A recent study reported that 51% of the children in a particular community were overweight or obese. Suppose a random sample of 400 public school children is taken from this community. Assume the sample was taken in such a way that the conditions for using the Central Limit Theorem are met. We are interested in finding the probability that the proportion of overweight/obese children in the sample will be greater than 0.47. Complete parts (a) and (b) below. a. Before doing any calculations, determine whether this probability is greater than 50% less than 50%. Why? O A. The answer should be greater than 50%, because 0.47 less than the population proportion of 0.51 and because the sampling distribution is approximately Normal. O B. The answer should be less than 50%, because 0.47 is less than the population proportion of 0.51 and because the sampling distribution is approximately Normal. Oc. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal. O D. The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal. b. Calculate the probability that 47% or more of the sample are overweight or obese. P(p20.47) = (Round to three decimal places as needed.)

Holt Mcdougal Larson Pre-algebra: Student Edition 2012
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Chapter11: Data Analysis And Probability
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A recent study reported that 51% of the children in a particular community were overweight or obese. Suppose a random sample of 400 public school children is taken from this community. Assume the sample was taken in such a way that the
conditions for using the Central Limit Theorem are met. We are interested in finding the probability that the proportion of overweight/obese children in the sample will be greater than 0.47. Complete parts (a) and (b) below.
a. Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why?
O A. The answer should be greater than 50%, because 0.47 is less than the population proportion of 0.51 and because the sampling distribution is approximately Normal.
O B. The answer should be less than 50%, because 0.47 is less than the population proportion of 0.51 and because the sampling distribution is approximately Normal.
O C. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal.
O D. The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal.
b. Calculate the probability that 47% or more of the sample are overweight or obese.
P(620.47) =|
(Round to three decimal places as needed.)
Transcribed Image Text:A recent study reported that 51% of the children in a particular community were overweight or obese. Suppose a random sample of 400 public school children is taken from this community. Assume the sample was taken in such a way that the conditions for using the Central Limit Theorem are met. We are interested in finding the probability that the proportion of overweight/obese children in the sample will be greater than 0.47. Complete parts (a) and (b) below. a. Before doing any calculations, determine whether this probability is greater than 50% or less than 50%. Why? O A. The answer should be greater than 50%, because 0.47 is less than the population proportion of 0.51 and because the sampling distribution is approximately Normal. O B. The answer should be less than 50%, because 0.47 is less than the population proportion of 0.51 and because the sampling distribution is approximately Normal. O C. The answer should be greater than 50%, because the resulting z-score will be positive and the sampling distribution is approximately Normal. O D. The answer should be less than 50%, because the resulting z-score will be negative and the sampling distribution is approximately Normal. b. Calculate the probability that 47% or more of the sample are overweight or obese. P(620.47) =| (Round to three decimal places as needed.)
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