A.) A Food Marketing Institute found that 31% of households spend more than $125

Question

A.) 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 93 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.23 and 0.5?

B.) Based on historical data, your manager believes that 28% of the company's orders come from first-time customers. A random sample of 96 orders will be used to estimate the proportion of first-time customers. What is the probability that the sample proportion is less than 0.32?

Explanation / Answer

Solution:-

A) Given that p = 0.31, n = 93 , q = 0.69

P(0.23 < p^ < 0.5)

= P((0.23-0.31)/sqrt(0.31*0.69/93) < (p^ - p )/sqrt(pq/n) < (0.5-0.31)/sqrt(0.31*0.69/93)

= P(?1.6681 < Z < 3.9618)

= 0.9525

B) p = 0.26, n = 96, q = 0.74

P(p^ > 0.32) = P(Z > (0.32-0.26)/sqrt(0.26*0.74/96))

= P(Z > 1.3402)

= 0.0901

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