Date | November 2013 | Marks available | 5 | Reference code | 13N.2.sl.TZ0.6 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The time taken for a student to complete a task is normally distributed with a mean of \(20\) minutes and a standard deviation of \(1.25\) minutes.
A student is selected at random. Find the probability that the student completes the task in less than \(21.8\) minutes.
The probability that a student takes between \(k\) and \(21.8\) minutes is \(0.3\). Find the value of \(k\).
Markscheme
Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.
attempt to standardize (M1)
eg \(z = \frac{{21.8 - 20}}{{1.25}},{\text{ 1.44}}\)
\({\text{P}}(T < 21.8) = 0.925\) A1 N2
[2 marks]
Note: There may be slight differences in answers, depending on whether candidates use tables or GDCs, or their 3 sf answers in subsequent parts. Do not penalise answers that are consistent with their working and check carefully for FT.
attempt to subtract probabilities (M1)
eg \({\text{P}}(T < 21.8) - {\text{P}}(T < k) = 0.3,{\text{ }}0.925 - 0.3\)
\({\text{P}}(T < k) = 0.625\) A1
EITHER
finding the \(z\)-value for \(0.625\) (A1)
eg \(z = 0.3186\) (from tables), \(z = 0.3188\)
attempt to set up equation using their \(z\)-value (M1)
eg \(0.3186 = \frac{{k - 20}}{{1.25}},{\text{ }} - 0.524 \times 1.25 = k - 20\)
\(k = 20.4\) A1 N3
OR
\(k = 20.4\) A3 N3
[5 marks]