Consider the probability density functions shown:
State the function p \left( x \right).
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
For each probability density function, find the expected value of a random variable X if it is distributed according to p \left( x \right).
p \left( x \right) = \dfrac{1}{40} when 10 \leq x \leq 50
p \left( x \right) = \dfrac{3}{160} \left(x^{2} + 4 x\right) when 0 \leq x \leq 4
Consider the following probability density functions:
Find the value of k.
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
\\ p \left( x \right)=\begin{cases}k x^{2} & \text{for } 4 \leq x \leq 7 \\ 0 & \text{otherwise} \end{cases}
p \left( x \right) = k \left(2 - x\right) \left(x - 6\right) when 2 \leq x \leq 6
p \left( x \right) = k x^{3} when 1 \leq x \leq 5
The length of a newborn baby, in centimeters, is a continuous random variable X which is defined by the probability density function p \left( x \right) = k \sin \left( \dfrac{\pi}{25} \left(x - 45\right)\right) when 45 \leq x \leq 70.
Find the value of k.
Using the product rule, find \dfrac{d}{d x}\left( - \dfrac{1}{2} x \cos \left( \dfrac{\pi}{25} \left(x - 45\right)\right) \right).
Hence, calculate the expected length of a randomly selected newborn.
Due to an outbreak of disease and a shortage of supply, grocers predict the prices of bananas, X, to be anywhere between \$5.90 and \$9.30 per kilogram in the next 6 months, with all prices equally likely. We are to model X as a continuous random variable.
Let p \left( x \right) be the probability density function for the random variable X. State the function defining this distribution.
Calculate the expected price of bananas in 6 months time.
If the predicted price is known to be at least \$6.90 per kilogram, calculate the probability that it’s at most \$7.60 per kilogram.
In 6 months time, the price of bananas at 10 grocers around the country will be studied. Find the probability that exactly 6 grocers are charging more than \$7.70 per kilogram. Round your answers to three decimal places.
Find the probability that more than 2 grocers of the 10 are charging more than \$7.70. Round your answers to three decimal places.
The time in minutes after 6 am that the recycle collection arrives on collection day can be represented by the random variable X, which is uniformly distributed over the domain \left[0, 54\right].
Let p \left( x \right) be the probability density function for the random variable X. State the function defining this distribution.
Calculate the probability, p, that the collection arrives between 6:12 am and 6:38 am.
Find the expected length of time residents must wait until their recycle is collected.
Out of 8 collection days, find the probability, q, that the collection arrives before 6:15 am at least twice. Round your answer to two decimal places.
Consider the probability density function p \left( x \right) = k e^{ - k x } when 0 \leq x.
Find the value of k given that the median value is 2 \ln 2.
Using the product rule, find \dfrac{d}{d x}\left( - x e^{ - 0.5 x } \right).
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
The length of time, in minutes, between customers calling a help centre can be modelled by the probability density function p \left( x \right) = k e^{ - k x } when 0 \leq x.
Find the value of k given that P \left( X \lt 1 \right) = 1 - e^{ - \frac{2}{3} }.
Using the product rule, find \dfrac{d}{d x}\left( - x e^{ - \frac{2}{3} x } \right).
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
Consider the probability density function\\ p \left(x \right)=\begin{cases}\dfrac{1}{20} & \text{for } 25 \leq x \leq 45 \\ 0 & \text{otherwise} \end{cases}
Find the expected value of p \left( x \right).
Find the variance of p \left( x \right).
Consider the probability density function p, where p \left( x \right) \gt 0 when 4 \leq x \leq 12 and p \left( x \right) = 0 otherwise. The graph of y = p \left( x \right) is shown
Find the expression that represents p \left( x \right) on the domain 4 \leq x \leq 12.
Find the expected value of p \left( x \right).
By performing an integration similar in part (b), we can find that E \left(X^{2}\right) = \dfrac{368}{5}.
Hence, calculate the variance, V \left(X\right).
The probability density function of a random variable X is given. Its non-zero values lie in the region 0 \leq x \leq k.
Calculate the value of k.
State the equation that defines the probability distribution function of X in the domain 0 \leq x \leq k.
Find the expected value of p \left( x \right).
By performing an integration similar in the part (c), we can find that \\ E \left(X^{2}\right) = \dfrac{98}{3}. Hence, calculate the variance, V \left(X\right).
Consider the probability density function defined by\\ p \left( x \right)=\begin{cases}\dfrac{1}{18} \left(x + 1\right) & \text{for } - 1 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases}
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
By performing an integration similar to the one in part (a), we can find that E \left( X^{2} \right) = 11. Hence, calculate the standard deviation \sigma of p \left( x \right). Round your answer to two decimal places.
Consider the probability density function defined by\\ p \left( x \right)=\begin{cases} \dfrac{3}{160} \left(x^{2} + 4 x\right) & \text{for } 0 \leq x \leq 4 \\ 0 & \text{otherwise} \end{cases}
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
By performing an integration similar to the one in part (a), we can find that \\ E \left( X^{2} \right) = \dfrac{216}{25}. Hence, calculate the variance V \left(X\right) of p \left( x \right).
Consider the probability density function defined by\\ p \left( x \right)=\begin{cases} k x^{2} & \text{for } 2 \leq x \leq 5 \\ 0 & \text{otherwise} \end{cases}
Find the value of k.
Find the expected value of a random variable X if it is distributed according to p \left( x \right).
By performing an integration similar to the one in part (b), we can find that \\ E \left(X^{2}\right) = \dfrac{1031}{65}. Hence, calculate the variance V \left(X\right) of p \left( x \right).
Mail gets delivered to an office any time between 10:15 am and 10:55 am each day. The probability that the mail arrives at any particular time in this period is the same. Let X be the time after 10:15 am that the mail is delivered:
Define the probability distribution function, p for the random variable X.
Find the probability that the mail is delivered to the office after 10:30 am.
Find the probability the mail is delivered before 10:50 am, given that it was delivered after 10:25 am.
Find the expected value of p \left( x \right).
Find the variance of p \left( x \right).
Driving from Brisbane to the Sunshine Coast takes somewhere between 40 and 120 minutes. A random variable X is the number of minutes (in excess of 40 minutes) that it takes to make the trip from Brisbane to the Sunshine Coast. Consider the probability density function p, where the values of p \left( x \right) are those shown below.\\ p \left( x \right)=\begin{cases} \dfrac{1}{1600}x; & 0 \leq x \leq 40 \\ -\dfrac{1}{1600}x+\dfrac{1}{20}; & 40 \lt x \leq 80 \\ 0; & \text{otherwise} \end{cases}
State the expected number of minutes the trip takes.
Find the probability that a trip from Brisbane to Sunshine Coast will take more than 100 minutes.
Calculate the probability that the trip will take between 57 and 73 minutes.
By integrating x^{2} p \left( x \right) over the entire domain, we can find that E \left(X^{2}\right) = \dfrac{5600}{3}. Hence, calculate the variance V \left(X\right) of p \left( x \right).
Consider the probability density function p defined by\\ p \left( x \right)=\begin{cases} k \cos \left( \dfrac{\pi}{4} x\right) & \text{for } 0 \leq x \leq 2 \\ 0 & \text{otherwise} \end{cases}
Find the value of k.
Using the product rule, find \dfrac{d}{d x}\left( x \sin \left( \dfrac{\pi}{4} x\right)\right).
Hence, find the expected value E \left(X\right) of p \left( x \right).
By performing an integration similar to the one in part (c), we can find that \\ E \left(X^{2}\right) = \dfrac{\pi^{2} - 8}{\dfrac{\pi^{2}}{4}}
Hence, calculate the standard deviation \sigma of p \left( x \right). Give your answer to two decimal places.
An online support centre receives requests for assistance through instant messaging. The time between requests for support on a Friday morning, in time t seconds can be modelled by the random variable T with a probability function p defined by\\ p \left (x \right)=\begin{cases} \dfrac{1}{10} e^{ - \frac{t}{10} } & \text{for all } t \geq 0\\ 0 & \text{otherwise} \end{cases}
Find the probability that the time between two requests is less than 5 seconds. Give your answer correct to two decimal places.
If the time between requests is more than 7 seconds, find the probability that is was less than 12 seconds. Give your answer correct to two decimal places.
Using the product rule, find \dfrac{d}{d t}\left( - t e^{ - \frac{t}{10} } \right).
Find the expected time between requests.
By performing an integration similar to the one in part (d), we can find that E \left(T^{2}\right) = 200. Hence, calculate the standard deviation \sigma of p \left( t \right).