Re: good random stats questions (please help, not difficult)

On May 30, 11:33 pm, Richard Ulrich <Rich.Ulr...@xxxxxxxxxxx> wrote:
On 30 May 2007 08:11:38 -0700, RandyT...@xxxxxxxxx wrote:





On May 25, 11:30 am, "Stratocaster" <sto...@xxxxxxxxxxx> wrote:
<RandyT...@xxxxxxxxx> wrote in message

news:1180105550.979900.230850@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

1) The weekly demand foro GM car sales follows a normal distribution
with a mean of 40,000 cars and a standard deviation of 12,000 cars. A)
there is a 5% chance that GM will sell more than what number of cars
during the next year

You know the formula for a z-score?

Find the appropriate Z value (from a table presumably), equate it to the
expression for the z-score, then plug and chug. You got this.

I am having a hard time finding the z value. I am given a STDV and a
mean, [break]

And "5%". What information do you see in a Normal table?

how can i find a Z- value from that? it seems like i am missing
something. would it be z= ((x-mean)/(STDV)). now it does let me know
there is a 5% chance that GM will sell more that a particular
number....is that the X value i am lookign for?

Yes.

--
Rich Ulrich, wpi...@xxxxxxxxxxxx://www.pitt.edu/~wpilib/index.html- Hide quoted text -

- Show quoted text -

what about the second part of the question. how can I use the "z" eqn
to solve for the probability of selling BETWEEN 2.0 and 2.3 million
cars. i normally have one number and one z value, here woudl i have
two numbers and two z values?

.

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