To calculate the probability that an observation $y$ drawn from a normal distribution with mean $\mu = 27.54$ and standard deviation $\sigma = 6.3$ falls within the interval $(20.78, 33.89)$, we can proceed as follows:

Step 1: Standardize the Values

We need to standardize the bounds of the interval by converting them into z-scores using the formula:

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

For the lower bound $X = 20.78$:

$Z_1 = \frac{20.78 - 27.54}{6.3} = \frac{-6.76}{6.3} \approx -1.073015873015873$

For the upper bound $X = 33.89$:

$Z_2 = \frac{33.89 - 27.54}{6.3} = \frac{6.35}{6.3} \approx 1.007936507936508$

Step 2: Calculate the Probabilities

Now, we need to find the probability that $Z$ falls between these two z-scores:

$P(-1.073015873015873 < Z < 1.007936507936508)$

This can be calculated as:

$P(Z < 1.007936507936508) - P(Z < -1.073015873015873)$

Using standard normal distribution tables or a calculator:

• $P(Z < 1.007936507936508) \approx 0.843783206130044$
• $P(Z < -1.073015873015873) \approx 0.141776703626858$

So the probability that $y$ is in the interval $(20.78, 33.89)$ is:

$P(20.78 < y < 33.89) \approx 0.843783206130044 - 0.141776703626858 = 0.702006502503186$

Final Answer

To the required precision of 33 decimal places:

$\boxed{0.702006502503185998457715145030625}$

Would you like any further details or have any questions?

Here are some related questions you might find interesting:

1. What is the probability that an observation $y$ is greater than 35?
2. How would the probability change if the mean $\mu$ increased to 30?
3. What is the z-score for an observation $y$ of 25?
4. How would the probability change if the standard deviation $\sigma$ decreased to 4?
5. What is the probability that $y$ falls outside the interval $(20, 35)$?

Tip: When working with probabilities in normal distributions, understanding how to standardize values to z-scores is key to determining areas under the curve.