A Food Marketing Institute found that 31% of households spend more than $125 a week on groceries. Assume the population proportion is 0.31 and a simple random sample of 81 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.35 and 0.5? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = (Enter your answer as a number accurate to 4 decimal places.)
Solution
Given that,
p = 0.31
1 - p = 1 - 0.31 = 0.69
n = 81
= p = 0.30
[p ( 1 - p ) / n] = [(0.31 * 0.69) / 81 ] = 0.0514
P( 0.35 < < 0.5 )
= P[(0.35 - 0.31) / 0.0514 < ( - ) / < (0.5 - 0.31) / ]0.0514
= P( 0.7782 < z < 3.70)
= P(z < 3.6965) - P(z < 0.7782 )
Using z table,
= 0.9999 - 0.7818
= 0.2181
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