Marcello the monkey is doing some calculations with his banana hoard. He has 200
ID: 2929792 • 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
Number of Banana = 200
total weight of 200 bananas = 25 kg = 25000 g
weight of an individual banana x = 25000/200 = 125 g
Standard deviation of an individal banana = 10g
(a) 95% confidence interval for = x +- Z95% (/n)
= 125 +- 1.96 * (10/200)
= (123.61 g , 126.39g)
(b) Here margin of error = 0.1 grams
Certainity Level = 99%
Critical test statitistic for 99% = 2.575
Let say there are n bananas
MArgin of error = Test Statistic * Standard error of the mean
0.1 = 2.575 * (10/n)
n = 66306 bananas
(c) Pudding requires three banans and an amount of custard to that of the bananas
So, here Weight of pudding weight of 3 bananas and same weight of 3 bananas.
so average weight of the pudding = 6 * 150 = 900 gm
Varaince of Pudding = Variance (sum of Three banans) + Variance (Custard) = 2 * Variance (sum of Three banans)
= 2 * [3 * 2] = 2 * 3 * 10 * 10 = 600
Standard deviation of weight of pudding = sqrt(600) = 24.50 gm