A small restaurant seats 20 diners, and is full every night. The chef knows from previous experience that 40% of the diners order steaks, so she always has 12 in her fridge at the beginning of the evening.
a) On what proportion of nights will the chef run out of steaks?
b) If a customer orders steak, what is the probability that he/she will receive one?
Answer:
a) 2.11%
b) 0.9978
I don't get the working to the questions. PLEASE HELP! THANKS :)
a) On what proportion of nights will the chef run out of steaks?
b) If a customer orders steak, what is the probability that he/she will receive one?
Answer:
a) 2.11%
b) 0.9978
I don't get the working to the questions. PLEASE HELP! THANKS :)
-
binomdist with n = 20, p = 0.4, q = 0.6
P[x] = nCx * p^x *q^(n-x), eg P[12] = 20c12 *.4^12*.6^28 = .0355
a) P[run out of steaks] = P[≥13] = 0.0210 or 2.10% <------
b) P[customer gets steak] = P[≤12] = 1 - 0.0210 = 0.9790 <------
for some mysterious reason, the answers are a bit different.
ok, let me run a check on wolfram alpha
a) sum C(20.k) *.4^k *.6^(20-k), k = 13 to 20 = .0210289
http://www.wolframalpha.com/input/?i=sum…
b)
P[x] = nCx * p^x *q^(n-x), eg P[12] = 20c12 *.4^12*.6^28 = .0355
a) P[run out of steaks] = P[≥13] = 0.0210 or 2.10% <------
b) P[customer gets steak] = P[≤12] = 1 - 0.0210 = 0.9790 <------
for some mysterious reason, the answers are a bit different.
ok, let me run a check on wolfram alpha
a) sum C(20.k) *.4^k *.6^(20-k), k = 13 to 20 = .0210289
http://www.wolframalpha.com/input/?i=sum…
b)