Question

It is reported that 45% of Canadians do online banking. A random sample of 20 Canadians across the country was drawn to investigate this.

a) Identify the "trial" here. Are trials independent of each other? Briefly explain in the context of the question.

b) Suppose the above situation satisfies all four requirements of a Binomial experiment. Define the Binomial random variable in this question. How many possible values does it have?

c) Find the mean of the Binomial random variable and provide an interpretation of this value in the context of the question.

d) What is the probability that at least 4 but less than 8 Canadians do online banking in this sample?

Answer 1

It is reported that 45% of Canadians do online banking. A random sample of 20 Canadians across the country was drawn to investigate this.

a) Identify the "trial" here. Are trials independent of each other? Briefly explain in the context of the question.

the trial is the subject of the experiment that is the conducted. here we want to know about the Canadians doing online banking.

Since or subjects are 'Canadians' they are our trials. There are two outcomes to this trial either the subject does online banking or they don't.

They are independent that is one's action of online banking does not affect other's action.

b) Suppose the above situation satisfies all four requirements of a Binomial experiment. Define the Binomial random variable in this question. How many possible values does it have?

There are two outcomes to this trial either the subject does online banking or they don't. The binomial event is for the successful outcome. Here it is 'Canadians doing online banking.

'Canadians doing online banking' is the random variable.

Since there are 20 Canadians investigated the number of successes can go from 0 - 20 that is from none to all Canadians in the experiment do online experiment.

There are possible 21 values (0 is inclusive)

c) Find the mean of the Binomial random variable and provide an interpretation of this value in the context of the question.

(n = 20 ,p = 0.45 )

E(X)= np

Mean = 9

On average 9 out of 20 Canadians do online banking.

d) What is the probability that at least 4 but less than 8 Canadians do online banking in this sample?

P( X =x) =

                = 20Cx * 0.45^x * 0.55^(20-x)

P(at least 4 means 4 or more than 4 but less than 8)= P(4 <= X < 8) = P(3 < x< 8) limits exclusive

= P( X = 4) + P( X = 5) + P( X = 6) + P( X = 7)

= 20C4 * 0.45⁴ * 0.55^(20-4) +......... 20C7 * 0.45⁷ * 0.55^(20-7)

Ans:0.24707