1. ## Check my Work please: probability

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 http://www.analyzemath.com/statistic...alculator.html
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: http://onlinestatbook.com/analysis_l...rmal_dist.html

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. ## Re: Check my Work please: probability

Are you sure about the correct answer for part a? I seem to be getting a different answer. Parts a and b are very related and my method seems to be getting the right answer for part b.

3. ## Re: Check my Work please: probability

Are you sure about the correct answer for part a? I seem to be getting a different answer. Parts a and b are very related and my method seems to be getting the right answer for part b.
That's the answer he gave us

a) .1896 and my answer was .2102

b) 3.42 and my answer was 41.798

4. ## Re: Check my Work please: probability

where

If we directly use the software (say R) to compute it, we get
0.1921899127810379

But I know how the answer come out. It is because that solution try
to convert to standard normal first, and note

And he round off here. say 2 d.p. for normal table, then

5. ## Re: Check my Work please: probability

Originally Posted by BGM
where

If we directly use the software (say R) to compute it, we get
0.1921899127810379

But I know how the answer come out. It is because that solution try
to convert to standard normal first, and note

And he round off here. say 2 d.p. for normal table, then

ooo I see

thank you that makes a lot of sense. Appreciate the help greatly

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