Question

The Gallup Poll once asked a random sample of 1600 adults “Do you happen to do step aerobics?” Suppose that in fact 26% of all Americans do step aerobics.

a) What is the mean of the sampling distribution of proportions who do step aerobics?

b) What is the standard deviation the sampling distribution of proportions who do step aerobics?

c) Justify the distribution: HINT: Check that np and n(1-p) are greater than or equal to 10.

d) What is the probability that proportion of adults that do step aerobics is between 24% and 30%?

e) What is the probability of the proportion of adults that do step aerobics is more than 28%?

Answer 1

a)

Mean = p = 0.26

b)

Standard deviation = sqrt [ p ( 1 - p) / n ]

= sqrt [ 0.26 ( 1 - 0.26) / 1600]

= 0.0110

c)

np = 1600 * 0.26 = 416 >= 10

n(1-p) = 1600 (1 - 0.26) = 1184 >= 10

Since np >= 10 and n( 1 - p) >= 10 ,

The sampling distribution of sample proportion is approximately normal.

d)

Using normal approximation,

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

= P(Z < ( - p) / SD)

So,

(0.24 < < 0.30) = P( < 0.30) - P( < 0.24)

= P(Z < ( 0.30 - 0.26) / 0.0110 ) - P(Z < ( 0.24 - 0.26) / 0.0110 )

= P(Z < 3.64) - P(Z < -1.82)

= 0.9999 - 0.0344 (From Z table)

= 0.9655

e)

P( > 0.28) = P(Z > ( 0.28 - 0.26) / 0.0110 )

= P(Z > 1.82)

= 1 - P(Z < 1.82)

= 1 - 0.9656 (From Z table)

= 0.0344