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The town of KnowWearSpatial, U.S.A. Operates a rubbish waste disposal facility that is overloaded if its 4749 households discard waste with weights having a mean that exceeds 27.21 lb/wk. For many different weeks, it is found that the samples of 4749 households have weights that are normally distributed with a mean of 26.86 lb and a standard deviation of 12.82 lb. What is the proportion of weeks in which the waste disposal facility is overloaded? P(M > 27.21) =

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Answer:

0.48911

Step-by-step explanation:

The formula for calculating a z-score is is z = (x-μ)/σ,

where x is the raw score, = 27.21 Ib

μ is the population mean = 26.86 lb

σ is the population standard deviation. = 12.82 lb

z = 0.027301

Probability value from Z-Table:

P(x<27.21) = 0.51089

P(x>27.21) = 1 - P(x<27.21) = 0.48911

The proportion of weeks in which the waste disposal facility is overloaded is

0.48911

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