Respuesta :
Answer:
The answer is 71.61%.
Step-by-step explanation:
Households discarding paper week is given to be normally distributed with a mean, [tex]\mu[/tex] = 9.4 lb.
The normally distributed distribution has a standard deviation given as [tex]\sigma[/tex] = 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 < [tex]\frac{x - \mu}{\sigma}[/tex] )
= 1 - p( Z < [tex]\frac{(7 - 9.4)}{4.2}[/tex] )
= 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%
