Normal Distributions and Transformation Formulas

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candice01

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Hi,
I am trying to use the transformation formula to solve this problem but I have to be missing something. The problem states: The production department of the newspaper has embarked on a quality improvement effort. Its first project relates the blackness of the newspaper print. Each day, a determination needs to be made concerning how black the newspaper is printed. Blackness is measured on a standard scale in which the target value is 1.0. Data collected over the past year indicate that the blackness is approximately normally distributed with a mean of 1.005 and a standard deviation of 0.10. Each day, one spot on the first newspaper printed is chosen, and the blackness of the spot is measured. The blackness of the newspaper is considered acceptable if the blackness of a spot is between 0.95 and 1.05.


Assume that the distribution has not changed from what it was in the past year, what is the probability that the blackness of the spot is :
a. less than 1.0?
b. between 0.95 and 1.0?
c. between 1.0 and 1.05?
d. less than 0.95 or greater than 1.05?

I thought i was supposed to use:

z= x-u(mean)/standard deviation which i said z=1.0-1.005/.10= -.05 but I don't know if this is right and what to do from here. Please help. Thank you very much. Candice
 
candice01 said:
Hi,
I am trying to use the transformation formula to solve this problem but I have to be missing something. The problem states: The production department of the newspaper has embarked on a quality improvement effort. Its first project relates the blackness of the newspaper print. Each day, a determination needs to be made concerning how black the newspaper is printed. Blackness is measured on a standard scale in which the target value is 1.0. Data collected over the past year indicate that the blackness is approximately normally distributed with a mean of 1.005 and a standard deviation of 0.10. Each day, one spot on the first newspaper printed is chosen, and the blackness of the spot is measured. The blackness of the newspaper is considered acceptable if the blackness of a spot is between 0.95 and 1.05.
Click to expand...


Assume that the distribution has not changed from what it was in the past year, what is the probability that the blackness of the spot is :
Click to expand...

a. less than 1.0?
Click to expand...

\(\displaystyle \overline{x}=1, \;\ \mu=1.005, \;\ \sigma=.1\)

\(\displaystyle z=\frac{\overline{x}-\mu}{\sigma}\)

The normal distribution goes from left to right. So, if we want <1, we have:

\(\displaystyle z=\frac{1-1.005}{.1}=-.05\)

Look this up in the table and we see that a z score of -.05 corresponds to .4801

A 48% probability that the blackness is less than 1.

b. between 0.95 and 1.0?
Click to expand...

We already found 1, it is .4801

\(\displaystyle z=\frac{.95-1.005}{.1}=-.55\)

In the table this is .2912

Since we are between, subtract the two values and get .4801-.2912=.1889

18.89% probability the blackness is between .95 and 1.

Now, try the others. Be careful with those that ask for 'greater'. Because the table runs left to right, you find your value and subtract from 1.
 
Thank you.
Ok it's starting to make a bit of sense.

for c. I have: z= 1.0-1.05/.10= -0.5 which equals 0.3085= 30%. Is this remotely right. Sorry I don't know how to add symbols to my posts yet.
 
Between 1 and 1.05

\(\displaystyle x=\frac{1.05-1.005}{.1}=.45\)

Looking in the table, this is .6736

We already know that 1 corresponds to .4801

.6736-.4801=.1935
 
Thanks a lot for your help. I think I'm beginning to confuse myself some so I'm going to take a break and try again in a little bit. Again thanks for your help I really appreciate it. I am very frustrated because I want to understand this a lot better.
 
Hello, I am now trying to find z for the sampling distribution of the mean.

The problem states:
Continuing its quality improvement effort first described in Chapter 6, the production department of the newspaper has been monitoring the blackness of the newspaper print. As before, blackness is measured on a standard scale in which the target value is 1.0. Data collected over the past year indicate that the blackness is approximately normally distributed, with a mean of 1.005 and a standard deviation of 0.10

Each day 25 spots on the first newspaper printed are chosen, and the blackness of the spots is measured. Assuming that the distribution has not changed from what it was the previous year, what is the probability that the mean blackness of the spots is:

a. less than 1.0
b. between 0.95 and 1.0
c. between 1.0 and 1.05
d. less than 0.95 or greater than 1.05
e. Suppose that the mean blackness of today's sample of 25 spots is 0.952. What conclusion can you reach about the blackness of today's newspaper based on this result? Explain.

I have 1.0-1.005/.10/5(which is the square root of 25). do you know how I can use excel and the formulas to figure this out also?

Thanks,

Candice
 
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