Respuesta :
Answer:
The probability that the disposal facility is overloaded in a randomly selected week is P=0.017.
Step-by-step explanation:
We know that a mean that exceeds 26.84 lb will overload the rubbish waste disposal facility.
The distribution of the households discard weights is normally distributed, with mean 26.46 lb and a standard deviation of 12.35 lb.
We have to calculate the probability of the sample mean of 4611 household discards. We have to calculate the sampling distribution, that has the following parameters:
[tex]\mu_s=\mu=26.46\\\\\sigma_s=\sigma/\sqrt{n}=12.35/\sqrt{4611}=12.35/67.90=0.18[/tex]
Then, we can calculate the probability that the house discards mean exceeds 26.84:
[tex]z=(X-\mu_s)/\sigma_s=(26.84-26.46)/0.18=0.38/0.18=2.11\\\\\\ P(X>26.84)=P(z>2.11)=0.01743[/tex]
The probability that the disposal facility is overloaded in a randomly selected week is P=0.017.
