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Date May 2010 Marks available 4 Reference code 10M.2.sl.TZ1.4
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 4 Adapted from N/A

Question

The following frequency distribution of marks has mean 4.5.


Find the value of x.

[4]
a.

Write down the standard deviation.

[2]
b.

Markscheme

\(\sum {fx = 1(2) + 2(4) +  \ldots  + 7(4)} \) , \(\sum {fx = 146 + 5x} \) (seen anywhere)     A1 

evidence of substituting into mean \(\frac{{\sum {fx} }}{{\sum f }}\)     (M1)

correct equation     A1

e.g. \(\frac{{146 + 5x}}{{34 + x}} = 4.5\) , \(146 + 5x = 4.5(34 + x)\)

\(x = 14\)     A1     N2 

[4 marks]

a.

\(\sigma = 1.54\)     A2     N2

[2 marks]

b.

Examiners report

Surprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of the frequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for a frequency. Unfortunately, candidates did not appreciate the unreasonableness of this result.

a.

Surprisingly, this question was not well done by many candidates. A good number of candidates understood the importance of the frequencies in calculating mean. Some neglected to sum the frequencies for the denominator, which often led to a negative value for a frequency. Unfortunately, candidates did not appreciate the unreasonableness of this result. In part (b), many candidates could not find the standard deviation in their GDC, often trying to calculate it by hand with no success. Further, many could not distinguish between the sample and the population standard deviation given in the GDC.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.1 » Presentation of data: frequency distributions (tables); frequency histograms with equal class intervals
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