Question 5 (of 6) 16.66 points A random sample of size n-225 is taken from a pop

Question

Question 5 (of 6) 16.66 points A random sample of size n-225 is taken from a population with population proportion P = 0.55. Use Table 1. a. Calculate the expected value and the standard error for the sampling distribution of the sample proportion. (Round "expected value" to 2 decimal places and "standard error" to 4 decimal places.) Expected value Standard error b. What is the probability that the sample proportion is between 0.50 and 0.60? (Round "2" value to 2 decimal places, and final answer to 4 decimal places.) c. What is the probability that the sample proportion is less than 0.50? (Round "z" value to 2 decimal places, and final answer to 4 decimal places.) References eBook & Resources

Explanation / Answer

a) Expected value of the sampling distribution of p(hat):
E(phat) = p = 0.55

Variance for the sampling distribution of P:
p(1-p) / n

0.55(1-0.55)/225= 0.55*0.45/225= 0.0011
Standard Error(SE) of the Sample Proportion:
(p(1-p) / n) by putting the values we get ,

Standard error= 0.03

b) mean= np= 225*0.55= 123.75

Variance= npq= 225*0.55*0.45= 55.69

Standard deviation= sqrt(55.69)= 7.46

P(0.50<X<0.60)= P (225*0.50<225*0.6) = P(112.5<x135)

Since =123.75 and =7.46 we have:

P ( 112.5<X<135 )=P ( 112.5123.75< X<135123.75 )=P ( 112.5123.757.46<X<135123.757.46)

Since Z=x , 112.5123.757.46=1.51 and135123.757.46=1.51 we have:

P ( 112.5<X<135 )=P ( 1.51<Z<1.51 )

By using the standard normal table to conclude that:

P ( 1.51<Z<1.51 )=0.87

c) P(X<0.50)= P(X<225*0.50)= P(X<112.5)

Since =123.75 and =7.46 we have:

P ( X<112.5 )=P ( X<112.5123.75 )=P (X<112.5123.757.46)

Since x=Z and 112.5123.757.46=1.51 we have:

P (X<112.5)=P (Z<1.51)

By using the standard normal table to conclude that:

P (Z<1.51)=0.0655

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