Question

The Long Term Evolution (LTE) standard for wireless communication has a goal of achieving 100 megabits per second (Mbps) data transfer rates between user and target application. That is a lofty goal, even for peak performance. In addition, the connection speed can vary due to the overall communication traffic in the area. Define random variable Xas the data transfer rate that a phone experiences. Assume Xis uniformly distributed on the interval 4.25 to 8.75 Mbps.

I. Determine the mean data transfer rate and the standard deviation for X.
II. Determine the probability that the data transfer rate is less than 5.5 Mbps at any given time.
III. Determine the probability that the data transfer rate is between 6.0 and 7.5 Mbps at any given tim.

Answer 1

I.
`E(X)=6.5`

σ = 1.299

II.
`P(X<5.5)=0.2777`

III.
`P(6.0<X<7.5)= 0.3333`

Step-by-step explanation:

Let's start defining the random variable X.

X : ''The data transfer rate that a phone experiences''

X is a continuous random variable.

X ~ U (4.25,8.75)

For a uniformly distributed random variable the probability density function is :

X ~ U [a,b]

`f(x)=(1)/(b-a)` if x ∈ (a,b)

`f(x)=0` if x ∉ (a,b)

In this exercise :

`f(x)=(1)/(8.75-4.25)=(1)/(4.5)` if x ∈ (4.25,8.75)

`f(x)=0` if x ∉ (4.25,8.75)

I. The mean of a uniformly distributed random variable is :

`E(X)=(a+b)/(2)`

`E(X)=(4.25+8.75)/(2)=6.5`

`E(X)=6.5`

The variance of a uniformly distributed random variable is :

`Var(X)=((b-a)^(2))/(12)=((8.75-4.25)^(2))/(12)=1.6875`

For the standard deviation :

`StandardDeviation=√(Var(X))=√(1.6875)=1.299`

σ = 1.299

II.
`P(X<5.5)`

If c = - ∞ and d = 5.5 ⇒
`P(X<5.5)=\int\limits^d_c {f(x)} \, dx`

`P(X<5.5)=(1)/(4.5).(5.5-4.25)=0.2777`

III.
`P(6.0<X<7.5)`

If e = 6.0 and f = 7.5 ⇒
`P(6.0<X<7.5)=\int\limits^f_e {f(x)} \, dx=(1)/(4.5).(7.5-6)=0.3333`

`P(6.0<X<7.5)=0.3333`