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Sue0113

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The hieght of women are approximately normally distributed with a mean of 65.4 inches and a standard deviation of 2.5 inches.
What percent of these women would be over 70 inches or more?

(70-64.5)/2.5 = 2.2 Is this correct?

How would I determine the hieghts of the shortest 8% of the women?
 
Sue0113 said:
The hieght of women are approximately normally distributed with a mean of 65.4 inches and a standard deviation of 2.5 inches.
What percent of these women would be over 70 inches or more?

(70-64.5)/2.5 = 2.2 Is this correct?
Click to expand...
Yes, that is the correct "standard z value", if that was what you meant. Now you need to use a table of standard normal values to find what percentage that corresponds to.

How would I determine the hieghts of the shortest 8% of the women?
Click to expand...
what z-value gives 8% in a table of standard normal values?
 
Therefore

I'm really having hard time understanding how to read the z table and to know which z table is correct 1 to use.
a) 2.2 would be 48.6%
b)8% would be .0319
Not sure if i'm over thinking or just not understanding
 
Sue0113 said:
I'm really having hard time understanding how to read the z table and to know which z table is correct 1 to use.


a) 2.2 would be 48.6% Don't use "would." That's akin to "equals." Now, you mean that a z-score of 2.20 would lead to about 48.6% (or an area
under the normal curve of about 0.4860). That is not possible, because 2.20 is a positive z-score. Those values indicate that the percent is
greater than 50%.



b)8% would be .0319 8% = .0800. The closest entry (area) in the negative z-score side of the table to .0800 is .0793.

What z-score corresponds to .0793?



Not sure if i'm over thinking or just not understanding
Click to expand...


Look at the right hand table in a textbook, for example. With a positive z-score
of 2.20. the entry inside the table that corresponds to that is .9861, 98.61%.

However, because you want to know the probability of heights of 70 inches or more,
you must subtract .9861 from 1. Then convert that result to a percent.


1 - 0.9861 = 0.0139


Then 1.39% of those women have heights of 70 inches or more.
 
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There's a very nice table at http://www.mathsisfun.com/data/standard-normal-distribution-table.html. You can see both z and percentage values by moveing the cursor over the table. When I move the cursor so that z= 2.2, I see that the percentage is 48.6%. But notice that only the right portion is colored blue. That tells you that 48.6 of the vallues are between 0 and 2.2. Since the entire (both positive and negative) area is 1.00, half of the area is 0.5 and the portion that is above z= 2.2 is .5- .486= 0.014 or 1.4%.

8% From the left side would be .5- .08= .42 from the center and I see that corresponds to z= -1.41. What height is that?
 
Still not getting this

So if .05-.08=.42 Z = -1.41 I have no idea what hieght this is cause I do not see a -1.41.
I not understanding this at all.
 
Sue0113 said:
So if .05-.08=.42 Z = -1.41 I have no idea what hieght this is cause I do not see a -1.41.
I not understanding this at all.
Click to expand...
Your z-score is "normalized." We subtract the mean of the actual data values from each data value and then divide by the standard deviation.

What does this operation do?

Well the new mean is zero. So if the z score is positive, it means in this case that the woman has above average height. If the z score is negative, the woman has below average height. Instead of the height being given directly, it is being given relatively. Good so far?

But what is the point?

The point is that if the population described is "normally" distributed, the value of the z-score let's us use a z table to determine approximately what percentage of woman have a height this extreme. If, for example, a woman has a z-score for height of 1.88, we know she is taller than average because her score is positive. BUT we can use the z-table to say 47.06% of women have a height that is less than hers but equal to or above the mean and another 50% have a height that is less than the mean so this woman is taller than 97.06% of women, or in the 2.94% of women with the greatest height.

All normal distributions look alike once you standardize or normalize for the mean and the standard deviation. So if you have your z table handy (as almost everyone does), you get more information from a z-score than from a mean.

Does this help?
 
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