# Thread: Homework - Probability Density Function

1. At my local sandwich bar, I have noticed that it always takes at least two minutes to serve a customer and it can take much longer to fulfil an unusual order. The time in minutes taken to serve a customer may be modelled by a continuous random variable T with probability density function f(t)= 64 t5 ,t≥ 2.
(i) Show that the c.d.f. of the random variable T is given by
F(t)= 1 − 16 t4 ,t≥ 2.

(ii) According to the model, what proportion of customers take more than four
minutes to serve?
Find the probability that it takes between five and ten minutes to serve a
customer.

(iii) Use the p.d.f. f(t) to calculate the mean and variance of the time taken to serve a customer.

(iv) Use Formula (4.1) of Unit 1 to calculate the mean time taken to serve a
custome.
Formula (4.1) = (see below)
(v) Simulate the time taken to serve a customer, using the random number
u=0.5536, which is an observation from the uniform distribution U(0,1).

foruma attached.

2. Originally Posted by wallace
At my local sandwich bar, I have noticed that it always takes at least two minutes to serve a customer and it can take much longer to fulfil an unusual order. The time in minutes taken to serve a customer may be modelled by a continuous random variable T with probability density function f(t)= 64 t5 ,t≥ 2.

(i) Show that the c.d.f. of the random variable T is given by

F(t)= 1 − 16 t4 ,t≥ 2.

(ii) According to the model, what proportion of customers take more than four

minutes to serve?

Find the probability that it takes between five and ten minutes to serve a

customer.

(iii) Use the p.d.f. f(t) to calculate the mean and variance of the time taken to serve a customer.

(iv) Use Formula (4.1) of Unit 1 to calculate the mean time taken to serve a

custome.

Formula (4.1) = (see below)

(v) Simulate the time taken to serve a customer, using the random number

u=0.5536, which is an observation from the uniform distribution U(0,1).

foruma attached.
(i) $F(t) = \int_{2}^{t} 64 u^{-5} \, du$.

(ii) $\Pr(T > 4) = \int_{4}^{+\infty} 64 t^{-5} \, dt$. Alternatively, calculate $1 - \Pr(T < 4) = 1 - F(4)$.

$\Pr(5 < T < 10) = \int_{5}^{10} 64 t^{-5} \, dt$. Alternatively, calculate $\Pr(T < 10) - \Pr(T < 5) = F(10) - F(5)$.

(iii) $E(T) = \int_{2}^{+\infty} t (64 t^{-5}) \, dt = \int_{2}^{+\infty} 64 t^{-4} \, dt$.

$Var(T) = E(T^2) - [E(T)]^2$ where $E(T^2) = \int_{2}^{+\infty} t^2 (64 t^{-5}) \, dt = \int_{2}^{+\infty} 64 t^{-3} \, dt$.

(iv) Calculate $\int_2^{+\infty} 1 - \left[ 1 - 16t^{-4}\right] \, dt$.

3. Thanks so much MR. F it has been the great help already.

here is the question on print screen.

4. Originally Posted by wallace
At my local sandwich bar, I have noticed that it always takes at least two minutes to serve a customer and it can take much longer to fulfil an unusual order. The time in minutes taken to serve a customer may be modelled by a continuous random variable T with probability density function f(t)= 64 t5 ,t≥ 2.
(i) Show that the c.d.f. of the random variable T is given by
F(t)= 1 − 16 t4 ,t≥ 2.

(ii) According to the model, what proportion of customers take more than four
minutes to serve?
Find the probability that it takes between five and ten minutes to serve a
customer.

(iii) Use the p.d.f. f(t) to calculate the mean and variance of the time taken to serve a customer.

(iv) Use Formula (4.1) of Unit 1 to calculate the mean time taken to serve a
custome.
Formula (4.1) = (see below)
(v) Simulate the time taken to serve a customer, using the random number
u=0.5536, which is an observation from the uniform distribution U(0,1).