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Math Help - Need help checking my work

  1. #1
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    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
    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
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  2. #2
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    Quote Originally Posted by armani View Post
    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
    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
    You're using the wrong random variable.
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