Question

Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider groups of 15 trucks and let X be the number of trucks in a group that have passed the inspection.

a. Verify that this is a binomial distribution
b. For what proportion of groups will between 8 and 10 trucks (inclusive) pass the inspection?
c. For what proportion of groups will exactly 3 trucks fail the inspection?
d. Find the mean and the standard deviation of the random variable X

Answer 1

a)For this case, we assume that we have independence bewteen the vents and each trial have a probability of success , so then the variable X follows a rdon variable

b)
`P(X=8)=(15C8)(0.75)^8 (1-0.75)^(15-8)=0.0566.`

`P(X=9)=(14C6)(0.75)^6 (1-0.75)^(159)=0.0566`

`P(X=10)=(15C7)(0.75)^7 (1-75)^(15-10)=0.0566`

And adding we got: 0.1698

c)
`P(X=3)=(15C3)(0.75)^3 (1-0.75)^(15-8)=0.0117`

d) The expected value is given by:

`E(X) = np = 15*0.75=11.25`

The deviation i given by:

`Sd(X) = \sqrt{15*0.75*(1-0.75)=1.677`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n p)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

Part a

For this case, we assume that we have independence bewteen the vents and each trial have a probability of success , so then the variable X follows a rdon variable

Part b

`P(X=8)=(15C8)(0.75)^8 (1-0.75)^(15-8)=0.0566.`

`P(X=9)=(14C6)(0.75)^6 (1-0.75)^(159)=0.0566`

`P(X=10)=(15C7)(0.75)^7 (1-75)^(15-10)=0.0566`

And adding we got: 0.1698

Part c

`P(X=3)=(15C3)(0.75)^3 (1-0.75)^(15-8)=0.0117`

Part d

The expected value is given by:

`E(X) = np = 15*0.75=11.25`

The deviation i given by:

`Sd(X) = \sqrt{15*0.75*(1-0.75)=1.677`