know, that:

$n = 80$ - the number of students

$\overline{x} = 45$ - 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:

$s = \overline{x} \times n = 45 \cdot 80 = 3600$

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:

$\overline{{x}_{c}} = {s}_{c} / n = \frac{3580}{80} = 44.75$

Finally we can write the answer:

The correct mean score is $\overline{{x}_{c}} = 44.75$