Question

The sample proportion p-bar provides an estimate for the population proportion p. The sampling distribution of the sample proportion is the probability distribution of the sample proportion.

Consider a population with a proportion p = 0.60, from which a sample of size n = 200 is drawn. What is the sampling distribution of the sample proportion p-bar?

Calculate the following probabilities using the distribution above:

P(? < 0.5)

P(? > 0.7)

P(0.5 < ? < 0.7)

Answer 1

Mean of sampling distribution of sample proportion = p = 0.6

Standard deviation of sampling distribution of sample proportion = sqrt [ p( 1 - p) / n ]

= sqrt [ 0.6 ( 1 - 0.6) / 200 ]

= 0.0364

Using normal approximation,

P( < p ) = P(Z < ( - p) / sqrt [ p( 1 - p) / n ]

a)

P( < 0.5) = P(Z < ( 0.5 - 0.6) / sqrt ( 0.6 * ( 1 - 0.6) / 200 ]

= P(Z < -2.89)

= 0.0019 (From Z table)

b)

P( > 0.7) = P(Z > ( 0.7 - 0.6) / sqrt ( 0.6 * ( 1 - 0.6) / 200 ]

= P(Z > 2.89)

= 1 - P(Z < 2.89)

= 1 - 0.9981 (From Z table)

= 0.0019

c)

P(0.5 < < 0.7) = P( < 0.7) - P( < 0.5)

= P(Z < ( 0.7 - 0.6) / sqrt ( 0.6 * ( 1 - 0.6) / 200 ] - P(Z < ( 0.5 - 0.6) / sqrt ( 0.6 * ( 1 - 0.6) / 200 ]

= P(Z < 2.89) - P(Z < -2.89)

= 0.9981 - 0.0019

= 0.9962