# 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
The text passage contains a mathematical problem involving normal distribution and provides the user's attempt at solving it, along with the correct answers for comparison.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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
user
The text passage contains a mathematical problem involving normal distribution and provides the user's attempt at solving it, along with the correct answers for comparison.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS