# 2019 May TZ1 Paper 2 - Q3

(ai)

The mean is:

\displaystyle \frac {8 + 4 + 7 + 6 + 10 + 9 + 7 + 3} 8 = \frac {27} 4

(aii)

The variance is:

\displaystyle \frac {8^2 + 4^2 + 7^2 + 6^2 + 10^2 + 9^2 + 7^2 + 3^2} 8 - \left(\frac {27} 4\right)^2 = \frac {79} {16}

So the standard deviation is:

\displaystyle \frac {\sqrt {79}} 4 = 2.222 \text { (3dp)}

(bi)

The mean will simply shift by +2, so the new mean is:

\displaystyle \frac {27} 4 + 2 = \frac {35} 4

(bii)

The standard deviation is unchanged, so is still:

\displaystyle \frac {\sqrt {79}} 4 = 2.222 \ldots

@q00 are standard deviations Bessel corrected for IB? Assumed not for this.

Sorted in ascending order, the students scored:

3, 4, 6, 7, 7, 8, 9, 10

The median here is the average of the 4 th and 5 th terms, ie. 7. If there are 9 scores, the median will be the 5 th score when sorted in ascending order.

If we add a ninth with score s \le 7, the student with 7 put 4 th will be “pushed up” to 5 th and the median will remain 7. (ie. there will still be a 7 in the 5 th position) If we add a ninth score with s > 7, 7 will still be in the 5 position. So the median remains unchanged with both cases. If you are having trouble seeing this, set say s = 6 then s = 8.

1 Like

@_gcx no it isn’t