I need help ASAP please help

Question

The town of KnowWearSpatial, U.S.A. operates a rubbish waste disposal facility that is overloaded if its 4937 households discard waste with weights having a mean that exceeds 27.07 lb/wk. For many different weeks, it is found that the samples of 4937 households have weights that are normally distributed with a mean of 26.83 lb and a standard deviation of 12.78 lb.

What is the proportion of weeks in which the waste disposal facility is overloaded?

P(M > 27.07) = 

Enter your answer as a number accurate to 4 decimal places. NOTE: Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Is this an acceptable level, or should action be taken to correct a problem of an overloaded system?

1. No, this is not an acceptable level because it is not unusual for the system to be overloaded.
2. Yes, this is an acceptable level because it is unusual for the system to be overloaded.

Answer 1

As always, we establish the conditions.

First, because the data on the ENTIRE population is known, not just a sample of the population, we assume we have correct population parameters.

Second, due to the Gausian nature of the changes in individual trash deposits, we can assume a normal distribution of the load of trash being delivered on a weekly basis. (note: the CLT, which deals with the measure of the mean is not sufficient to support this assumption)

Third, the amount of waste delivered by any household in any week is independent of the other households. This is a necessary condition.

Fourth, The proportion of weeks that the trash is considered to be in an overload condition assumes that the probability of having an overload condition is constant and that the proportion of weeks in an overload condition (o) to the total number of weeks (w) being examined is o/w and equals p , which is the probability that M exceeds 27.070 lbs. (i.e. P(M>12.070) M-> N[26.830,12.780] )

Problem parameters given. µ= 26.830 σ= 12.780 r.v. is M (mean mass delivered in week to trash by each household)

Calculate P(M>27.070)

Since the problem specified that a standardized normal distribution (Z distribution) was to be used the formula for cumulative distribution of a standard normal distribution Φ(z) will be used.

Calculate Z z = (27.070-26.830)/12.780 or 0.01877934

Calculate Φ(0.01877934) = 0.5744807 (note: this is left tail)

Calculate P(z>0.01877934) = 1 - 0.5744807 or 0.4255193

Rounding to 4 decimal places = .4255

Alternatively this could have been done based on the average weekly load at the waste disposal site (132,460 lb mean and standard deviation of 898.0 lbs, overload = 133,644.6 lb) , but the results would still be the same.

The question of whether the situation requires action or if the system is overloaded too often is not a scientific question but a political one and which would require the consensus of the people being serviced as defined by those people placed in positions to represent them. The water and waste board might just as well say that since the system , on the average, is not overloaded, that things are ok. From an engineering standpoint it should be perfectly possible to construct a system where one could expect at least 2 sigma confidence (97.7% one sided confidence) So I would personally say:

No, this is not an acceptable level because it [should be] unusual for the system to be overloaded.