| Date | November 2011 | Marks available | 3 | Reference code | 11N.3sp.hl.TZ0.3 |
| Level | HL only | Paper | Paper 3 Statistics and probability | Time zone | TZ0 |
| Command term | Find | Question number | 3 | Adapted from | N/A |
Question
The random variable X represents the lifetime in hours of a battery. The lifetime may be assumed to be a continuous random variable X with a probability density function given by \(f(x) = \lambda {{\text{e}}^{ - \lambda x}}\), where \(x \geqslant 0\).
Find the cumulative distribution function, \(F(x)\), of X.
Find the probability that the lifetime of a particular battery is more than twice the mean.
Find the median of X in terms of \(\lambda \).
Find the probability that the lifetime of a particular battery lies between the median and the mean.
Markscheme
\(\int {\lambda {{\text{e}}^{ - \lambda t}}{\text{d}}t = - {{\text{e}}^{ - \lambda t}}{\text{ }}( + c)} \) A1
\( \Rightarrow F(x) = \left[ { - {{\text{e}}^{ - \lambda t}}} \right]_0^x\) (M1)
\( = 1 - {{\text{e}}^{ - \lambda t}}{\text{ }}(x \geqslant 0)\) A1
[3 marks]
\(1 - F\left( {\frac{2}{\lambda }} \right)\) M1
\( = {{\text{e}}^{ - 2}}\,\,\,\,\,( = 0.135)\) A1
[2 marks]
\(F(m) = \frac{1}{2}\) (M1)
\( \Rightarrow {{\text{e}}^{ - \lambda m}} = \frac{1}{2}\) A1
\( \Rightarrow - \lambda m = \ln \frac{1}{2}\)
\( \Rightarrow m = \frac{1}{\lambda }\ln 2\) A1
[3 marks]
\(F\left( {\frac{1}{\lambda }} \right) - F\left( {\frac{{\ln 2}}{\lambda }} \right)\) M1
\( = \frac{1}{2} - {{\text{e}}^{ - 1}}\,\,\,\,\,( = 0.132)\) A1
[2 marks]
Examiners report
For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.
For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.
For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.
For most candidates the question started well, but many did not appear to understand how to find the cumulative distribution function in (b). Many were able to integrate \(\lambda {{\text{e}}^{ - \lambda x}}\), but then did not know what to do with the integral. Parts (c), (d) and (e) were relatively well done, but even candidates who successfully found the cumulative distribution function often did not use it. This resulted in a lot of time spent integrating the same function.
