Date | May 2008 | Marks available | 7 | Reference code | 08M.1.hl.TZ1.9 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Find | Question number | 9 | Adapted from | N/A |
Question
The random variable T has the probability density function
f(t)=π4cos(πt2), −1⩽t⩽1.
Find
(a) P(T = 0) ;
(b) the interquartile range.
Markscheme
(a) Any consideration of ∫00f(x)dx (M1)
0 A1 N2
(b) METHOD 1
Let the upper and lower quartiles be a and −a
π4∫1acosπt2dt=0.25 M1
⇒[π4×2πsinπt2]1a=0.25 A1
⇒[12sinπt2]1a=0.25
⇒[12−12sinπa2]=0.25 A1
⇒12sinπa2=14
⇒sinπa2=12
πa2=π6
a=13 A1
Since the function is symmetrical about t = 0 ,
interquartile range is 13−(−13)=23 R1
METHOD 2
π4∫a−acosπt2dt=0.5=π2∫a0cosπt2dt M1A1
⇒[sinaπ2]=0.5 A1
⇒aπ2=π6
⇒a=13 A1
The interquartile range is 23 R1
[7 marks]
Examiners report
All but the best candidates struggled with part (a). The vast majority either did not attempt it or let t = 1 . There was no indication from any of the scripts that candidates wasted an undue amount of time in trying to solve part (a). Many candidates attempted part (b), but few had a full understanding of the situation and hence were unable to give wholly correct answers.