(can someone please help me answer this question i will make you brainliast)
1 answer:
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Answer:
The answer is 71.61%.
Step-by-step explanation:
Households discarding paper week is given to be normally distributed with a mean, = 9.4 lb.
The normally distributed distribution has a standard deviation given as = 4.2 lb.
The proportion of households that throw out at least 7 lb of paper a week is to be found.
Therefore we have to find p(X ≥ 7) which can be written as
p(X ≥ 7) = 1 - p(X < 7)
= 1 - p( Z < )
= 1 - p( Z < )
= 1 - p( Z < -0.571 )
= 1 - 0.2839
= 0.7161
Therefore the percentage of households that throw out at least 7 lb paper a week is 71.61%