We'll now put together all we have learned thus far about continuous random variables and begin to apply it to real-world applications.
In each of the following practice questions, we'll be exploring a different probability distribution function, from among the functions we have been exploring in this course.
Before we begin, let's review some of the important information we need from previous chapters that will be useful to recall here.
A probability density function, $f(x)$f(x), must satisfy the following two properties:
- $f(x)\ge0$f(x)≥0 for all $x$x (since probability values are non-negative)
-
$\int_{-\infty}^{+\infty}\ f(x)\ dx=1$∫+∞−∞ f(x) dx=1 (This is because the sum of all the probabilities is $1$1).
Note: Often our probability function occurs between two specific values, on an interval $[a,b]$[a,b] and can be defined as $0$0 elsewhere, thus from the second property above we would have $\int_b^a\ f(x)\ dx=1$∫ab f(x) dx=1.
For a continuous random variable $X$X then the cumulative distribution function (CDF) is:
$F\left(x\right)=P\left(X\le x\right)$F(x)=P(X≤x) for all x.
This means that:
$F(x)=\int_{-\infty}^x\ f(t)\ dt$F(x)=∫x−∞ f(t) dt where $f\left(t\right)$f(t) is the probability density function.
Note: if $f\left(t\right)>0$f(t)>0 on an interval $[a,b]$[a,b], and $0$0 elsewhere, then $F(x)=\int_{-\infty}^x\ f(t)\ dt=\int_a^x\ f(t)\ dt$F(x)=∫x−∞ f(t) dt=∫xa f(t) dt.
The mean, $\mu$μ or $E\left(X\right)$E(X), of a PDF is calculated by:
$E(X)=\int_a^b\ xf\left(x\right)\ dx$E(X)=∫ba xf(x) dx
The median of a PDF is calculated by determining the value of $x$x over the interval $\left[a,b\right]$[a,b] such that the area under $f\left(x\right)$f(x) is equal to $0.5$0.5:
$\int_a^x\ f(x)\ dx=0.5$∫xa f(x) dx=0.5
The variance of a PDF is calculated by:
| $Var\left(X\right)$Var(X) | $=$= | $E\left(\left(x-\mu\right)^2\right)$E((x−μ)2) |
| $=$= | $\int_a^b\ x^2\ f(x)dx-\mu^2$∫ba x2 f(x)dx−μ2 |
It's useful to note that now we've explored both discrete random variables and continuous random variables, we often see a reappearance of discrete random variables, and in particular the binomial distribution, appearing as subparts in the problems we come across. This is a good time to go back and review the binomial distribution before venturing further!
Practice questions
question 1
The time $t$t, in hours, that an owl spends hunting each night can be modelled by a continuous random variable with probability density function given below.
| $f\left(t\right)$f(t) | $=$= |  |
$\frac{k}{32}t\left(4-t\right)$k32t(4−t), $0\le t\le4$0≤t≤4 |
| $0$0 otherwise |
Determine the value of $k$k.
Calculate the probability the owl spends more than $3$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.
Round your answer to two decimal places.
Question 2
Due to an outbreak of disease and a shortage of supply, growers predict the prices of bananas, $X$X, to be anywhere between $\$5.90$$5.90 and $\$9.30$$9.30 per kilogram in the next $6$6 months, with all prices equally likely. We are to model $X$X as a continuous random variable.
Let $p\left(x\right)$p(x) be the probability density function for the random variable $X$X.
State the function defining this distribution.
$p\left(x\right)$p(x) $=$= 
$\editable{}$ if $5.90\le x\le9.30$5.90≤x≤9.30 $\editable{}$ for all other values of $x$x Calculate the expected price of bananas in $6$6 months time.
If the predicted price is known to be at least $\$6.90$$6.90 per kilogram, calculate the probability that it’s at most $\$7.60$$7.60 per kilogram.
In $6$6 months time, the price of bananas at $10$10 grocers around the country will be studied. What is the probability that exactly $6$6 grocers are charging more than $\$7.70$$7.70 per kilogram? Round your answer to three decimal places.
What is the probability that at least $2$2 grocers of the $10$10 are charging more than $\$7.70$$7.70? Round your answer to three decimal places.
question 3
The time $t$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.
| $F\left(t\right)$F(t) | $=$= | |
$0$0, $t<0$t<0 |
|
$\frac{1}{288}\left(6t^2-\frac{t^3}{3}\right)$1288(6t2−t33), $0\le t\le12$0≤t≤12 |
|||
| $1$1, $t>12$t>12 |
Calculate the probability that the cow grazes for less than $4$4 hours in a given day.
Calculate the probability that the cow grazes for at least $7$7 hours in a given day.
Calculate the median time $m$m, in hours, that the cow spends grazing.