To compute the probability that an observation $y$ drawn from a normal distribution with mean $\mu = 46.44$ and standard deviation $\sigma = 7.38$ is greater than 53.97, we follow these steps:

Step 1: Standardize the value to the Z-distribution

We convert the value $53.97$ to its corresponding Z-score using the formula: $Z = \frac{X - \mu}{\sigma}$ where $X = 53.97$, $\mu = 46.44$, and $\sigma = 7.38$.

$Z = \frac{53.97 - 46.44}{7.38} = \frac{7.53}{7.38} \approx 1.0204$

Step 2: Find the cumulative probability

We now find the cumulative probability corresponding to this Z-score using the standard normal distribution table or a calculator.

$P(Z \leq 1.0204) \approx 0.8468$

Step 3: Calculate the probability that $y$ is greater than 53.97

The probability that $y$ is greater than 53.97 is given by: $P(y > 53.97) = 1 - P(Z \leq 1.0204)$ $P(y > 53.97) = 1 - 0.8468 = 0.1532$

Thus, the probability that an observation $y$ drawn from the distribution is greater than 53.97 is 0.153 (to three decimal places).

Would you like more details or have any questions?

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Here are 5 related questions:

1. How would you find the probability that $y$ is less than 40 in this distribution?
2. Can you calculate the probability that $y$ lies between 50 and 60 in the same distribution?
3. What is the Z-score interpretation for a value of $y$ below the mean?
4. How does changing the mean of the distribution affect the probability calculation?
5. What is the relationship between the standard deviation and the spread of the normal distribution?

Tip: To find the probability of an observation being greater than a certain value in a normal distribution, you can always subtract the cumulative probability from 1.