Question

A Food Marketing Institute found that 29% of households spend more than $125 a week on groceries. Assume the population proportion is 0.29 and a simple random sample of 132 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.25 and 0.4?

Answer =  (Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Solution

Given that,

p = 0.29

1 - p = 1 - 0.29 = 0.71

n = 132

= p = 0.29

  [p ( 1 - p ) / n] = [(0.29 * 0.71) / 132 ] = 0.0395

P( 0.25 < < 0.4 )

= P[(0.25 - 0.29) / 0.0395 < ( - ) / < (0.4 - 0.29) / 0.0395 ]

= P(-1.01 < z < 2.78)

= P(z < 2.78) - P(z < -1.01)

Using z table,   

= 0.9973 - 0.1562

= 0.8411