Date | May 2012 | Marks available | 3 | Reference code | 12M.2.sl.TZ2.4 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
The heights of a group of seven-year-old children are normally distributed with mean \(117{\text{ cm}}\) and standard deviation \(5{\text{ cm}}\). A child is chosen at random from the group.
Find the probability that this child is taller than \(122.5{\text{ cm}}\).
The heights of a group of seven-year-old children are normally distributed with mean \(117{\text{ cm}}\) and standard deviation \(5{\text{ cm}}\). A child is chosen at random from the group.
The probability that this child is shorter than \(k{\text{ cm}}\) is \(0.65\). Find the value of k .
Markscheme
evidence of appropriate method (M1)
e.g. \(z = \frac{{122.5 - 117}}{5}\) , sketch of normal curve showing mean and \(122.5\), \(1.1\)
\({\rm{P}}(Z < 1.1) = 0.8643\) (A1)
\(0.135666\)
\({\rm{P(H}} > 122.5) = 0.136\) A1 N3
[3 marks]
\(z = 0.3853\) (A1)
set up equation (M1)
e.g. \(\frac{{X - 117}}{5} = 0.3853\) , sketch
\(k = 118.926602\)
\(k = 199\) A1 N3
[3 marks]
Examiners report
There were many completely successful attempts at this question, with good use of formulae and calculator features.
There were many completely successful attempts at this question, with good use of formulae and calculator features.
However, in part (b) some candidates did not recognize the need to find the standardized value and set their equation equal to the probability given in the question, thus earning only one mark.