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Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.

x 0 1 2 3 4
P(x) 0.2 0.3 0.1 0.1 0.3

Requried:
a. Find P(3)
b. Find P( Greater than or equal to 3).
c. Find the probability that all four tires have low air pressure.
d. Find the probability that no tires have low air pressure.
e. Compute the mean µX.
f. Compute the standard deviation σX.

Respuesta :

Answer:

a) [tex] P(X=3) = 0.1[/tex]

b) [tex] P(X\geq 3) =1-P(X<3) = 1-P(X\leq 2) = 1-[P(X=0) +P(X=1)+P(X=2)][/tex]

And replacing we got:

[tex] P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4[/tex]

c) [tex] P(X=4) = 0.3[/tex]

d) [tex] P(X=0) = 0.2[/tex]

e) [tex] E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2[/tex]

f) [tex] E(X^2)= \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4[/tex]

And the variance would be:

[tex] Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4[/tex]

And the deviation:

[tex] \sigma =\sqrt{2.4} = 1.549[/tex]

Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

[tex] P(X=3) = 0.1[/tex]

Part b

We want this probability:

[tex] P(X\geq 3) =1-P(X<3) = 1-P(X\leq 2) = 1-[P(X=0) +P(X=1)+P(X=2)][/tex]

And replacing we got:

[tex] P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4[/tex]

Part c

For this case we want this probability:

[tex] P(X=4) = 0.3[/tex]

Part d

[tex] P(X=0) = 0.2[/tex]

Part e

We can find the mean with this formula:

[tex] E(X)= \sum_{i=1}^n X_i P(X_i)[/tex]

And replacing we got:

[tex] E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2[/tex]

Part f

We can find the second moment with this formula

[tex] E(X^2)= \sum_{i=1}^n X^2_i P(X_i)[/tex]

And replacing we got:

[tex] E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4[/tex]

And the variance would be:

[tex] Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4[/tex]

And the deviation:

[tex] \sigma =\sqrt{2.4} = 1.549[/tex]

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