Question

the random sample of 400 people is selected and the populatin is .29

a. describe the sample distribution of the sample proportion

b. find the probability of obtaining the sample proportion is greater than .33

c. find the sample portion that has the top 7%

Answer 1

a.

the mean proportion is p=0.29

standard deviation of proportion = [ p*(1-p)/n ]^0.5

for this distribution = [ 0.29*(1-0.29)/400 ]^0.5 = 0.0227

b.

z = (p-mean)/SD

z for 0.33

z = (0.33-0.29)/(0.0227) = 1.76

P(p>0.33) = P(z>Z)

P(z>Z) = 1 - P(z<Z)

P(z<Z) = 0.9608 {from table given below}

P(p>0.33) = 1 - 0.9608

P(p>0.33) = 0.0392

c.

top 7%

P(z>Z) = 0.07

P(z<Z) = 1 - P(z>Z) = 1 - 0.07 = 0.93

Z = 1.48 {from the table}

p = mean + Z*SD

= 0.29 + 1.48*0.0227

= 0.3236

the top 7% is the sample portion with proportion p > 0.3236

(please UPVOTE)

Table for P(z<Z) :