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borishaifa [10]
4 years ago
8

Least to greatest 7/12,0.75,5/6

Mathematics
2 answers:
irina [24]4 years ago
8 0
Hello there.

<span>Least to greatest 7/12,0.75,5/6

7/12 ,5/6, 0.75</span>
blagie [28]4 years ago
8 0
0.75, 5/6, 7, 12.

that should be the correct answer.
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A recent article in the paper claims that government ethics are at an all-time low. Reporting on a recent sample,
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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 = <u><em>Number of constituents who believe their representative possesses low ethical standards</em></u>

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}

= <u>0.417890</u>

<u></u>

Hence, the required probability is 0.417890.

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