5.03a Continuous random variables: pdf and cdf

The continuous random variable \(X\) takes values in the interval \(0 \leqslant x \leqslant 3\) only. For \(0 \leqslant x \leqslant 3\) the graph of its probability density function f consists of two straight line segments meeting at the point \(( 1 , k )\), as shown in the diagram. Find \(k\) and hence show that the distribution function F is given by $$\mathrm { F } ( x ) = \begin{cases} 0 & x \leqslant 0 , \\ \frac { 1 } { 3 } x ^ { 2 } & 0 < x \leqslant 1 , \\ x - \frac { 1 } { 2 } - \frac { 1 } { 6 } x ^ { 2 } & 1 < x \leqslant 3 , \\ 1 & x > 3 . \end{cases}$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\). Find

1. the probability density function of \(Y\),
2. the median value of \(Y\).

5 The continuous random variable \(X\) has probability density function f given by $$\mathrm { f } ( x ) = \begin{cases} 0.01 \mathrm { e } ^ { - 0.01 x } & x \geqslant 0 \\ 0 & x < 0 \end{cases}$$

1. State the value of \(\mathrm { E } ( X )\).
2. Find the median value of \(X\).
3. Find the probability that \(X\) lies between the median and the mean.

7 The waiting time, \(T\) minutes, before a customer is served in a restaurant has distribution function F given by $$\mathrm { F } ( t ) = \begin{cases} 1 - \mathrm { e } ^ { - \lambda t } & t \geqslant 0 \\ 0 & t < 0 \end{cases}$$ where \(\lambda\) is a positive constant. The standard deviation of \(T\) is 8 . Find

1. the value of \(\lambda\),
2. the probability that a customer has to wait between 5 and 10 minutes before being served,
3. the median value of \(T\).

9 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 a } & - a \leqslant x \leqslant a \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a positive constant. Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = \mathrm { e } ^ { X }\). Find the distribution function of \(Y\). Given that \(a = 4\), find the value of \(k\) for which \(\mathrm { P } ( Y \geqslant k ) = 0.25\).

6 The random variable \(X\) has distribution function F given by $$\mathrm { F } ( x ) = \begin{cases} 1 - \mathrm { e } ^ { - 0.6 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$ Identify the distribution of \(X\) and state its mean. Find

1. \(\mathrm { P } ( X > 4 )\),
2. the median of \(X\).

7 A random sample of 80 observations of the continuous random variable \(X\) was taken and the values are summarised in the following table. It is required to test the goodness of fit of the distribution having probability density function f given by $$f ( x ) = \begin{cases} \frac { 3 } { x ^ { 2 } } & 2 \leqslant x < 6 \\ 0 & \text { otherwise. } \end{cases}$$ Show that the expected frequency for the interval \(2 \leqslant x < 3\) is 40 and calculate the remaining expected frequencies. Carry out a goodness of fit test, at the \(10 \%\) significance level.

7 The random variable \(T\) is the lifetime, in hours, of a randomly chosen decorative light bulb of a particular type. It is given that \(T\) has a negative exponential distribution with mean 1000 hours.

1. Write down the probability density function of \(T\).
2. Find the probability that a randomly chosen bulb of this type has a lifetime of more than 2000 hours. A display uses 10 randomly chosen bulbs of this type, and they are all switched on simultaneously. Find the greatest value of \(t\) such that the probability that they are all alight at time \(t\) hours is at least 0.9 .

9 Cotton cloth is sold from long rolls of cloth. The number of flaws on a randomly chosen piece of cloth of length \(a\) metres has a Poisson distribution with mean \(0.8 a\). The random variable \(X\) is the length of cloth, in metres, between two successive flaws.

1. Explain why, for \(x \geqslant 0 , \mathrm { P } ( X > x ) = \mathrm { e } ^ { - 0.8 x }\).
2. Find the probability that there is at least one flaw in a 4 metre length of cloth.
3. Find
   1. the distribution function of \(X\),
   2. the probability density function of \(X\),
   3. the interquartile range of \(X\).

9 The continuous random variable \(X\) has probability density function given by $$\mathrm { f } ( x ) = \begin{cases} 0 & x < 2 \\ a \mathrm { e } ^ { - ( x - 2 ) } & x \geqslant 2 \end{cases}$$ where \(a\) is a constant. Show that \(a = 1\). Find the distribution function of \(X\) and hence find the median value of \(X\). The random variable \(Y\) is defined by \(Y = \mathrm { e } ^ { X }\). Find

1. the probability density function of \(Y\),
2. \(\mathrm { P } ( Y > 10 )\).

8 The random variable \(X\) has probability density function f given by $$\mathrm { f } ( x ) = \begin{cases} 2 \mathrm { e } ^ { - 2 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\).
2. Find the median value of \(X\). The random variable \(Y\) is defined by \(Y = \mathrm { e } ^ { X }\).
3. Find the probability density function of \(Y\).

6 The random variable \(T\) is the lifetime, in hours, of a randomly chosen battery of a particular type. It is given that \(T\) has a negative exponential distribution with mean 400 hours.

1. Write down the probability density function of \(T\).
2. Find the probability that a battery of this type has a lifetime that is less than 500 hours.
3. Find the median of the distribution.

10 The continuous random variable \(X\) has probability density function given by $$\mathrm { f } ( x ) = \begin{cases} 0 & x < 0 , \\ \frac { a } { 2 ^ { x } } & x \geqslant 0 , \end{cases}$$ where \(a\) is a positive constant. By expressing \(2 ^ { x }\) in the form \(\mathrm { e } ^ { k x }\), where \(k\) is a constant, show that \(X\) has a negative exponential distribution, and find the value of \(a\). State the value of \(\mathrm { E } ( X )\). The variable \(Y\) is related to \(X\) by \(Y = 2 ^ { X }\). Find the distribution function of \(Y\) and hence find its probability density function.

6 The continuous random variable \(X\) has probability density function f given by $$\mathrm { f } ( x ) = \begin{cases} 0 & x < 1 \\ \frac { 1 } { 2 } & 1 \leqslant x \leqslant 3 \\ 0 & x > 3 \end{cases}$$ Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find

1. the probability density function of \(Y\),
2. the expected value and variance of \(Y\).

7 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 2 } { 15 } x & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Show that the distribution function G of \(Y\) is given by $$\mathrm { G } ( y ) = \begin{cases} 0 & y < 1 \\ \frac { 1 } { 15 } \left( y ^ { \frac { 2 } { 3 } } - 1 \right) & 1 \leqslant y \leqslant 64 \\ 1 & y > 64 \end{cases}$$ Find

1. the median value of \(Y\),
2. \(\mathrm { E } ( Y )\).

6 The random variable \(X\) has probability density function f given by $$\mathrm { f } ( x ) = \begin{cases} \frac { 1 } { 6 } \mathrm { e } ^ { - \frac { 1 } { 6 } x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$ Find

1. the distribution function of \(X\),
2. the probability that \(X\) lies between the median and the mean.

The continuous random variable \(X\) takes values in the interval \(0 \leqslant x \leqslant 5\) only. For \(0 \leqslant x \leqslant 5\) the graph of its probability density function f consists of two straight line segments, as shown in the diagram. Find \(k\) and show that f is given by $$f ( x ) = \begin{cases} \frac { 1 } { 8 } x & 0 \leqslant x \leqslant 2 \\ \frac { 1 } { 4 } & 2 < x \leqslant 5 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is given by \(Y = X ^ { 2 }\).

1. Find the probability density function of \(Y\).
2. Show that \(\mathrm { E } ( Y ) = 10.25\).
3. Show that the median of \(Y\) is the square of the median of \(X\).

6 The random variable \(T\) is the time, in suitable units, between two successive arrivals in a hospital casualty department. The probability density function of \(T\) is f , where $$\mathrm { f } ( t ) = \begin{cases} 0.2 \mathrm { e } ^ { - 0.2 t } & t \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$ State the expected value of \(T\). Write down the distribution function of \(T\) and find \(\mathrm { P } ( T > 10 )\).

Guided tours of a museum begin every 60 minutes. A randomly chosen tourist arrives \(X\) minutes after the start of a tour. The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { ( x - 20 ) ^ { 2 } } { 24000 } & 0 < x < 60 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(T\) is the time that the tourist has to wait for the next tour to begin. Show that the distribution function G of \(T\) is given by $$\mathrm { G } ( t ) = \begin{cases} 0 & t \leqslant 0 \\ \frac { 8 } { 9 } - \frac { ( 40 - t ) ^ { 3 } } { 72000 } & 0 < t < 60 \\ 1 & t \geqslant 60 \end{cases}$$ Find the median and the mean of \(T\).

7 The time, \(T\) seconds, between successive cars passing a particular checkpoint on a wide road has probability density function f given by $$\mathrm { f } ( t ) = \begin{cases} \frac { 1 } { 100 } \mathrm { e } ^ { - 0.01 t } & t \geqslant 0 \\ 0 & \text { otherwise } . \end{cases}$$

1. State the expected value of \(T\).
2. Find the median value of \(T\). Sally wishes to cross the road at this checkpoint and she needs 20 seconds to complete the crossing. She decides to start out immediately after a car passes. Find the probability that she will complete the crossing before the next car passes.

10 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 2 } & 1 \leqslant x \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find the distribution function of \(Y\). Sketch the graph of the probability density function of \(Y\). Find the probability that \(Y\) lies between its median value and its mean value.

7 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 21 } x ^ { 2 } & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 2 }\). Show that \(Y\) has probability density function given by $$\operatorname { g } ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16 \\ 0 & \text { otherwise } \end{cases}$$ Find

1. the median value of \(Y\),
2. the expected value of \(Y\).

5 The distance, \(X \mathrm {~km}\), completed by a new car before any mechanical fault occurs has distribution function F given by $$\mathrm { F } ( x ) = \begin{cases} 1 - \mathrm { e } ^ { - a x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$ where \(a\) is a positive constant. The mean value of \(X\) is 10000 . Find

1. the value of \(a\),
2. the probability that a new car completes less than 15000 km before any mechanical fault occurs. The probability that a new car completes at least \(d \mathrm {~km}\) before any mechanical fault occurs is 0.75 .
3. Find the value of \(d\).

7 The random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\). The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Find
2. the probability density function of \(Y\),
3. the value of \(k\) for which \(\mathrm { P } ( Y \geqslant k ) = \frac { 7 } { 12 }\).

7 The random variable \(X\) has probability density function f given by $$\mathrm { f } ( x ) = \begin{cases} 0.2 \mathrm { e } ^ { - 0.2 x } & x \geqslant 0 \\ 0 & \text { otherwise } \end{cases}$$

1. Find the distribution function of \(X\).
2. Find \(\mathrm { P } ( X > 2 )\).
3. Find the median of \(X\).

7 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 21 } x ^ { 2 } & 1 \leqslant x \leqslant 4 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 2 }\).

1. Show that \(Y\) has probability density function given by $$g ( y ) = \begin{cases} \frac { 1 } { 42 } y ^ { \frac { 1 } { 2 } } & 1 \leqslant y \leqslant 16 \\ 0 & \text { otherwise } \end{cases}$$
2. Find the median value of \(Y\).
3. Find the expected value of \(Y\).