User interface language: English | Español

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.

[2]
a.

The probability that a student takes between \(k\) and \(21.8\) minutes is \(0.3\). Find the value of \(k\).

[5]
b.

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]

a.

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]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 5 - Statistics and probability » 5.9 » Normal distributions and curves.
Show 55 related questions

View options