Question

A Food Marketing Institute found that 28% of households spend more than $125 a week on groceries. Assume the population proportion is 0.28 and a simple random sample of 123 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.34 and 0.45? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

Solution

Given that,

p = 0.28

1 - p = 1 - 0.28 = 0.72

n = 123

= p = 0.28

=  [p ( 1 - p ) / n] = [(0.28 * 0.72) / 123 ] = 0.0405

P( 0.34 < < 0.45)

= P[(0.34 - 0.28) / 0.0405< ( - ) / < (0.45 - 0.28) / 0.0405]

= P( 1.48< z < 4.20)

= P(z < 4.20) - P(z < 1.48)

Using z table,   

= 1 - 0.9306

= 0.0694