The time t, in hours, that an owl spends hunting each night can be modelled by a continuous random variable with probability density function given below:f \left( t \right) = \begin{cases} \dfrac{k}{32}t \left(4-t \right) & 0\leq t\leq 4 \\ 0 & \text{ otherwise } \end{cases}
Find the value of k.
Calculate the probability the owl spends more than 3 hours hunting during one night.
Calculate the expected time the owl spends hunting in hours.
Calculate the standard deviation of the time the owl spends hunting in hours.
The amount of coffee used by a cafe each week is modelled by a continuous random variable X with a mean of 14.5 \text{ kg} and a standard deviation of 1.5 \text{ kg}. If the coffee costs \\ C = 24 X + 15 in dollars (due to cost per kilogram and weekly delivery fee), find:
The expected weekly cost in dollars.
The standard deviation of the weekly cost in dollars.
Noah always arrives at school between 7:50 am and 8:55 am. The probability distribution function which models the time at which Noah arrives at school is graphed below, where t is the time in minutes after 7:50 am:
State the value of a in minutes.
Find the value of k.
Find the probability, p, that Noah arrives at school after 8:05 am.
Given that Noah arrived before 8:30 am, find the probability, q, that he arrived after 8:25 am.
Calculate the median value, m, of the random variable T, which represents the time, in minutes, Noah arrives to school after 7:50 am. Round your answer to two decimal places.
Use calculus to calculate the expected number of minutes after 7:50 am at which Noah arrives at school.
A written French examination is worth a total of 180 marks. The results of the examination can be modelled by a continuous random variable X where the expected value E \left( X \right) is 117 and the variance V \left( X \right) is 15.
If the results were recorded as percentages, find the mean result.
If the marks were recorded as percentages, find the standard deviation of the results. Round your percentage to one decimal place.
Compared to previous years' exams, this exam was deemed too difficult so it is decided that the marks should be scaled. All raw marks will be multiplied by 1.25 and then 11 marks will be added. Calculate the new mean mark after scaling.
An artist taking commissions on cartoons takes between 2 and 5 hours to complete pieces. The time varies and can be approximately modelled by the following probability density function:f\left(x\right) = \begin{cases} \dfrac{10}{3x^2} & \text{ for } 2\leq x\leq 5 \\ 0 & \text{ otherwise } \end{cases}
Find the mean time, rounded to two decimal places, taken for the artist to create a cartoon in hours.
Find the variance, rounded to two decimal places, for the time taken for the artist to create a cartoon in hours.
If the artist charges a consignment fee of \$30 plus \$50 per hour, what would be the expected price, rounded to the nearest cent, of a piece in dollars?
Calculate the variance, rounded to the nearest whole number, in the price of her cartoons.
The time taken to service a certain brand of washing machine varies between 0 and 3 hours and can be modelled by the following probability density function:f \left(t \right) = \begin{cases} \dfrac{4}{27} \left(x^3-6x^2+9x\right) & \text{ for } 0\leq x\leq 3 \\ 0 & \text{ otherwise } \end{cases}
Find the mean time to service a washing machine in hours.
Find the standard deviation for the time in hours.
If an appliance repair business charges a \$60 callout fee plus \$90 per hour, what would be the expected service charge for this brand in dollars?
What would be the standard deviation for the service charge for this brand in dollars?
A produce merchant sells chicken feed by the kilogram. Over time, they find that their sales range from 2 \text{ kg} to 70 \text{ kg}, with most sales being approximately 20 \text{ kg}.
Sketch a possible probability density function of the weights sold.
Write a function that describes the probability density function in part (a).
Find the probability that the next sale is between 30 and 40 \text{ kg}.
Calculate the expected mass of the sale.
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.
When calling a Health Insurance Provider, the length of time (measured in minutes) that a person is put on hold can be modelled by an exponential distribution. The probability distribution function has the form f \left( t \right) = 2 e^{ - 2 t } for t \geq 0 and f \left( t \right) = 0 elsewhere.
Find the mean length of time spent waiting for an operator while on hold.
Calculate m, the median length of time spent waiting for an operator on hold.
Explain why the median length of time spent waiting on hold is less than the mean time spent.
Find the probability that a randomly selected caller had to wait more than 0.5 minutes to talk to an operator.
For quality control, the experiences of 16 callers are selected at random. Find the probability that at most 5 callers had to wait more than 0.5 minutes to talk to an operator. Round your answer to two decimal places.
Tricia has kept note of the time it takes her to work. The most common time is 22 minutes, with a minimum of 17 minutes and a maximum of 31 minutes.
Sketch a possible probability density function of the time taken to work.
Write a function that describes the triangular probability density function for the time it takes her to get to work.
Find the probability that she takes between 18 and 20 minutes to get to work tomorrow.
Find the probability of Tricia taking less than 25 minutes.
In the next 12 workdays, find the probability, rounded to two decimal places, that on at most 9 of these days she will take less than 25 minutes.
The time t, in hours, that a cow grazes in a paddock per day can be modelled by a continuous random variable with a cumulative distribution function given below:F(t) = \begin{cases} 0, & t \lt 0 \\ \dfrac{1}{288}\left(6t^2-\dfrac{t^3}{3}\right), & 0 \leq t \lt 12 \\ 1, & t \gt 12 \end{cases}
Calculate the probability that the cow grazes for:
Less than 4 hours in a given day.
At least 7 hours in a given day.
Calculate the median time m, in hours, that the cow spends grazing.
The time t, in minutes, that it takes a randomly selected student to complete a puzzle is modelled by a random variable with a probability density function given below:f \left(t \right) = \begin{cases} \dfrac{20t - t^2}{1125}, & 5\leq t\leq 20 \\ 0, & \text{elsewhere} \end{cases}
Write a function that describes the cumulative distribution.
Calculate the probability that a student takes less than 10 minutes to complete the puzzle.
If 30 students are completing the puzzle, calculate the probability that more than 12 of them take less than 10 minutes to complete the puzzle. Round your answer to two decimal places.
Buses go into the city every 10 minutes, so Jenny doesn't bother looking at the timetable before catching the bus.
Construct a probability density function b \left( t \right) where t is the number of minutes for the next bus when she reaches the stop.
Find the probability of Jenny getting a bus within 3 minutes of arriving at the stop.
Find the expected wait time for Jenny.
Calculate the standard deviation of wait times for Jenny.
Of the next 8 times that Jenny catches a bus from this stop, find the probability that, for the first 3 times, she will have to wait for more than 7 minutes.
A computer does a regular virus check every 30 minutes.
Construct a probability density function v \left( t \right) where t is the number of minutes until the next virus check at a randomly chosen time.
Find the probability that a virus check occurs within 5 minutes of a randomly chosen time.
Find the expected wait time for the next virus check.
Calculate the standard deviation of wait times.
In 30 randomly chosen times, find the probability that the wait time until the next virus check is more than 20 minutes on at least 12 occasions. Round your answer to two decimal places.
A busy telephone switchboard with several operators handles, on average, 5 phone calls per minute. Consider the time interval in seconds between two telephone calls arriving at the same switchboard.
This situation can be modelled by an exponential density function of the form \dfrac{1}{k} e^{ - \frac{t}{k} }, t \geq 0 and 0 otherwise.
Find k, the average number of seconds between phone calls.
Calculate the probability of the time between calls being more than 1 minute.
Calculate the expected value of the distribution.
Calculate the expected time between calls in minutes.
Find the probability, rounded to two decimal places, that, out of the next 6 calls, the time between calls was more than 1 minute for exactly 1 pair of calls.
Sean wants to sell a dune buggy on an online auction. He aims to place this up for sale between 9:00 am and 5:00 pm tomorrow. Past experience shows that there is an average of 10 bids per hour. Consider the time interval in minutes between two bids.
This situation can be modelled by an exponential density function of the form \dfrac{1}{k} e^{ - \frac{t}{k} }, t \geq 0 and 0 otherwise.
Calculate the exact probability that for the next 15 bids, the time between bids was more than 2 minutes for the first 5 bids.
A group of people are practicing their accuracy at an archery centre. Each person, in turn, takes a shot at one of the several targets pinned onto the backboard. The horizontal position of the shot from the left side of the backboard is modelled by the following probability density function in metres:f\left(x\right) = \begin{cases} \dfrac{1}{6}\left(1+\cos\pi x\right), & 0\leq x \leq 6 \\ 0, & \text{elsewhere} \end{cases}
Calculate the probability that a person's shot is within 2 metres from the left side of the backboard.
Calculate the expected horizontal location of a person's shot. Give your answer in metres from the left side of the backboard.
Calculate the standard deviation, rounded to two decimal places, of the horizontal location of a person's shot from the left side of the backboard.
Calculate the probability that, of the next 14 shots, the last 5 are within 2 metres from the left side of the backboard.
Aicia observed that the average rate of visits to her website is 2 per day. The time interval in hours between two visits can be modelled by an exponential distribution with a cumulative function given below:F \left( t \right) = \begin{cases} 1-e^{-\frac{t}{k}}, & t \geq 0 \\ 0, & \text{otherwise} \end{cases}
Find k, the average number of hours between visits.
Calculate the probability of the time between two visits being less than 2 hours.
Calculate the probability of the time between two visits being more than 4 hours.
Calculate the median time m, rounded to two decimal places, between visits in hours.
The average number of defects in a 1 \text{ km} strip of optic fibre cable is 4. Consider the distance in metres between two defects. This situation can be modelled by an exponential distribution with a cumulative function given below:F\left(x\right) = \begin{cases} 1-e^{-\frac{x}{k}}, & x \geq 0 \\ 0, & \text{otherwise} \end{cases}
Calculate the median distance m, rounded to two decimal places, between defects in metres.
The price of a particular stock in dollars at the end of the day can be modelled by the following probability density function:f\left(x\right) = \begin{cases} \dfrac{1}{10}\left(1+\sin\left(\pi x-\dfrac{\pi}{2}\right)\right), & 0\leq x \leq 10 \\ 0, & \text{elsewhere} \end{cases}
The prices at the end of each day are independent. Find the probability that, out of 12 days, at least 5 of them saw a stock price exceeding \$4 at the end of the day. Round your answer to two decimal places.