The question asks to find the bone density test scores that separate the lowest 5% and the highest 5% of a normal distribution, with a mean ($\mu$) of 0 and a standard deviation ($\sigma$) of 1.

To find the cutoff values, we can determine the Z-scores that correspond to the bottom 5% and top 5% in a standard normal distribution:

1. For the bottom 5%, we need the Z-score for the 5th percentile. In a standard normal table, the Z-score approximately equals $-1.645$.
2. For the top 5%, we need the Z-score for the 95th percentile. This Z-score approximately equals $+1.645$.

Thus, the cutoff values are approximately:

• Lower cutoff: $-1.645$
• Upper cutoff: $+1.645$

Solution for the Options

The correct graph should display shaded regions at both ends, with labels at $-1.645$ and $+1.645$. Based on the image, option B is correct as it shows these two cut-off areas shaded on both tails.

Final Answer

The bone density scores are: $-1.65, 1.65$ (rounded to two decimal places).

Would you like further details on this calculation?

Related Questions:

1. What does the Z-score represent in a normal distribution?
2. How is the 5th percentile of a normal distribution calculated?
3. Why do we use Z-scores to find cutoff values in a normal distribution?
4. Can these cutoff values be used in other contexts of standard normal distributions?
5. What would be the cutoff values if the standard deviation was different from 1?

Tip:

When working with standard normal distributions, Z-scores simplify finding probabilities and cutoff values, as they relate directly to standard deviations from the mean.