5.03b Solve problems: using pdf

9 Every weekday Jonathan takes an underground train to work. On any weekday the time in minutes that he has to wait at the station for a train is modelled by the continuous uniform distribution over \([ 0,5 ]\).

1. Find the probability that Jonathan has to wait at least 3 minutes for a train. The total time that Jonathan has to wait on two days is modelled by the continuous random variable \(X\) with probability density function given by \(\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 25 } x & 0 \leqslant x \leqslant 5 , \\ \frac { 1 } { 25 } ( 10 - x ) & 5 < x \leqslant 10 , \\ 0 & \text { otherwise } . \end{cases}\)
2. Find the probability that Jonathan has to wait a total of at most 6 minutes on two days. Jonathan's friend suggests that the total waiting time for 5 days, \(T\) minutes, will almost certainly be less than 18 minutes. In order to investigate this suggestion, Jonathan constructs the simulation shown in Fig. 9. All of the numbers in the simulation have been rounded to 2 decimal places. \begin{table}[h]

   	A	B	C	D	E	F
   1	Mon	Tue	Wed	Thu	Fri	Total T
   2	1.78	4.36	2.74	3.88	4.64	17.41
   3	0.95	1.30	4.83	4.29	1.81	13.18
   4	4.27	4.90	4.57	1.41	3.66	18.81
   5	0.80	0.06	3.20	1.76	0.35	6.17
   6	0.03	4.82	1.26	3.53	0.13	9.77
   7	3.88	4.73	1.19	3.75	1.29	14.84
   8	4.11	3.54	4.33	0.77	4.50	17.25
   9	3.54	0.11	3.85	2.86	1.58	11.94
   10	1.87	1.82	3.00	3.53	1.83	12.05
   11	4.00	2.98	4.59	1.73	1.76	15.06
   12	1.91	3.85	2.08	1.72	2.82	12.38
   13	0.10	4.86	2.51	0.52	2.17	10.15
   14	1.24	4.26	0.95	1.33	1.78	9.57
   15	2.99	0.69	3.85	3.41	2.42	13.36
   16	4.67	1.76	2.13	3.48	3.10	15.14
   17	1.94	1.07	0.91	0.63	3.34	7.89
   18	0.11	2.29	0.71	4.21	0.86	8.18
   19	0.43	4.58	4.89	1.86	2.84	14.60
   20	4.23	0.88	2.71	4.88	4.20	16.91
   21	3.72	4.58	3.11	4.89	3.18	19.49

   \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{table}
3. Use the simulation to estimate \(\mathrm { P } ( T > 18 )\).
4. Explain how Jonathan could obtain a better estimate. Jonathan thinks that he can use the Central Limit Theorem to provide a very good approximation to the distribution of \(T\).
5. Find each of the following.
   Jonathan travels to work on 200 days in a year.
6. Find the probability that the total waiting time for Jonathan in a year is more than 510 minutes.
   [0pt] [3]

10 The probability density function of the continuous random variable \(X\) is given by \(f ( x ) = \begin{cases} k x ^ { m } & 0 \leqslant x \leqslant a , \\ 0 & \text { otherwise, } \end{cases}\) where \(a , k\) and \(m\) are positive constants.

1. Show that \(k = \frac { m + 1 } { a ^ { m + 1 } }\).
2. Find the cumulative distribution function of \(X\) in terms of \(x , a\) and \(m\).
3. Given that \(\mathrm { P } \left( \frac { 1 } { 4 } a < X < \frac { 1 } { 2 } a \right) = \frac { 1 } { 10 }\),
   1. show that \(2 p ^ { 2 } - 10 p + 5 = 0\), where \(p = 2 ^ { m }\),
   2. find the value of \(m\). \section*{END OF QUESTION PAPER}

12 The continuous random variable \(X\) has cumulative distribution function given by $$F ( x ) = \begin{cases} 0 & x < 0 \\ k \left( a x - 0.5 x ^ { 2 } \right) & 0 \leqslant x \leqslant a \\ 1 & x > a \end{cases}$$ where \(a\) and \(k\) are positive constants.

1. Determine the median of \(X\) in terms of \(a\).
2. Given that \(a = 10\), determine the probability that \(X\) is within one standard deviation of its mean.

10 The continuous random variable \(X\) has probability density function given by \(f ( x ) = \begin{cases} \frac { 4 } { 15 } \left( \frac { a } { x ^ { 2 } } + 3 x ^ { 2 } - \frac { 7 } { 2 } \right) & 1 \leqslant x \leqslant 2 , \\ 0 & \text { otherwise, } \end{cases}\) where \(a\) is a positive constant.

1. Find the cumulative distribution function of \(X\) in terms of \(a\).
2. Hence or otherwise determine the value of \(a\).
3. Show that the median value \(m\) of \(X\) satisfies the equation $$8 m ^ { 4 } - 28 m ^ { 2 } + 9 m - 4 = 0 .$$
4. Verify that the median value of \(X\) is 1.74, correct to \(\mathbf { 2 }\) decimal places.
5. Find \(\mathrm { E } ( X )\).
6. Determine the mode of \(X\).

4 An archer fires arrows at a circular target of radius 50 cm . The distance in cm that an arrow lands from the centre of the target is modelled by the random variable \(X\), with probability density function given by \(f ( x ) = \begin{cases} a x & 0 \leqslant x \leqslant 50 , \\ 0 & \text { otherwise, } \end{cases}\) where \(a\) is a constant.

1. Determine the value of \(a\).
2. Determine the probability that an arrow will land within 5 cm of the centre of the target.
3. Determine the median distance from the centre of the target that an arrow will land.

11 The discrete random variable \(X\) has a uniform distribution over the set of all integers between 25 and \(n\) inclusive, where \(n\) is a positive integer with \(n > 25\).

1. Determine \(\mathrm { P } \left( \mathrm { X } < \frac { \mathrm { n } + 25 } { 2 } \right)\) in each of the following cases.

11 The continuous random variable \(X\) has probability density function given by \(f ( x ) = \begin{cases} a x ^ { 2 } & 0 \leqslant x < 2 , \\ b ( 3 - x ) ^ { 2 } & 2 \leqslant x \leqslant 3 , \\ 0 & \text { otherwise } \end{cases}\) where \(a\) and \(b\) are positive constants.

1. Given that \(\mathrm { E } ( X ) = 2\), determine the values of \(a\) and \(b\).
2. Determine the median value of \(X\).
3. A random sample of 50 observations of \(X\) is selected. Given that \(\operatorname { Var } ( X ) = 0.2\), determine an estimate of the probability that the mean value of the 50 observations is less than 1.9.

2. The smallest angle \(\theta\), in degrees, of a right-angled triangle with hypotenuse 8 cm , is uniformly distributed across all possible values. \includegraphics[max width=\textwidth, alt={}, center]{8f47b2ff-f954-42ec-8ecc-fc64313a7b89-04_419_696_479_687}

1. Find the mean and standard deviation of \(\theta\).
2. The shortest side of the triangle is of length \(X \mathrm {~cm}\). Find the probability that \(X\) is greater than 5 .

4. The continuous random variable, \(X\), has the following probability density function $$f ( x ) = \begin{cases} k x & \text { for } 0 \leqslant x < 1 \\ k x ^ { 3 } & \text { for } 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$ where \(k\) is a constant. \includegraphics[max width=\textwidth, alt={}, center]{4ecf99c5-c4b3-41b7-a8df-a7c2ca7fcd6a-3_851_935_826_678}

1. Show that \(k = \frac { 4 } { 17 }\).
2. Determine \(\mathrm { E } ( X )\).
3. Calculate \(\mathrm { E } ( 3 X - 1 )\) and \(\operatorname { Var } ( 3 X - 1 )\).

2. Emlyn aims to produce podcast episodes that are a standard length of time, which he calls the 'target time'. The time, \(X\) minutes, above or below the target time, which he calls the 'allowed time', can be modelled by the following cumulative distribution function. $$F ( x ) = \begin{cases} 0 & x < - 2 \\ \frac { x + 2 } { 5 } & - 2 \leqslant x < 1 \\ \frac { x ^ { 2 } - x + 3 } { 5 } & 1 \leqslant x \leqslant 2 \\ 1 & x > 2 \end{cases}$$

1. Calculate the upper quartile for the 'allowed time'.
2. Find \(f ( x )\), the probability density function, for all values of \(x\).
3. 1. Calculate the mean 'allowed time'.
   2. Interpret your answer in context.

2 The continuous random variable \(Y\) has cumulative distribution function defined by $$\mathrm { F } ( y ) = \left\{ \begin{array} { c c } 0 & y < 0 \\ \frac { y ^ { 2 } } { 36 } & 0 \leq y \leq 6 \\ 1 & y > 6 \end{array} \right.$$ Find the value of \(\mathrm { P } ( Y > 4 )\) Circle your answer. \(\frac { 4 } { 9 }\) \(\frac { 5 } { 9 }\) \(\frac { 16 } { 27 }\) \(\frac { 11 } { 27 }\)

6 The random variable \(T\) can take the value \(T = - 2\) or any value in the range \(0 \leq T < 12\) The distribution of \(T\) is given by \(\mathrm { P } ( T = - 2 ) = c , \mathrm { P } ( 0 \leq T \leq t ) = 225 k - k ( 15 - t ) ^ { 2 }\) 6

1. 1. Show that \(1 - c = 216 k\) [0pt] [3 marks] 6
      1. (ii) Given that \(c = 0.1\), find the value of \(\mathrm { E } ( T )\) [0pt] [3 marks]
         6
   2. Show that \(\mathrm { E } ( \sqrt { | T | } ) = \frac { 5 \sqrt { 2 } + 52 \sqrt { 3 } } { 50 }\) [0pt] [3 marks]

1. The continuous random variable \(X\) has probability density function

$$f ( x ) = \begin{cases} \frac { 1 } { 18 } ( 11 - 2 x ) & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

1. Find \(\mathrm { P } ( \mathrm { X } < 3 )\)
2. State, giving a reason, whether the upper quartile of \(X\) is greater than 3, less than 3 or equal to 3 Given that \(\mathrm { E } ( \mathrm { X } ) = \frac { 9 } { 4 }\)
3. use algebraic integration to find \(\operatorname { Var } ( \mathrm { X } )\) The cumulative distribution function of \(X\) is given by $$F ( x ) = \left\{ \begin{array} { l r } 0 & x < 1 \\ \frac { 1 } { 18 } \left( 11 x - x ^ { 2 } + c \right) & 1 \leqslant x \leqslant 4 \\ 1 & x > 4 \end{array} \right.$$
4. Show that \(\mathrm { c } = - 10\)
5. Find the median of \(X\), giving your answer to 3 significant figures. \section*{Q uestion 2 continued}

4 The continuous random variable \(X\) has cumulative distribution function given by $$\mathrm { F } ( x ) = \left\{ \begin{array} { c c } 0 & x \leqslant 0 \\ k \left( x ^ { 3 } - \frac { 3 } { 8 } x ^ { 4 } \right) & 0 < x \leqslant 2 \\ 1 & x > 2 \end{array} \right.$$ where \(k\) is a constant.

1. Show that \(k = \frac { 1 } { 2 }\)
2. Showing your working clearly, use calculus to find
   1. \(\mathrm { E } ( X )\)
   2. the mode of \(X\)
3. Describe, giving a reason, the skewness of the distribution of \(X\)

5 \begin{figure}[h] \end{figure} The random variable \(X\) has probability density function \(\mathrm { f } ( x )\) and Figure 1 shows a sketch of \(\mathrm { f } ( x )\) where $$f ( x ) = \left\{ \begin{array} { c c } k ( 1 - \cos x ) & 0 \leqslant x \leqslant 2 \pi \\ 0 & \text { otherwise } \end{array} \right.$$

1. Show that \(k = \frac { 1 } { 2 \pi }\) The random variable \(Y \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)\) and \(\mathrm { E } ( Y ) = \mathrm { E } ( X )\) The probability density function of \(Y\) is \(g ( y )\), where $$g ( y ) = \frac { 1 } { \sigma \sqrt { 2 \pi } } e ^ { - \frac { 1 } { 2 } \left( \frac { y - \mu } { \sigma } \right) ^ { 2 } } \quad - \infty < y < \infty$$ Given that \(\mathrm { g } ( \mu ) = \mathrm { f } ( \mu )\)
2. find the exact value of \(\sigma\)
3. Calculate the error in using \(\mathrm { P } \left( \frac { \pi } { 2 } < Y < \frac { 3 \pi } { 2 } \right)\) as an approximation to \(\mathrm { P } \left( \frac { \pi } { 2 } < X < \frac { 3 \pi } { 2 } \right)\)

8 A circle, centre \(O\), has radius \(x \mathrm {~cm}\), where \(x\) is an observation from the random variable \(X\) which has a rectangular distribution on \([ 0 , \pi ]\)

1. Find the probability that the area of the circle is greater than \(10 \mathrm {~cm} ^ { 2 }\)
2. State, giving a reason, whether the median area of the circle is greater or less than \(10 \mathrm {~cm} ^ { 2 }\) The triangle \(O A B\) is drawn inside the circle with \(O A\) and \(O B\) as radii of length \(x \mathrm {~cm}\) and angle \(A O B x\) radians.
3. Use algebraic integration to find the expected value of the area of triangle \(O A B\). Give your answer as an exact value.

1. The continuous random variable \(X\) has cumulative distribution function given by

$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r } 0 & x < 2 \\ 1.25 - \frac { 2.5 } { x } & 2 \leqslant x \leqslant 10 \\ 1 & x > 10 \end{array} \right.$$

1. Find \(\mathrm { P } ( \{ X < 5 \} \cup \{ X > 8 \} )\)
2. Find the median of \(X\).
3. Find \(\mathrm { E } \left( X ^ { 2 } \right)\)
4. 1. Sketch the probability density function of \(X\).
   2. Describe the skewness of the distribution of \(X\).

1. The continuous random variable \(X\) has cumulative distribution function given by

$$\mathrm { F } ( x ) = \left\{ \begin{array} { c r } 0 & x < 1 \\ 1.5 x - 0.25 x ^ { 2 } - 1.25 & 1 \leqslant x \leqslant 3 \\ 1 & x > 3 \end{array} \right.$$

1. Find the exact value of the median of \(X\)
2. Find \(\mathrm { P } ( X < 1.6 \mid X > 1.2 )\) The random variable \(Y = \frac { 1 } { X }\)
3. Specify fully the cumulative distribution function of \(Y\)
4. Hence or otherwise find the mode of \(Y\)

1. The continuous random variable \(X\) has cumulative distribution function given by

$$F ( x ) = \left\{ \begin{array} { c r } 0 & x < 0 \\ k \left( x - a x ^ { 2 } \right) & 0 \leqslant x \leqslant 4 \\ 1 & x > 4 \end{array} \right.$$ The values of \(a\) and \(k\) are positive constants such that \(\mathrm { P } ( X < 2 ) = \frac { 2 } { 3 }\)

1. Find the exact value of the median of \(X\)
2. Find the probability density function of \(X\)
3. Hence, deduce the value of the mode of \(X\), giving a reason for your answer.

1. A continuous random variable \(X\) has probability density function

$$f ( x ) = \left\{ \begin{array} { c l } a x ^ { - 2 } - b x ^ { - 3 } & 2 \leqslant x < \infty \\ 0 & \text { otherwise } \end{array} \right.$$ where \(a\) and \(b\) are constants. Given that \(\mathrm { P } ( X \leqslant 4 ) = \frac { 3 } { 8 }\)

1. use algebraic integration to show that \(a = 3\) Show your working clearly.
2. Find the exact value of the median of \(X\)

4 The discrete random variable \(X\) has the distribution \(\mathrm { U } ( n )\).

1. Use the results \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\) and \(\mathrm { E } ( X ) = \frac { n + 1 } { 2 }\) to show that \(\operatorname { Var } ( X ) = \frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)\). It is given that \(\mathrm { E } ( X ) = 13\).
2. Find the value of \(n\).
3. Find \(\mathrm { P } ( X < 7.5 )\). It is given that \(\mathrm { E } ( a X + b ) = 10\) and \(\operatorname { Var } ( a X + b ) = 117\), where \(a\) and \(b\) are positive.
4. Calculate the value of \(a\) and the value of \(b\).

7 The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = \begin{cases} \frac { 1 } { 4 } x \left( 4 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise } \end{cases}$$

1. Show that \(\mathrm { f } ( x )\) satisfies the conditions for a probability density function.
2. Find the value of \(a\) such that \(\mathrm { P } ( X < a ) = \frac { 15 } { 16 }\).

4 A survey is carried out into the length of time for which customers wait for a response on a telephone helpline. A statistician who is analysing the results of the survey starts by modelling the waiting time, \(x\) minutes, by an exponential distribution with probability density function (PDF) $$\mathrm { f } ( x ) = \begin{cases} \lambda \mathrm { e } ^ { - \lambda x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$ \section*{(i) In this question you must show detailed reasoning.} The mean waiting time is found to be 5.0 minutes. Show that \(\lambda = 0.2\).
(ii) Use the model to calculate the probability that a customer has to wait longer than 20 minutes for a response. In practice it is found that no customer waits for more than 15 minutes for a response. The statistician constructs an improved model to incorporate this fact.
(iii) On the diagram in the Printed Answer Booklet, sketch the following, labelling the curves clearly:

1. the PDF of the model using the exponential distribution,
2. a possible PDF for the improved model.