Respuesta :
Answer:
Probability distribution
P(x=0)=0.42
P(x=1)=0.50
P(x=2)=0.08
Mean=0.66
Step-by-step explanation:
First, we state the possible values that the variable can take.
- It can take the value x=0, when the 3 altimeters selected are correctly calibrated.
- It can take the value x=1, when only one of the three altimeters is not correctly calibrated.
- It can take the value x=2, when the 2 altimeters that are not correctly calibrated are selected. This is the maximum value for x, because the others altimeters are correctly calibrated
Calculation of propabilities
a) x=0
[tex]P(x=0)=P(1=c)*P(2=c)*P(3=c)=\frac{7}{9}*\frac{6}{8}*\frac{5}{7} =\frac{210}{504}= 0.42[/tex]
b) x=1
In this case, we have three ways of getting one altimeter not correrctly calibrated (in first, second or third place). The three ways have the same probability of occurrence.
[tex]P(x=1)=3*P(1=n)*P(2=c)*P(3=c)\\\\P(x=1)=3*\frac{2*7*6}{9*8*7} =\frac{252}{504}= 0.5[/tex]
c) x=2
In this case, we have three ways of getting the two altimeters not correrctly calibrated (first and second, first and third, and second and third). The three ways have the same probability of occurrence.
[tex]P(x=2)=3*P(1=n)*P(2=n)*P(3=c)\\\\P(x=2)=3*\frac{2*1*7}{9*8*7} =\frac{84}{504} =0.08[/tex]
As it is the complement of P(x=0) and P(x=1) it can also have been calculated as:
[tex]P(x=2)=1-P(x=0)-P(x=1)[/tex]
The mean is equal to the expected value:
[tex]E(x)=P(0)*0+P(1)*1+P(2)*2=0+0.50*1+0.08*2=0.50+0.16=0.66[/tex]
