To calculate the probability that an observation $y$ drawn from a normal distribution with mean $\mu = 37.41$ and standard deviation $\sigma = 8.83$ is greater than 52.79, we will use the Z-score method.

Step 1: Convert the observation to a Z-score

The Z-score formula is:

$Z = \frac{y - \mu}{\sigma}$

Where:

• $y = 52.79$
• $\mu = 37.41$
• $\sigma = 8.83$

$Z = \frac{52.79 - 37.41}{8.83} = \frac{15.38}{8.83} \approx 1.7429$

Step 2: Find the CDF value corresponding to the Z-score

We need the cumulative probability up to $Z = 1.7429$. The CDF for $Z = 1.7429$ can be looked up in a standard normal distribution table or calculated using a statistical tool. The approximate value is:

$\Phi(1.7429) \approx 0.9597$

Step 3: Compute the probability for the observation being greater than 52.79

The probability of the observation being greater than 52.79 is:

$P(y > 52.79) = 1 - \Phi(1.7429) = 1 - 0.9597 = 0.0403$

Final Answer

The probability that an observation drawn from this normal distribution is greater than 52.79 is approximately 0.040300000000000000000000000000000.

Would you like more details, or do you have any questions?

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1. What is the significance of the Z-score in normal distribution?
2. How do you find the probability for values less than a certain observation?
3. How would the probability change if the standard deviation was larger?
4. What is the interpretation of the CDF in this context?
5. Can you explain how the normal distribution curve is used to find these probabilities?

Tip: For extremely precise calculations, using statistical software or a calculator designed for high-precision operations is essential.