Problem Breakdown:

1. Data Overview:

   • Mean ($\mu$) = 121
   • Standard Deviation ($\sigma$) = 16
   • You are finding the percentage of people with readings below 130.

2. Step 1: Calculate Z-Score for 130 The formula for the z-score is: $z = \frac{x - \mu}{\sigma}$ Substituting: $z = \frac{130 - 121}{16} = \frac{9}{16} \approx 0.56$

3. Step 2: Use the Z-Score Table From the table provided:

   • For $z = 0.5$, the percentile is 69.15.
   • For $z = 0.6$, the percentile is 72.57.

4. Step 3: Interpolate to Approximate $z = 0.56$ Since $z = 0.56$ is between $z = 0.5$ and $z = 0.6$, linear interpolation will give: $\text{Percentile} = 69.15 + (0.56 - 0.5) \times (72.57 - 69.15)$ $\text{Percentile} \approx 69.15 + 0.06 \times 3.42 \approx 69.15 + 0.2052 \approx 69.36$

5. Final Answer: The percentage of people with readings below 130 is approximately 69.36%.

Would you like further details or explanations?

Here are 5 follow-up questions to expand your understanding:

1. How is linear interpolation applied in z-score tables?
2. Why is the z-score formula necessary for normal distributions?
3. How do we verify interpolation accuracy?
4. What happens if $x$ is below the mean?
5. What would the result be for readings above 130 instead?

Tip: Always check the z-score table range carefully; rounding errors can impact results.