The length of time it takes to fill an order at a local sandwich shop is normally distributed with a mean of 4.1 minutes and a standard deviation of 1.3 minutes.
a) what is the probability that the average waiting time for a random sample of 10 customers is between 4.0 and 4.2 minutes?
X_1, X_2, ..., X_10 are idependent, normal distributed random variables with the mean and sd given.
Y = X_1 + ... + X_10 is a normally distributed random variable with mean 4.1*12 = 49.2 and variance (1.3)^2*12 = 20.28, so the sd is sqrt(20.28) = 4.5
P(48<y<50.6) = .2102
I used Normal Distribution Probability Calculator
the 2nd calculator to plug it all in.
The correct answer is .1896. what did I do wrong?
b) the probability is 95% that the average waiting time for a random sample of 10 customers is greater than how many minutes?
mean = 49.2
sd = 4.5
shaded area 0.95 ABOVE
I used this calculator: Inverse Normal Distribution
This gives 41.798 -- so the probability that 10 random customers wait at least 41.798 is 95% (so each customer has a 95% probability of waiting at least 4.180 minutes)
correct answer is 3.42
appreciate any help, thank you