## 3.45

• Part (a)
The key is in chapter 2, where it explains that the standard deviation is measured in the same units as the data presented in the experiments.
For instance, in the normal distribution $N(\mu, \sigma)$,  2.5 standard deviations below the mean will be at $\mu - 2.5 \times \sigma$.
In the particular case of the standard normal distribution $N(0,1)$, the $z$-score is at $0 - 2.5 \times 1 = -2.5$.
Now, we only need to look up Table A for the $z$-score –2.5. Notice that the corresponding percentage is 0.00621 (that is our 0.6%)
• Part (b)
According to the wording of part (b), older women have a mean BMD “two units smaller than that of young adults”; therefore, as the standard deviation is also the same, we can think of the normal distribution of BMD for women as $N(2,1)$.
In this case, the z-score that indicates the percentage of older women with osteoporosis will be $2 - 2.5 \times 1 = -0.5$.  We look up Table A for –0.5, and the corresponding percentage is 0.3085 (that is, 30.85%)
1. September 8, 2010 at 8:24 pm

Thank you so much!

2. September 9, 2010 at 9:19 am

Fransisco, thank you. But I am a little confused. I understand the reasoning, but isn’t the formula for z score x – mean divided by deviation. However, you subtracted the mean from the x, the opposite from the formula? Can you please explain why you are subtracting the mean from x, instead of using the formula to subtract the x from the mean?? I am confused why sometimes, in the earlier exaples, we subtracted x from mean, but here in this problem you are subtracting the mean from x? Can you please explain? Thank you…

• September 9, 2010 at 9:36 am

Not exactly: the formula that you mention helps you find the z-score in a situation $N(\mu, \sigma)$ where the mean is not zero and/or the std is not one: they will give you the $x$, the values of $\mu$ and $\sigma$, and then you apply that formula to find the z-score.

Notice that for example, in the first part, they are telling us directly what the z-score is: 2.5 standard deviations below the mean.

Since they are offering your neither $x$, nor mean ($\mu$) nor std ($\sigma$), we are playing directly with the $N(0,1)$, so no adjustment is necessary. The trick there was to realize that the standard deviation is measured in the same units as the variables are measured.

3. September 23, 2010 at 11:51 am

Fransisco-

I have been studying, and I am a little confused and was wondering if you could clarify something for me – how do we know the z-scores for the 68-95-99.7 rule? I mean, what z-scores reflect these 3 percentages undert the cuve? What z-scores correspond to the 68-95-99.7 rule? Would you mind helping me on that? I am sorry, but I am confused on this!

Scott

• September 23, 2010 at 2:34 pm

The best way to go about it, in my opinion, is by visualization of the graph of the normal distribution, pointing up where the cuts for the 68%, 95% and 99.7% are.

The image below is a generic normal distribution $N(\mu, \sigma)$.

In the case of the standard normal distribution $(\mu=0$ and $\sigma=1)$, the corresponding $z-$scores would be:
For 50%: 0
For (50 + 34)%: 1.
For (50 – 34)%: -1
For (50 + 34 + 13.6)%: 2.
For (50 – 34 – 13.6)%: -2
and so on…

• September 24, 2010 at 11:44 am

Thank you so much Fransisco – this really helped me understand this concept! I appreciate all your help