Question

A recent article in the paper claims that government ethics are at an all-time low. Reporting on a recent sample,
the paper claims that 43% of all constituents believe their representative possesses low ethical standards.
Suppose 20 of a representative's constituents are randomly and independently sampled. Assuming the paper's
claim is correct, find the probability that more than eight but fewer than 12 of the 20 constituents sampled believe
their representative possesses low ethical standards. Round to six decimal places.

A. 0.273335
B. 0.734089
C. 0.417890
D. 0.646919

Answer 1

Answer:

Probability that more than eight but fewer than 12 of the 20 constituents sampled believe  their representative possesses low ethical standards is 0.417890.

Step-by-step explanation:

We are given that the paper claims that 43% of all constituents believe their representative possesses low ethical standards.

Suppose 20 of a representative's constituents are randomly and independently sampled.

The above situation can be represented through binomial distribution;

$P(X=r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 20 constituents

            r = number of success = more than eight but fewer than 12

            p = probability of success which in our question is probability that

                  all constituents believe their representative possesses low

                  ethical standards, i.e; p = 43%

Let X = Number of constituents who believe their representative possesses low ethical standards

So, X ~ Binom(n = 20 , p = 0.43)

Now, Probability that more than eight but fewer than 12 of the 20 constituents sampled believe  their representative possesses low ethical standards is given by = P(8 < X < 12)

P(8 < X < 12)  =  P(X = 9) + P(X = 10) + P(X = 11)

= $\binom{20}{9} \times 0.43^{9} \times (1-0.43)^{20-9}+ \binom{20}{10} \times 0.43^{10} \times (1-0.43)^{20-10}+\binom{20}{11} \times 0.43^{11} \times (1-0.43)^{20-11}$

= $167960 \times 0.43^{9} \times 0.57^{11}+ 184756 \times 0.43^{10} \times 0.57^{10}+167960 \times 0.43^{11} \times 0.57^{9}$

= 0.417890

Hence, the required probability is 0.417890.

Answer 2

The probability that more than eight but fewer than 12 of the 20 constituents sampled believe their representative possesses low ethical standards is; C. 0.417890

How to calculate binomial probabilities?

We are given;

proportion; p = 43% = 0.43

sample size; n = 20

P(X = x) = ⁿCₓ * pˣ * (1 - p)ⁿ ⁻ ˣ

Thus the probability that more than eight but fewer than 12 of the 20 constituents sampled believe their representative possesses low ethical standards is;

P(8 < X < 12)  =  P(X = 9) + P(X = 10) + P(X = 11)

From online binomial probability calculator, we have;

P(8 < X < 12)  =  0.17419889 + 0.14455452 + 0.09913627

P(8 < X < 12)  = 0.417890

Read more about binomial probability at; https://brainly.com/question/15246027