The mass of garbage put out for weekly curbside pickup, per household, in a cert
ID: 3244764 • Letter: T
Question
The mass of garbage put out for weekly curbside pickup, per household, in a certain community, is reported to have a mean of 6.1 kg and a standard deviation of 2.2 kg. Based on this information: a) Assume that the population is very large. For a sample of 100 households from tiles community, calculate, to the nearest %, the probability that the sample mean will be between 5.9 and 6.3 kg. b) Redo Part (a), for a sample size of 1,000 households instead of 100. c) Redo Part (a) (ie, n = 100), but this time assume a population of 400 households. d) Assuming again that the population is very large, let us now suppose that 5 consecutive samples of 100 households are taken, and each one of them has a sample mean greater than 6.6 kg. Calculate the probability of obtaining this result. e) Based on your answer for Part (d), is there reason to believe that the mean weekly garbage output per household is not actually around 6.1 kg? If so, in which direction has the mean garbage output shifted? Explain your answers using your previous calculations for this question, plus your knowledge of statistics fundamentals.Explanation / Answer
a)
P( 5.9 < X < 6.3) = P(X<6.3) - P(X<5.9)
= P(Z < 6.3 - 6.1/2.2/sqrt(100)) - P(Z < 5.9-6.1/2.2/sqrt(100))
= P(Z < 0.9091) - P(Z < -0.9091)
= 0.8184 - 0.1816
= 0.6368 or 63.68% or 64%
b)
P( 5.9 < X < 6.3) = P(X<6.3) - P(X<5.9)
= P(Z < 6.3 - 6.1/2.2/sqrt(1000)) - P(Z < 5.9-6.1/2.2/sqrt(1000))
= P(Z < 2.8748) - P(Z < -2.8748)
= 0.998 - 0.002
= 0.996 or 99.6%
c)
similarly do for n=400
d)
P(X> 6.6) = P(Z > 6.6-6.1/2.2/sqrt(100))
= P(Z > 2.2727)
= 0.0115
= 1.15%
e)
no since only 1.15% household have mean garbage greater than 6.6
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