| 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.
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.
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]
(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]
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.
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.
