Put some air in your tires: 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) 0.1 0.2 0.3 0.3 0.1
A) Find P (1).
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 uy.
F) Compute the standard deviationo Round the answer to at least three decimal places.

Respuesta :

Answer:

A) P(1) = 0.2

B) P(greater than or equal 3) = 0.4

C) P(4) = 0.1

D) P(0) = 0.1

E) [tex]\mu_y = 0.2[/tex]

F) [tex]\sigma = 0.089[/tex]

Step-by-step explanation:

A)

to find P(1) we see in the table the value of P(x) when x is equal to 1.

when x=1, we have that P(x) = 0.2, therefore P(1) = 0.2

B)

To find P(greather than or equal 3), we just need to find P(3) and P(4), and sum their values. Seeing in the table, we have that P(3) = 0.3 and P(4) = 0.1, so:

P(greater than or equal 3) = P(3) + P(4) = 0.3 + 0.1 = 0.4

C)

All four tires having low air pressure is represented by the value of x equal 4, so we have to find P(4). Seeing in the table, we have that P(4) = 0.1

D)

No tires having low air pressure is represented by the value of x equal 0, so we have to find P(0). Seeing in the table, we have that P(0) = 0.1

E)

To find the mean of P(x) we just need to sum all the values and divide by the number of values:

[tex]\mu_y = (0.1 + 0.2 + 0.3 + 0.3 + 0.1)/5 = 1/5 = 0.2[/tex]

F)

The formula to calculate the standard deviation is:

[tex]\sigma = \sqrt{\frac{1}{N}\sum(x_i-\mu)^2}[/tex]

[tex]\sigma = \sqrt{\frac{1}{5}((0.1-0.2)^2+(0.2-0.2)^2+(0.3-0.2)^2+(0.3-0.2)^2+(0.1-0.2)^2)}[/tex]

[tex]\sigma = \sqrt{\frac{1}{5}(0.01+0+0.01+0.01+0.01)}[/tex]

[tex]\sigma = 0.089[/tex]