You operate a restaurant. You read that a sample survey by the National Restaura

Question

You operate a restaurant. You read that a sample survey by the National Restaurant Association shows that 40% of adults are commited to eating nutritious food when eating away from home. To help plan your menu, you decide to conduct a sample survey in your own area. You will use random digit dialing to contact an SRS of 200 households by phone. You find 100 of your 200 respondents concerned about nutrition. What is the probability that X is 100 or larger if p=0.4 is true?

options:

0.0026

0.0125

0.0542

0.1120

0.1674

Explanation / Answer

solution=

The numbers 200 and 0.4 are large enough such that you should use the normal approximation to the binomial distribution, which states that:
where X ~ B(n,p)
then X ~ N(np,sqrt npq)
so X ~ N(80,6.928)

P(X > 100)
= 1 - P(X <= 100)
= 1 - P(Z <= (100 - 80)/6.928)
= 1 - P(Z <=2.89 )
= 1 - 0.998074
= 0.002
It's entirely possible for 100 respondents to be concerned about nutrition and still have the probability as 0.4

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