Marcello the monkey is doing some calculations with his banana hoard. He has 200
ID: 3335777 • Letter: M
Question
Marcello the monkey is doing some calculations with his banana hoard. He has 200 bananas, which have a total weight of 25 kilograms. Assume the weight of a banana can be represented as a random variable with mean and standard deviation . Marcello remembers from his freshman course in bananology that is equal to 10 grams.
a. Construct a 95% confidence interval for , the average weight of an individual banana.
b. Being a precision oriented monkey, Marcello would like to know to within a margin of 0.1 grams with 99% certainty. How many bananas would he need in order to accomplish this?
c. Marcello’s banana pudding calls for three bananas and an amount of custard whose weight is exactly equal to that of the bananas. What is the standard deviation of the weight of a recipe of banana pudding?
Explanation / Answer
a. Construct a 95% confidence interval for , the average weight of an individual banana.
Solution:
We are given,
Population standard deviation = = 10 gm
Sample size = n = Total number of bananas = 200, total weight = 25 kilograms
Xbar = Average weight of banana = 25/200 = 0.125 kg = 125 gm
Confidence level = 95%
Critical z value = 1.96 (by using z-table)
Confidence interval = Xbar -/+ Z*/sqrt(n)
Confidence interval = 125 -/+ 1.96*10/sqrt(200)
Confidence interval = 125 -/+ 1.3859
Lower limit = 125 - 1.3859 = 123.61
Upper limit = 125 + 1.3859 = 126.39
Confidence interval = (123.61 gm, 126.39 gm)
b. Being a precision oriented monkey, Marcello would like to know to within a margin of 0.1 grams with 99% certainty. How many bananas would he need in order to accomplish this?
Solution:
We are given,
E = 0.1, C = 0.99, = 10
Z = 2.5758 (by using z-table)
Sample size formula is given as below:
n = (Z*/E)^2
n = (2.5758*10/0.1)^2 = 66348.97
Required sample size = 66349
c. Marcello’s banana pudding calls for three bananas and an amount of custard whose weight is exactly equal to that of the bananas. What is the standard deviation of the weight of a recipe of banana pudding?
Solution:
We know that the estimate for the standard deviation is given as the standard error of population mean.
Standard error of population mean is given as below:
SE = /sqrt(n)
We have = 10
We consider three bananas plus equal weight of custard, this means n = 3+3 = 6
Estimate for standard deviation = 10/sqrt(6) = 4.082483