Part 1: Probability of a Bone Density Test Score Greater than -1.63

Given:

• The test scores are normally distributed.
• Mean ($\mu$) = 0
• Standard deviation ($\sigma$) = 1
• We need to find the probability that the test score is greater than -1.63.

Step 1: Visualize the Problem

To understand this, let's plot the standard normal distribution curve. The curve is bell-shaped, centered at 0, and spans the range of possible bone density scores. We want to calculate the area under this curve to the right of $x = -1.63$.

Step 2: Calculate the Probability

The probability corresponds to the area under the normal distribution curve. To find this, we look for the cumulative distribution function (CDF) value for $x = -1.63$, then subtract it from 1 since we are interested in the area greater than -1.63.

We calculate the CDF value for $x = -1.63$ using standard normal distribution tables or a calculator:

$P(X > -1.63) = 1 - P(X \leq -1.63)$

Using a standard normal table or calculator: $P(X \leq -1.63) \approx 0.0516$

Thus: $P(X > -1.63) = 1 - 0.0516 = 0.9484$

Step 3: Graph

The graph of the standard normal distribution will have a shaded region to the right of $x = -1.63$, covering approximately 94.84% of the area under the curve.

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Would you like a detailed graph of the normal distribution curve showing this probability?

Relative Questions:

1. What is the z-score in a normal distribution?
2. How do you interpret the area under a normal distribution curve?
3. What does a negative z-score indicate in the context of bone density?
4. How would the result change if the mean were not 0?
5. Can you explain why we subtract from 1 when finding probabilities greater than a certain value?

Tip:

To find probabilities for any normal distribution, you can standardize using z-scores, which simplifies comparisons to the standard normal table.