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Korvikt [17]
2 years ago
14

Which of the following is NOT true of the confidence level of a confidence​ interval? Choose the correct answer below. A. The co

nfidence level is also called the degree of confidence. B. There is a 1-a ​chance, where a is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample. C. The confidence level gives us the success rate of the procedure used to construct the confidence interval. D. The confidence level is often expressed as the probability or area 1​-a, where a is the complement of the confidence level.
Mathematics
1 answer:
Grace [21]2 years ago
8 0

Answer:

<em>There is a 1-a ​chance, where a is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample.</em>  ( B )

Step-by-step explanation:

Confidence level depicts the probability that the confidence interval actually contains the values of p ( true values of P ) hence

<em>There is a 1-a ​chance, where a is the complement of the confidence​ level, that the true value of p will fall in the confidence interval produced from our sample</em> Is a complete misinterpretation of the confidence interval therefore it is NOT true

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a) The expected value is 2.680642

b) The minimun number of newspapers the manager should order is 6.

Step-by-step explanation:

a) Lets call X the demanded amount of newspapers demanded, and Y the amount of newspapers sold. Note that 4 newspapers are sold when at least four newspaper are demanded, but it can be <em>more</em> than that.

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we obtain:

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PY(1) = ε^(-3)*3^1/1! = 3*ε^(-3) = 0.14936

PY(2) = ε^(-3)*3^2/2! = 4.5*ε^(-3) = 0.22404

PY(3) = ε^(-3)*3^3/3! = 4.5*ε^(-3) = 0.22404

PY(4) = 1- (ε^(-3)*(1+3+4.5+4.5)) = 0.352768

E(Y) = 0*PY(0)+1*PY(1)+2*PY(2)+3*PY(3)+4*PY(4) =  0.14936 + 2*0.22404 + 3*0.22404+4*0.352768 = 2.680642

The store is <em>expected</em> to sell 2.680642 newspapers

b) The minimun number can be obtained by applying the cummulative distribution function of X until it reaches a value higher than 0.95. If we order that many newspapers, the probability to have a number of requests not higher than that value is more 0.95, therefore the probability to have more than that amount will be less than 0.05

we know that FX(3) = PX(0)+PX(1)+PX(2)+PX(3) = 0.04978+0.14936+0.22404+0.22404 = 0.647231

FX(4) = FX(3) + PX(4) = 0.647231+ε^(-3)*3^4/4! = 0.815262

FX(5) = 0.815262+ε^(-3)*3^5/5! = 0.91608

FX(6) = 0.91608+ε^(-3)*3^6/6! = 0.966489

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I hope this helped you!

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