Date | May 2013 | Marks available | 3 | Reference code | 13M.2.sl.TZ2.2 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The random variable X is normally distributed with mean 20 and standard deviation 5.
Find P(X≤22.9) .
Given that P(X<k)=0.55 , find the value of k .
Markscheme
evidence of appropriate approach (M1)
eg z=22.9−205
z=0.58 (A1)
P(X≤22.9)=0.719 A1 N3
[3 marks]
z-score for 0.55 is 0.12566… (A1)
valid approach (must be with z-values) (M1)
eg using inverse normal, 0.1257=k−205
k=20.6 A1 N3
[3 marks]
Examiners report
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and z-scores while others used the GDC directly; the GDC approach earned full marks more often than the z-score approach.
The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and z-scores while others used the GDC directly; the GDC approach earned full marks more often than the z-score approach. A common error in part (b) was to set the expression for z-score equal to the probability. Many candidates had difficulty giving answers correct to three significant figures; this was particularly an issue if no working was shown.