Careful observation of the coffee machine in the SAC building has revealed the f
ID: 3359930 • Letter: C
Question
Careful observation of the coffee machine in the SAC building has revealed the following. The amount of coffee dispensed for a “large” coffee is normally distributed with a mean of 10.0 oz. and the standard deviation is 0.4 oz. The cup holds 11.0 oz.
a)What is the probability of an overflow when a large coffee is ordered – i.e., more than 11.0 oz. is dispensed?
b)Give the probability mass function for the random variable X, the number of customers who purchase a large coffee from this machine until one experiences a cup that overflows. Assume that the amount of coffee dispensed for one cup is independent of the amount dispensed for any previous cup. In your pmf, assign appropriate values to any parameters.
c)Give the probability mass function for the random variable Y, the number who experience an overflowed cup, out of the next ten customers who purchase a large coffee from this machine. Again assume that the amount dispensed for any one cup is independent of the amount dispensed for any prior cup. Also, be sure to give specific values for any parameters.
d)It has also been observed that one in five customers who put a dollar bill in the coffee machine are ripped off by being given too little in change. What is the probability that a customer who puts in a dollar to get a large coffee will have the “wonderful” experience of both being ripped off and having the cup overflow? State any assumptions you need to figure this out.
Explanation / Answer
(a) Given mean = 10 oz, st.dev = 0.4 oz , capacity = 11 oz
We will have to find the area under the standard normal curve which is greater than 11 oz to find the probability that the cup will overflow, so first fining the z value we use the formula
z=(x-mean)/st. dev. = (11-10)/0.4 = 2.5
Therefore referring to a standard normal dist. table we find the probabilty of z>2.5 is
1-z(2.5) = 1-0.9937 = 0.00621
Hence there is a 0.00621 or 0.62% chance of the cup overflowing
(b) Now say n different people order the large cup. Two outcomes are possible on each trial
Either the cup overflows or it doesnt. proabilty of a success denoted by p and failure =1-p
The amount dispensed is independent of the previous dispensation
This therfore follows a binomial distribution where the pmf will be
f(x) = (nCx)*p^x*(1-p)^(n-x)
where f(x) is the probability of x successes in n trials and p is the proability of success i.e. cup not overflowing
So in 5 trials if we want to find that that the fifth person will get an overflowed cup, then
f(x)=(5C4)*(0.9937)^4*(0.00621)^(5-4)
(c) The probability that a partiular number will experience overflowed cup out of 10 trials is given by the pmf
f(x) = (u^x) * (e^-u) /( Factorial(x)) where f(x) = probability of x occurences in an interval, u = mean number of occurences in 10 trials and e = 2.718 so for example if on an average 3 people get overflowed cup in 10 people, then the probability that 4 people will get overflowed cups in the next 10 customers will be
f(x) = (3^4) * (e^-3) / (Factorial(4))
(d) probability of getting ripped off = 1/5 =0.2 and probability of cup overflowing = 0.00621 hence probability of getting both ripped and having the cup overflow = 0.2*0.00621 = 0.0012