| Date | November 2015 | Marks available | 3 | Reference code | 15N.3sp.hl.TZ0.1 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Calculate | Question number | 1 | Adapted from | N/A |
Question
It is known that the standard deviation of the heights of men in a certain country is \(15.0\) cm.
One hundred men from that country, selected at random, had their heights measured.
The mean of this sample was \(185\) cm. Calculate a \(95\% \) confidence interval for the mean height of the population.
A second random sample of size \(n\) is taken from the same population. Find the minimum value of \(n\) needed for the width of a \(95\% \) confidence interval to be less than \(3\) cm.
Markscheme
valid attempt to use \(\bar x \pm z\frac{\sigma }{{\sqrt n }}\) (M1)
\([182,{\text{ }}188]\) A1A1
Note: Accept answers that round to the correct \(3\) sf.
[3 marks]
\(1.96 \times \frac{{15.0}}{{\sqrt n }} < 1.5\) M1A1
\(n > {\left( {\frac{{15.0}}{{1.5}} \times 1.96} \right)^2}\) (M1)
Note: Award M1 for attempting to solve the inequality.
Note: Allow the use of \( = \).
minimum value \(n = 385\) A1
[4 marks]
Total [7 marks]
