SOLUTION: Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 75 women are randomly selected, find the probability

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Question 78633This question is from textbook Elementary Statistics
: Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean height between 63 and 65 inches. This question is from textbook Elementary Statistics

Found 2 solutions by Edwin McCravy, stanbon:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean height between 63 and 65 inches.


I'm sorry that statistics course are in a 
confusing state of transition between the
two types of tables and the use of the
technology, and that you are a victim of
circumstances. 

It depends on what textbook you are using and whether 
your teacher is old-fashioned or modern.  Different 
textbooks have different kinds of normal distribution 
tables.  Some teachers require you to use tables and 
others allow you to use a graphing calculator such as 
the TI-83 or TI-84.

I'll try to cover all three methods below:

The answer by calculator is .9811659853. The
answer by tables is .9812

Here's how to find that on a TI-83 or TI-84.

normalcdf(63,65,63.6,2.5/Ö(75))

Then press ENTER. You will get the above answer
immediately.

Here's how to get that on your TI calculator:

To get:

normalcdf(

1. press 2ND 
2. press VARS
3. press 2

There is a key for the comma just above the 7 key.
The square root Ö( is gotten by pressing 2ND then x²

If you are required to use the tables, then first you
must calculate the z-scores for the left and right 
bounds  x = 53 and x = 65 by the formula:

     x - m
z = -------
      s/Ön

where m = 63.6, s = 2.5, n = 75.

z-score for 63 is calculated thusly:

     63 - 63.6
z = ----------- = -2.078460969 
      2.5/Ö75

To use tables you will have to round that off
to -2.08 

and the z-score for 65: 

     65 - 63.6
z = ----------- = 4.849742261 
      2.5/Ö75

You would round that off to 4.85

But that will be too large to be on any table.

Some tables have negative numbers and
and some don't.

If your table has negative z-scores, then:

then you will take the area to the left of 
4.85 to be 1.

Then you will look up the area to the left
of -2.08 and find it to be .0188.

So you will subtract 1 - .0188 and get .9812

If your table does not have negative z-scores
then you ignore the sign, look up 2.08, and 
find .4812. Then you must add .5 to that and 
get .9812.
 
 
Edwin

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 75 women are randomly selected, find the probability that they have a mean height between 63 and 65 inches.
---------------------
Find the z-score of 63 and of 65:
z(63)=(63-63.6)/[2.5/sqrt(75)]=-2.07846...
z(65)=(65-63.6)/[2.5/sqrt(75)]=4.8497...
P(-2.07846<=z<=4.8497)=0.9812
Cheers,
Stan H.