Question

A population proportion is 0.59. Suppose a random sample of 661 items is sampled randomly from this population.

a. What is the probability that the sample proportion is greater than 0.60?

b. What is the probability that the sample proportion is between 0.57 and 0.60?

c. What is the probability that the sample proportion is greater than 0.57?

d. What is the probability that the sample proportion is between 0.53 and 0.56?

e. What is the probability that the sample proportion is less than 0.49?

Answer 1

Using normal approximation

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

a)

P( > 0.60) = P(Z > ( 0.6 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ]

= P(Z > 0.52)

= 0.3015

b)

P(0.57 < < 0.60) = P( < 0.60) - P( < 0.57)

= P(Z < ( 0.6 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ] - P(Z < ( 0.57 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ]

= P(Z < 0.52) - P(Z < -1.05)

= 0.6985 - 0.1469

= 0.5516

c)

P( > 0.57) = P(Z > ( 0.57 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ]

= P(Z > -1.05)

= P(Z < 1.05)

= 0.8531

d)

P(0.53 < < 0.56) = P( < 0.56) - P( < 0.53)

= P(Z < ( 0.56 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ] - P(Z < ( 0.53 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ]

= P(Z < -1.57) - P(Z < -3.14)

= 0.0582 - 0.0008

= 0.0574

e )

P( < 0.49) = P(Z > ( 0.49 - 0.59) / sqrt [ 0.59 ( 1 - 0.59 ) / 661 ]

= P(Z > 5.23)

= 0