Question

Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabilities: x 01 2 3 4 5 0.1 0.15 0.25 0.25 0.15 0.1 Determine the following: (a) F(x) (b) Mean and variance (c) P(X 2) (d) P(X 3.5)

Answer 1

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: number of bars of service on your cell phone.

The posible values of this variable are {0, 1, 2, 3, 4, 5}

a. F(X) is the cummulative distribution function P(X≤x₀) you can calculate it by adding all the point probabilities for each value:

X: 0; 1; 2; 3; 4; 5

P(X): 0.15; 0.25; 0.25; 0.15; 0.1; 0.1

F(X): 0.15; 0.4; 0.65; 0.8; 0.9; 1

The maximum cumulative probability of any F(X) is 1, knowing this, you can calculate the probability of X=5 as: 1 - F(4)= 1 - 0.9= 0.1

b.

The mean of the sample is:

E(X)= ∑Xi*Pi= (0*0.15)+(1*0.25)+(2*0.25)+(3*0.15)+(4*0.1)+(5*0.1)

E(X)= 2.1

V(X)= E(X²)-(E(X))²

V(X)= ∑Xi²*Pi- (∑Xi*Pi)²= 6.7- (2.1)²= 2.29

∑Xi²*Pi= (0²*0.15)+(1²*0.25)+(2²*0.25)+(3²*0.15)+(4²*0.1)+(5²*0.1)= 6.7

c.

P(X<2)

You can read this expression as the probability of having less than 2 bars of service. This means you can have either one or zero service bars, you can rewrite it as:

P(X<2)= P(X≤1)= F(1)= 0.4

d.

P(X≤3.5)

This expression means "the probability of having at most (or less or equal to) 3.5 service bars", this variable doesn't have the value 3.5 in its definition range so you have to look for the accumulated probability until the lesser whole number. This expression includes the probabilities of X=0, X=1, X=2, and X=3, you can express it as the accumulated probability until 3, F(3):

P(X≤3.5)= P(X≤3)= F(3)= 0.80

I hope it helps!