Question

Q4. A simple random sample of size n=180 is obtained from a population whose size=20,000 and whose population proportion with a specified characteristic is p=0.45. Determine whether the sampling distribution has an approximately normal distribution. Show your work that supports your conclusions.

Q5. Using the values in Q4, calculate the probability of obtaining x=72 or more individuals with a specified characteristic.

Answer 1

To determine whether the sampling distribution has an approximately normal distribution, it should satisfied the condition

nP and nQ >= 5

np = 180 * 0.45 = 81

nQ = 180 * ( 1 - 0.45 ) = 99

Since both satisfied the condition, we can say that sampling distribution has an approximately normal distribution.

Question 5

Mean = n * P = ( 180 * 0.45 ) = 81
Variance = n * P * Q = ( 180 * 0.45 * 0.55 ) = 44.55
Standard deviation = = 6.6746

P ( X >= 72 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 72 - 0.5 ) =P ( X > 71.5 )

P ( X > 71.5 ) = 1 - P ( X < 71.5 )
Standardizing the value

Z = ( 71.5 - 81 ) / 6.6746
Z = -1.42

P ( Z > -1.42 )
P ( X > 71.5 ) = 1 - P ( Z < -1.42 )
P ( X > 71.5 ) = 1 - 0.0778
P ( X > 71.5 ) = 0.9222