According to insurance records a car with a certain protection system will be recovered 8787% of the time. find the probability
1 answer:
Let X be the number of cars recovered from stolen cars. Let n be the total number of stolen cars and p be the probability that car will be recovered.
Given: n=10, x=5
It is given that a car with a certain protection system will be recovered 87% of the time. It means probability of recovering a car is p=0.87
Probability that 5 out of 10 stolen cars will be recovered is
P(X=x ) =
P(X=5) =
P(X=5) = 0.0047
The probability that exactly 5 out of 10 stolen cars will be recovered is 0.0047
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We need to 'standardise' the value of X = 14.4 by first calculating the z-score then look up on the z-table for the p-value (which is the probability)
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%
The formula for z-score:
z = (X-μ) ÷ σ
Where
X = 14.4
μ = the average mean = 18
σ = the standard deviation = 1.2
Substitute these value into the formula
z-score = (14.4 - 18) ÷ 1..2 = -3
We are looking to find P(Z < -3)
The table attached conveniently gives us the value of P(Z < -3) but if you only have the table that read p-value to the left of positive z, then the trick is to do:
1 - P(Z<3)
From the table
P(Z < -3) = 0.0013
The probability of the runners have times less than 14.4 secs is 0.0013 = 0.13%