Question

According to a survey, 18% of the car owners said that they get the maintenance service done on their cars according to the schedule recommended by the auto company. Suppose that this result is true for the current population of car owners. a. Let x be a binomial random variable that denotes the number of car owners in a random sample of 12 who get the maintenance service done on their cars according to the schedule recommended by the auto company. What are the possible values that x can assume

Answer 1

Answer:

a) For this case we define the random variable X="number of car owners in a random sample of 12 who get the maintenance service done on their cars according to the schedule recommended by the auto company". And for this case the distribution for X is:

$X \sim Binom (n =12, p=0.18)$

And the possible values for the random variable are X=0,1,2,3,4,5,6,7,8,9,10,11,12

b) Using the probability mass function we got:

$P(X=3)=(12C3)(0.18)^3 (1-0.18)^{12-3}=0.21506$

Step-by-step explanation:

Previous concepts

A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.

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".

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:  

$nCx=\frac{n!}{(n-x)! x!}$  

Solution to the problem

Part a

For this case we define the random variable X="number of car owners in a random sample of 12 who get the maintenance service done on their cars according to the schedule recommended by the auto company". And for this case the distribution for X is:

$X \sim Binom (n =12, p=0.18)$

And the possible values for the random variable are X=0,1,2,3,4,5,6,7,8,9,10,11,12

Part b

Find to 3 decimal places the probability that exactly 3 car owners in a random sample of 12 get the maintenance service done on their cars according to the schedule recommended by the auto company.

Using the probability mass function we got:

$P(X=3)=(12C3)(0.18)^3 (1-0.18)^{12-3}=0.21506$