text: # Thread: Need help checking my work
  
  1. ## Need help checking my work
  
  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?
  
  My work:
  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
  
  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)
  
  appreciate any help, thank you
  
  2. Originally Posted by armani
  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?
  
  My work:
  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? Mr F says: The correct random variable is W = Y/10 ....
  
  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