Question #a7009

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Answer 1 · 2016-07-15T08:14:41

The correct mean is $¯ ¯¯ ¯ x_{c} = 44.75$ $bar(x_c)=44.75$

From the data given we know, that:

$n = 80$ $n=80$ - the number of students

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

$x_{m} = 92$ $x_m=92$ - wrong value

$x_{c} = 72$ $x_c=72$ - correct value

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

$s = ¯ x \times n = 45 \cdot 80 = 3600$ $s=bar(x)xxn=45*80=3600$

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

$s_{c} = 3600 - 92 + 72 = 3580$ $s_c=3600-92+72=3580$

Now we can calculate the correct mean as:

$¯ ¯¯ ¯ x_{c} = \frac{s_{c}}{n} = \frac{3580}{80} = 44.75$ $bar(x_c)=s_c/n=3580/80=44.75$

Finally we can write the answer:

The correct mean score is $¯ ¯¯ ¯ x_{c} = 44.75$ $bar(x_c)=44.75$