Respuesta :
The value is P(p'-p > 0.17) = 0.09
From the question we are told that
The population proportion is p=0.19
The sample size is n = 479
Generally given that the sample size is large enough , i.e n > 30 then the mean of this sampling distribution is mathematically represent
[tex]u_{x} =p=0.19[/tex]
Generally the standard deviation is mathematically represented as
σ[tex]=\sqrt{ \frac{p(1-p)}{n}[/tex]
σ= [tex]\sqrt{ \frac{0.19(1-0.19)}{479}[/tex]
σ=0.017
Generally the the probability that the sample proportion will differ from the population proportion by greater than 17% is mathematically represented as
[tex]P(p'-p > 0.17) =P (z > \frac{0.17}{0.017} )[/tex]
[tex]P(p'-p > 0.17) =P (z > 10 )[/tex]
[tex]P(p'-p > 0.17) =P (z > 10 ) - P (z > -10 )[/tex]
From the z table the area under the normal curve to the left corresponding to 10 and - 10 is
[tex]P(p'-p > 0.17) = 0.97042 - 0.87450\\P(p'-p > 0.17) = 0.09[/tex]
So, the value of probability is greater than 17% is P(p'-p > 0.17) = 0.09
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The probability that the sample proportion will be greater than 17% is 0.8677 or 86.77%.
The true proportion of the sample or the mean of the sample = 19%, that is, μ = 19% or 0.19.
The sample size (n) = 479.
The sample proportion is to be calculated at the point 17% or 0.17.
The standard error (s) = √{μ(1 - μ)/n} = √{0.19(1 - 0.19)/479} = 0.0179.
We are asked to calculate the probability that the sample proportion is greater that 17% or 0.17.
This is written as P(X > 0.17) = P(Z > {(0.17-0.19)/0.0179}) = P(Z > -1.1173) = 1 - P(Z ≤ -1.1173) = 1 - 0.1304 = 0.8696 or 86.96%.
To calculate this using the calculator, we use the calculator function:
Normalcdf(0.17,10000000,0.19,0.0179) = 0.8677 or 86.77%.
Thus, the probability that the sample proportion will be greater than 17% is 0.8677 or 86.77%.
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