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=21.8−201.25, 1.44
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 P(T<21.8)−P(T<k)=0.3, 0.925−0.3
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=k−201.25, −0.524×1.25=k−20
k=20.4 A1 N3
OR
k=20.4 A3 N3
[5 marks]