5.03c Calculate mean/variance: by integration

4 A continuous random variable \(X\) has a probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { k } & a \leqslant x \leqslant b \\ 0 & \text { otherwise } \end{cases}$$ where \(b > a > 0\).

1. 1. Prove that \(k = b - a\).
   2. Write down the value of \(\mathrm { E } ( X )\).
   3. Show, by integration, that \(\mathrm { E } \left( X ^ { 2 } \right) = \frac { 1 } { 3 } \left( b ^ { 2 } + a b + a ^ { 2 } \right)\).
   4. Hence derive a simplified formula for \(\operatorname { Var } ( X )\).
2. Given that \(a = 4\) and \(\operatorname { Var } ( X ) = 3\), find the numerical value of \(\mathrm { E } ( X )\).
   \includegraphics[max width=\textwidth, alt={}]{34517557-011e-4956-be7d-b26fe5e64d0a-08_1347_1707_1356_153}

2 The continuous random variable \(X\) has probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { k } & a \leqslant x \leqslant b \\ 0 & \text { otherwise } \end{cases}$$

1. Write down, in terms of \(a\) and \(b\), the value of \(k\).
2. 1. Given that \(\mathrm { E } ( X ) = 1\) and \(\operatorname { Var } ( X ) = 3\), find the values of \(a\) and \(b\).
   2. Four independent values of \(X\) are taken. Find the probability that exactly one of these four values is negative.
      [0pt] [3 marks]

6 The continuous random variable \(X\) has the cumulative distribution function $$\mathrm { F } ( x ) = \begin{cases} 0 & x < 0 \\ \frac { 1 } { 2 } x - \frac { 1 } { 16 } x ^ { 2 } & 0 \leqslant x \leqslant 4 \\ 1 & x > 4 \end{cases}$$

1. Find the probability that \(X\) lies between 0.4 and 0.8 .
2. Show that the probability density function, \(\mathrm { f } ( x )\), of \(X\) is given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } - \frac { 1 } { 8 } x & 0 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$
3. 1. Find the value of \(\mathrm { E } ( X )\).
   2. Show that \(\operatorname { Var } ( X ) = \frac { 8 } { 9 }\).
4. The continuous random variable \(Y\) is defined by $$Y = 3 X - 2$$ Find the values of \(\mathrm { E } ( Y )\) and \(\operatorname { Var } ( Y )\).

6. Patients suffering from 'flu are treated with a drug. The number of days, \(t\), that it then takes for them to recover is modelled by the continuous random variable \(T\) with the probability density function $$\begin{array} { l l } \mathrm { f } ( t ) = \frac { 3 t ^ { 2 } ( 4 - t ) } { 64 } & 0 \leq t \leq 4 \\ \mathrm { f } ( t ) = 0 & \text { otherwise. } \end{array}$$

1. Find the mean and standard deviation of \(T\).
2. Find the probability that a patient takes more than 3 days to recover.
3. Two patients are selected at random. Find the probability that they both recover within three days.
4. Comment on the suitability of the model.

2. The random variable \(X\), which can take any value in the interval \(1 \leq X \leq n\), is modelled by the continuous uniform distribution with mean 12.

1. Show that \(n = 23\) and find the variance of \(X\).
2. Find \(\mathrm { P } ( 10 < X < 14 )\).

4. A continuous random variable \(X\) has probability density function $$\begin{array} { l l } \mathrm { f } ( x ) = 0 & x < 1 , \\ \mathrm { f } ( x ) = k x & 1 \leq x \leq 4 , \\ \mathrm { f } ( x ) = 0 & x > 4 . \end{array}$$

1. Sketch a graph of \(\mathrm { f } ( x )\), and hence find the value of \(k\).
2. Calculate the mean and the variance of \(X\). \section*{STATISTICS 2 (A)TEST PAPER 4 Page 2}

7. The fraction of sky covered by cloud is modelled by the random variable \(X\) with probability density function $$\begin{array} { l l } \mathrm { f } ( x ) = 0 & x < 0 \\ \mathrm { f } ( x ) = k x ^ { 2 } ( 1 - x ) & 0 \leq x \leq 1 , \\ \mathrm { f } ( x ) = 0 & x > 1 . \end{array}$$

1. Find \(k\) and sketch the graph of \(\mathrm { f } ( x )\).
2. Find the mean and the variance of \(X\).
3. Find the cumulative distribution function \(\mathrm { F } ( x )\).
4. Given that flying is prohibited when \(85 \%\) of the sky is covered by cloud, show that cloud conditions allow flying nearly \(90 \%\) of the time.

7. In a competition at a funfair, participants have to stay on a log being rotated in a pool of water for as long as possible. The length of time, in tens of seconds, that the competitors stay on the log is modelled by the random variable \(T\) with the following probability density function: $$\mathrm { f } ( t ) = \begin{cases} k ( t - 3 ) ^ { 2 } , & 0 \leq t \leq 3 \\ 0 , & \text { otherwise } \end{cases}$$

1. Show that \(k = \frac { 1 } { 9 }\).
2. Sketch f \(( t )\) for all values of \(t\).
3. Show that the mean time that competitors stay on the \(\log\) is 7.5 seconds. When the competition is next run the organisers decide to make it easier at first by spinning the log more slowly and then increasing the speed of rotation. The length of time, in tens of seconds, that the competitors now stay on the log is modelled by the random variable \(S\) with the following probability density function: $$f ( s ) = \begin{cases} \frac { 1 } { 12 } \left( 8 - s ^ { 3 } \right) , & 0 \leq s \leq 2 \\ 0 , & \text { otherwise } \end{cases}$$
4. Find the change in the mean time that competitors stay on the log.

6. The continuous random variable \(X\) has the following probability density function: $$f ( x ) = \begin{cases} \frac { 1 } { 16 } x , & 2 \leq x \leq 6 \\ 0 , & \text { otherwise } \end{cases}$$

1. Sketch \(\mathrm { f } ( x )\) for all values of \(x\).
2. Find \(\mathrm { E } ( X )\).
3. Show that \(\operatorname { Var } ( X ) = \frac { 11 } { 9 }\).
4. Define fully the cumulative distribution function \(\mathrm { F } ( x )\) of \(X\).
5. Show that the interquartile range of \(X\) is \(2 ( \sqrt { } 7 - \sqrt { 3 } )\). END

6. \begin{figure}[h] \end{figure} Figure 1 shows a square of side \(t\) and area \(t ^ { 2 }\) which lies in the first quadrant with one vertex at the origin. A point \(P\) with coordinates ( \(X , Y\) ) is selected at random inside the square and the coordinates are used to estimate \(t ^ { 2 }\). It is assumed that \(X\) and \(Y\) are independent random variables each having a continuous uniform distribution over the interval \([ 0 , t ]\).
[0pt] [You may assume that \(\mathrm { E } \left( X ^ { n } Y ^ { n } \right) = \mathrm { E } \left( X ^ { n } \right) \mathrm { E } \left( Y ^ { n } \right)\), where \(n\) is a positive integer.]

1. Use integration to show that \(\mathrm { E } \left( X ^ { n } \right) = \frac { t ^ { n } } { n + 1 }\). The random variable \(S = k X Y\), where \(k\) is a constant, is an unbiased estimator for \(t ^ { 2 }\).
2. Find the value of \(k\).
3. Show that \(\operatorname { Var } S = \frac { 7 t ^ { 4 } } { 9 }\). The random variable \(U = q \left( X ^ { 2 } + Y ^ { 2 } \right)\), where \(q\) is a constant, is also an unbiased estimator for \(t ^ { 2 }\).
4. Show that the value of \(q = \frac { 3 } { 2 }\).
5. Find Var \(U\).
6. State, giving a reason, which of \(S\) and \(U\) is the better estimator of \(t ^ { 2 }\). The point \(( 2,3 )\) is selected from inside the square.
7. Use the estimator chosen in part (f) to find an estimate for the area of the square.

6. Emily is monitoring the level of pollution in a river. Over a period of time she has found that the amount of pollution, \(X\), in a 100 ml sample of river water has a continuous distribution with probability density function \(\mathrm { f } ( x )\) given by $$f ( x ) = \left\{ \begin{array} { c c } \frac { 2 x } { a ^ { 2 } } & 0 \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{array} \right.$$ where \(a\) is a constant. Emily takes a random sample \(X _ { 1 } , X _ { 2 } , X _ { 3 } , \ldots , X _ { n }\) to try to estimate the value of \(a\).

1. Show that \(\mathrm { E } ( \bar { X } ) = \frac { 2 a } { 3 }\) and \(\operatorname { Var } ( \bar { X } ) = \frac { a ^ { 2 } } { 18 n }\) The random variable \(S = p \bar { X }\), where \(p\) is a constant, is an unbiased estimator of \(a\).
2. Write down the value of \(p\) and find \(\operatorname { Var } ( S )\). Felix suggests using the statistic \(M = \max \left\{ X _ { 1 } , X _ { 2 } , X _ { 3 } , \ldots , X _ { n } \right\}\) as an estimator of \(a\).
   He calculates \(\mathrm { E } ( M ) = \frac { 2 n } { 2 n + 1 } a\) and \(\operatorname { Var } ( M ) = \frac { n } { ( n + 1 ) ( 2 n + 1 ) ^ { 2 } } a ^ { 2 }\)
3. State, giving your reasons, whether or not \(M\) is a consistent estimator of \(a\). The random variable \(T = q M\), where \(q\) is a constant, is an unbiased estimator of \(a\).
4. Write down, in terms of \(n\), the value of \(q\) and find \(\operatorname { Var } ( T )\).
5. State, giving your reasons, which of \(S\) or \(T\) you would recommend Emily use as an estimator of \(a\). Emily took a sample of 5 values of \(X\) and obtained the following:
   5.3
   4.3 \(\begin{array} { l l } 5.7 & 7.8 \end{array}\) 6.9
6. Calculate the estimate of \(a\) using your recommended estimator from part (e).
7. Find the standard error of your estimate, giving your answer to 2 decimal places.

6. A random sample \(X _ { 1 } , X _ { 2 } , X _ { 3 } , \ldots , X _ { 2 n }\) is taken from a population with mean \(\frac { \mu } { 3 }\) and variance \(3 \sigma ^ { 2 }\). A second random sample \(Y _ { 1 } , Y _ { 2 } , Y _ { 3 } , \ldots , Y _ { n }\) is taken from a population with mean \(\frac { \mu } { 2 }\) and variance \(\frac { \sigma ^ { 2 } } { 2 }\), where the \(X\) and \(Y\) variables are all independent. \(A\), \(B\) and \(C\) are possible estimators of \(\mu\), where $$\begin{aligned} & A = \frac { X _ { 1 } + X _ { 2 } + X _ { 3 } + Y _ { 1 } + Y _ { 2 } } { 2 } \\ & B = \frac { 3 X _ { 1 } } { 2 } + \frac { 2 Y _ { 1 } } { 3 } \\ & C = \frac { 3 X _ { 1 } + 4 Y _ { 1 } } { 3 } \end{aligned}$$

1. Show that two of \(A , B\) and \(C\) are unbiased estimators of \(\mu\) and find the bias of the third estimator of \(\mu\).
2. Showing your working clearly, find which of \(A\), \(B\) and \(C\) is the best estimator of \(\mu\). The estimator $$D = \frac { 1 } { k } \left( \sum _ { i = 1 } ^ { 2 n } X _ { i } + \sum _ { i = 1 } ^ { n } Y _ { i } \right)$$ is an unbiased estimator of \(\mu\).
3. Find \(k\) in terms of \(n\).
4. Show that \(D\) is also a consistent estimator of \(\mu\).
5. Find the least value of \(n\) for which \(D\) is a better estimator of \(\mu\) than any of \(A\), \(B\) or \(C\).

6. A random sample of size \(n\) is taken from the random variable \(X\), which has a continuous uniform distribution over the interval [ \(0 , a\) ], \(a > 0\) The sample mean is denoted by \(\bar { X }\)

1. Show that \(Y = 2 \bar { X }\) is an unbiased estimator of \(a\) The maximum value, \(M\), in the sample has probability density function $$f ( m ) = \left\{ \begin{array} { c c } \frac { n m ^ { n - 1 } } { a ^ { n } } & 0 \leqslant m \leqslant a \\ 0 & \text { otherwise } \end{array} \right.$$
2. Find E(M)
3. Show that \(\operatorname { Var } ( M ) = \frac { n a ^ { 2 } } { ( n + 2 ) ( n + 1 ) ^ { 2 } }\) The estimator \(S\) is defined by \(S = \frac { n + 1 } { n } M\) Given that \(n > 1\)
4. state which of \(Y\) or \(S\) is the better estimator for \(a\). Give a reason for your answer.

1. The continuous random variable \(X\) has probability density function \(\mathrm { f } ( x )\)

$$f ( x ) = \left\{ \begin{array} { c c } \frac { x } { 2 \theta ^ { 2 } } & 0 \leqslant x \leqslant 2 \theta \\ 0 & \text { otherwise } \end{array} \right.$$ where \(\theta\) is a constant.

1. Use integration to show that \(\mathrm { E } \left( X ^ { N } \right) = \frac { 2 ^ { N + 1 } } { N + 2 } \theta ^ { N }\)
2. Hence
   1. write down an expression for \(\mathrm { E } ( X )\) in terms of \(\theta\)
   2. find \(\operatorname { Var } ( X )\) in terms of \(\theta\) A random sample \(X _ { 1 } , X _ { 2 } , \ldots , X _ { n }\) where \(n \geqslant 2\) is taken to estimate the value of \(\theta\) The random variable \(S _ { 1 } = q \bar { X }\) is an unbiased estimator of \(\theta\)
3. Write down the value of \(q\) and show that \(S _ { 1 }\) is a consistent estimator of \(\theta\) The continuous random variable \(Y\) is independent of \(X\) and is uniformly distributed over the interval \(\left[ 0 , \frac { 2 \theta } { 3 } \right]\), where \(\theta\) is the same unknown constant as in \(\mathrm { f } ( x )\). The random variable \(S _ { 2 } = a X + b Y\) is an unbiased estimator of \(\theta\) and is based on one observation of \(X\) and one observation of \(Y\).
4. Find the value of \(a\) and the value of \(b\) for which \(S _ { 2 }\) has minimum variance.
5. Show that the minimum variance of \(S _ { 2 }\) is \(\frac { \theta ^ { 2 } } { 11 }\)
6. Explain which of \(S _ { 1 }\) or \(S _ { 2 }\) is the better estimator for \(\theta\)

3 The probability density function of the continuous random variable \(X\) is given by $$\mathrm { f } ( x ) = \begin{cases} c + x & - c \leqslant x \leqslant 0 \\ c - x & 0 \leqslant x \leqslant c \\ 0 & \text { otherwise } \end{cases}$$ where \(c\) is a positive constant.

1. (A) Sketch the graph of the probability density function.
   (B) Show that \(c = 1\).
2. Find \(\mathrm { P } \left( X < \frac { 1 } { 4 } \right)\).
3. Find

4 The random variable \(X\) has a continuous uniform distribution on [ 0,10 ].

1. Find \(\mathrm { P } ( 3 < X < 6 )\).
2. Find each of the following.
   Marisa is investigating the sample mean, \(Y\), of 8 independent values of \(X\). She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of \(X\) together with the value of \(Y\). All of the values in the spreadsheet have been rounded to 2 decimal places. \begin{table}[h]

   1	A	B	C	D	E	F	G	H	I	J
   1	\(X _ { 1 }\)	\(X _ { 2 }\)	\(X _ { 3 }\)	\(X _ { 4 }\)	\(X _ { 5 }\)	\(X _ { 6 }\)	\(X _ { 7 }\)	\(X _ { 8 }\)	\(Y\)
   2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
   3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
   4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
   5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
   6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
   7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
   8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
   9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
   10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
   11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
   12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
   13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
   14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
   15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
   16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
   17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
   18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
   19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
   20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
   21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
   22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
   23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
   24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
   25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
   26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
   27

   \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{table}
3. Use the spreadsheet to estimate \(\mathrm { P } ( 3 < Y < 6 )\).
4. Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i). Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\).
5. Calculate an estimate of \(\mathrm { P } ( 3 < Y < 6 )\) using the Central Limit Theorem. A Normal probability plot of the 25 simulated values of \(Y\) is shown in Fig. 4.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
6. Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate \(\mathrm { P } ( 3 < Y < 6 )\). Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of \(Y\) between 3 and 6. This value is 808.
7. Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\). Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.
8. Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.

2 The continuous random variable \(X\) has cumulative distribution function given by \(F ( x ) = \begin{cases} 0 & x < a , \\ \frac { x - a } { b - a } & a \leqslant x \leqslant b , \\ 1 & x > b , \end{cases}\) where \(a\) and \(b\) are constants with \(0 < \mathrm { a } < \mathrm { b }\).

1. Find \(\mathrm { P } \left( \mathrm { X } < \frac { 1 } { 2 } ( \mathrm { a } + \mathrm { b } ) \right)\).
2. Sketch the graph of the probability density function of \(X\).
3. Find the variance of \(X\) when \(a = 2\) and \(b = 8\).

6 The probability density function of the continuous random variable \(X\) is given by \(f ( x ) = \begin{cases} 2 ( 1 + a x ) & 0 \leqslant x \leqslant 1 , \\ 0 & \text { otherwise } , \end{cases}\) where \(a\) is a constant.

1. Show that \(a = - 1\).
2. Find the cumulative distribution function of \(X\).
3. Find \(\mathrm { P } ( X < 0.5 )\).
4. Show that \(\mathrm { E } ( X )\) is greater than the median of \(X\).

3 At a factory, flour is packed into bags. A model for the mass in grams of flour packed into each bag is \(1500 + X\), where \(X\) is a continuous random variable with probability density function $$f ( x ) = \left\{ \begin{array} { c c } k x ( 6 - x ) & 0 \leq x \leq 6 \\ 0 & \text { elsewhere, } \end{array} \right.$$ where \(k\) is a constant.

1. Show that \(k = \frac { 1 } { 36 }\).
2. Find the probability that a randomly selected bag of flour contains 1505 grams of flour or more.
3. Find

6 A lottery has tickets numbered 1 to \(n\) inclusive, where \(n\) is a positive integer. The random variable \(X\) denotes the number on a ticket drawn at random.

1. Determine \(\mathrm { P } \left( \mathrm { X } \leqslant \frac { 1 } { 4 } \mathrm { n } \right)\) in each of the following cases.
   1. \(n\) is a multiple of 4 .
   2. \(n\) is of the form \(4 k + 1\), where \(k\) is a positive integer. Give your answer as a single fraction in terms of \(n\).
2. Given that \(n = 101\), find the probability that \(X\) is within one standard deviation of the mean.

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]

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. 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.