Asanitation department is interested in estimating the mean amount of garbage per bin for all bins in the city. assume the amount of garbage per bin is normally distributed. in a random sample of 40 bins, the sample mean amount is 52.8 pounds and the population standard deviation is 3.9 pounds. construct 95% and 99% confidence intervals for the mean amount of garbage per bin for all bins in the city. don't use 68/95/99.7 rule; instead use r to compute the quantities such as quantiles.

a) 51.592 b) 54.008 c) 51.213 d) 54.387

Step-by-step explanation:

a) lower limit of 95% interval

Lower limit = 52.8 - (1.96) (3.9/sqrt 40) = 51.592

b) upper limit of 95% interval

upper limit = 52.8 + (1.96) (3.9/sqrt 40) = 54.008

c) lower limit of 99% interval

Lower limit = 52.8 - (2.576) ( 3.9 / sqrt 40)= 51.213

d) upper limit of 99% interval

Upper limit = 52.8 + (2.576) ( 3.9 / sqrt 40) = 54.387

40 minutes

step-by-step explanation:

add the time and multiply by 2

for a it would be 24 children.

for b i don't understand the question. is it saying for each individual person they pick 5 reds to 7 greens or is it the whole group. clarify and i hope this can you.

step-by-step explanation: