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Date November 2008 Marks available 3 Reference code 08N.2.sl.TZ0.7
Level SL only Paper 2 Time zone TZ0
Command term Find Question number 7 Adapted from N/A

Question

The scores of a test given to students are normally distributed with a mean of 21. \(80\% \) of the students have scores less than 23.7.

Find the standard deviation of the scores.

[3]
a.

A student is chosen at random. This student has the same probability of having a score less than 25.4 as having a score greater than b.

(i)     Find the probability the student has a score less than 25.4.

(ii)    Find the value of b.

 

[4]
b(i) and (ii).

Markscheme

evidence of approach     (M1)

e.g. finding \(0.84 \ldots \), using  \(\frac{{23.7 - 21}}{\sigma }\)

correct working     (A1)

e.g. \(0.84 \ldots = \frac{{23.7 - 21}}{\sigma }\) , graph     A1

\(\sigma = 3.21\)

[3 marks]

a.

(i) evidence of attempting to find \({\text{P}}(X < 25.4)\)     (M1)

e.g. using \(z = 1.37\)   

\({\text{P}}(X < 25.4) = 0.915\)     A1     N2

(ii) evidence of recognizing symmetry     (M1)

e.g. \(b = 21 - 4.4\) , using \(z = - 1.37\)     A1     N2

[4 marks]

b(i) and (ii).

Examiners report

Candidates who clearly understood the nature of normal probability answered this question cleanly. A common misunderstanding was to use the value of 0.8 as a z-score when finding the standard deviation.

a.

Many correctly used their GDC to find the probability in part (b). Fewer used some aspect of the symmetry of the curve to find a value for b.

b(i) and (ii).

Syllabus sections

Topic 5 - Statistics and probability » 5.9 » Normal distributions and curves.
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