Question

5. It is reported that 20% of Canadian are on CERB (Canada Emergency Response Benefit). In order to further investigate this topic, a researcher selects a random sample of 10 Canadian across the country. Assuming that this is a proper Binomial experiment, what is the probability that at least 3 Canadian are on CERB in the sample? [3 Marks]. 5. It is reported that 20 % of Canadian are on CERB ( Canada Emergency Response Benefit ) . In order to further investigate this topic , a researcher selects a random sample of 10 Canadian across the country . Assuming that this is a proper Binomial experiment , what is the probability that at least 3 Canadian are on CERB in the sample ? [ 3 Marks ] .​

Answer 1

Using the binomial distribution, it is found that there is a 0.3222 = 32.22% probability that at least 3 Canadian are on CERB in the sample.

What is the binomial distribution formula?

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

The values of the parameters are:

n = 10, p = 0.2.

The probability that at least 3 Canadian are on CERB in the sample is:

$P(X \geq 3) = 1 - P(X < 3)$

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2).

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.2)^{0}.(0.8)^{10} = 0.1074$

$P(X = 1) = C_{10,1}.(0.2)^{1}.(0.8)^{9} = 0.2684$

$P(X = 2) = C_{10,2}.(0.2)^{2}.(0.8)^{8} = 0.3020$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1074 + 0.2684 + 0.3020 = 0.6778.

$P(X \geq 3) = 1 - P(X < 3) = 1 - 0.6778 = 0.3222$

0.3222 = 32.22% probability that at least 3 Canadian are on CERB in the sample.

More can be learned about the binomial distribution at brainly.com/question/24863377

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