Question

Check 0/10 ptsRetries 2 A Food Marketing Institute found that 32% of households spend more than $125 a week on groceries. Assume the population proportion is 0.32 and a simple random sample of 175 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.37 and 0.5?

Answer 1

Solution

Given that,

p = 0.32

1 - p = 1 - 0.32 = 0.68

n = 175

= p = 0.32

=  [p ( 1 - p ) / n] = [(0.32 * 0.68) / 175 ] = 0.0353

P(0.37 < < 0.5 )

= P[(0.37 - 0.32) /0.0353 < ( - ) / < (0.5 - 0.32) / 0.0353 ]

= P(1.42 < z < 5.10)

= P(z < 5.10) - P(z < 1.42)

Using z table,   

= 1 - 0.9222

= 0.0778