Question

the true proportion of fish freaky fish on a farm is known to be 30%, what is the chance of randomly catching a bucket of 20 fish on this farm and finding the proportion of freaky fish in the bucket is between 25% and 35%?(ie. the sample proportion is within 5% of the true proportion,P)

Answer 1

Population Proportion = P = 0.30

So,

Q = 1 - P = 0.70

Sample Size = n = 20

SE =

To find P(0.25 < <0.35):

Case 1: for from 0.25 to mid value:

Z = (0.25 - 0.30)/0.1025 = - 0.4878

Table of Area Under Standard Normal Curve gives area = 0.1879

Case 2: for from mid value:to 0.35:

Z = (0.35 - 0.30)/0.1025 = 0.4878

Table of Area Under Standard Normal Curve gives area = 0.1879

So,

P(0.25 < < 0.35) = 0.1879 X 2 = 0.3758

So,

Answer is:

0.3758