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
explanation: The document is a forum post where the user is seeking help to check their work on a statistics problem involving normally distributed waiting times at a sandwich shop. The post contains two questions and the user's attempted solutions, along with the correct answers provided for comparison. The user is asking for help to identify where they went wrong in their calculations.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a forum post where the user is seeking help to check their work on a statistics problem involving normally distributed waiting times at a sandwich shop. The post contains two questions and the user's attempted solutions, along with the correct answers provided for comparison. The user is asking for help to identify where they went wrong in their calculations.
  
  Label: multiple-question-and-answer