Question

The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.17 and a sample of 600 households will be selected from the population.

(1)What is the probability that the sample proportion will be within ±0.02 of the population proportion? (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

p = 0.17

1 - p = 0.83

n = 600

= p = 0.17  

= (p*(1-p))/n =  (0.17*0.83)/600 = 0.01534

P(0.15 < <0.19 ) = P((0.15-0.17)/0.01534 ) < ( - ) /  < (0.19-0.17) /0.01534 ) )

= P(-1.30 < z < 1.30 )

= P(z < 1.30) - P(z < -1.30)

= 0.9032 - 0.0968

Probability = 0.8064