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anonymous

  • 5 years ago

the center for disease control reports that 24.8 percent of californians are obese. a sample of 112 california residents is taken. what is the probability that over 30 percent of the sample is obese?

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  1. anonymous
    • 5 years ago
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    so basically i know the mean is 27.78 and the std deviation is 4.57...now what

  2. dumbcow
    • 5 years ago
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    30% of 112 is 33.6 now you want to find out how many std dev's away this is from the mean z = (33.6-27.78)/4.57 = 1.27 now you have to look in a stats z table for probabilities of the standard normal distribution 1 - P(Z<= 1.27)

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