Normal Distribution Question

  • Thread starter splitendz
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Hi Guy's,

I have problems answering questions like this....(i'll just make up a question)

The time it takes to bake a cake in a bakery shop is a random variable that has a normal distribution with a mean of 4.5 minutes and standard deviation of 1 minute.

Lets suppose this bakery has been given an order to make 78 cakes. What is the probability that the baker will have in less than 380 minutes all 78 cakes ready?

My first instinct is to treat this as a poisson distribution but if I was to evaulate P[N < 79] it would simply take too long.

Any ideas would be great :)
The sample mean is distributed about normally. In this case it is distributed exactly normally, because the observations have the normal distribution. The sample mean has the same mean as the population mean, and its standard deviation is the population's standard deviation divided by the square root of the sample size. With u the mean baking time for his sample of 78 cakes, the mean of u is 4.5 and the standard deviation of u is 1/sqrt(78). From the distribution of the mean, you can easily determine the distribution of the sum.
BT - why talk about the sample mean?

The sum of n normal distributions(mu,sigma) is distributed normally with mean equal to n * mu, standard deviation = sqrt(n)*sigma. That's a fairly standard result due to the fact that when adding two independent random variables with normal distributions, sum their means, and add their variances for the distribution of their sum.
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Or, you could do it your way.

I was talking about the sample mean because maybe his pie baking times are _approximately_ normal, and not _exactly_ normal. By the Central Limit Theorem you can say that the sample mean is still about normal. Not that it makes a difference in this case, but that's why I approached it by the sample mean.
How will knowing the sample distribution of the sum assist in determining the probability of the baker making 78 cakes in under 380 minutes?
The total time it takes to bake the 78 cakes is the sum of the sample of the 78 baking times of each of the cakes.
Ah, I think I get it. The mean for a population and a sample are the same and the standard deviation is given by 1/sqrt(N) as you said. Once you have the stanard deviation and the mean for the sample distribution you can then standardise the normal distribution and find the probability accordingly. In this case it would be the probability that the time it takes to bake a cake is less than 380/78 given mu = 4.5 and a standard deviation of 1/sqrt(78) = 0.11. Right?

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