Yes, that's correct. Of course, that reduces to simply .
Hello I need help with the following as I am not sure of what I am doing is right.
A pizza company advertises that it puts 0.5 lb of cheese on each of its pizzas. In fact, the amount of cheese on a pizza is normally distributed with mean
0.5 lb and standard deviation 0.025 lb.
i) calculate the probability the amount of cheese on a pizza is between 0.525 and 0.550 lb
I know how to do this using
P (0.525 -0.5 /(0.025) < Z < (0.050 - 0.5) /(0.025) ) and then using tables to get the answer
ii) Three pizzas are selected at random. Calculate the probability that all three have AT LEAST 0.475 lb of cheese
For this question do I find the probability the following way?? :
1- P ( Z < 0.474 - 0.5) /0.025)??
which gives a probability of 0.8508
and knowing this probability then i use binomial distribution?
P ( X = 3) = 3C3 p^(3) q^(3-3) ??
substituting p with the probability obtained form the normal distribution above.
Is this right?