This problem is driving me crazy!

The length of time to complete a door assembly on an automobile factory assembly line is normally distributed with a mean of 6.7 minutes and standard deviation of 2.2 minutes. For a door selected at random, what is the probability the assembly line time will be:
a. 5 minutes or less?
b. 10 minutes or more?
c. between 5 and 10 minutes?

For each of these, compute the appropriate z scores for the standard normal distribution, and then determine the probabilities for the z scores using the normal distribution table.

z = (x - mu) / s

(a) z = (5 - 6.7) / 2.2 --> then find the proportion of area under the curve that is below this z score

(b) z = (10 - 6.7) / 2.2 --> here you find the proportion of area above this z-score

(c) here you need two z scores since you need to find the probability between two values - since you've already found the area below 5 and above 10, the area between 5 and 10 is just 1 minus the sum of (a) and (b)

If you look in the Examples section, there is a post entitled "The Vaunted Normal Distribution" (http://www.talkstats.com/examples/147-vaunted-normal-distribution.html) which talks about how to solve these types of problems.

Thanks for your help John. :) I have another problem I need help with, if you could please.

A large vat of mixed commercial chemicals is supposed to have a mean pH of 6.3 with a standard deviation of 1.9. Assume a normal distribution for pH values. If a random sample of ten readings in the vat is taken and the mean is computed, find each of the following:
a. P(5.2<xbar)
b.P(xbar<7.1)
c.P(5.2<x<7.1)

You follow the same process as the previous problem, except z is calculated a bit differently since we're now talking about the probability of seeing certain sample means, not just individual items.

z = (x - mu) / (s / sqrt(n))

So, for (a), we need:

P(xbar > 5.2)

z = (5.2 - 6.3)/(1.9 / sqrt(10)) = -1.83

So, P(xbar > 5.2) is the same as determining P(z > -1.83)

Thanks John:wave:

Can you tell me if I got the right answers

The manufacturer of a new compact car claims the miles per gallon for the gasoline comsumption is mound shaped an symmetric with a mean of 25.9 mpg and a standard deviation of 9.5 mpg. If 30 such cars are tested, what is the probability the average mpg is :
A. less than 23 mpg? answer -.07= .4721
B. more than 28 mpg ? answer .05 = .5198
C. between 23 and 28 mpg? answer .0478

z = (x - mu) / (s / sqrt(n))

(A) z = -1.672
P(x < 23) = P(z < -1.672)

(B) z = 1.211
P(x > 28) = P(z > 1.211)

(C) add up the answers in (A) and (B), then subtract from 1.0

thank you John