To find the probability that a randomly selected value is greater than 200.9 in a normal distribution with a mean ($\mu$) of 65 and a standard deviation ($\sigma$) of 59.1, we will use the z-score formula:

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

Where:

• $X = 200.9$
• $\mu = 65$
• $\sigma = 59.1$

Step 1: Calculate the z-score

$z = \frac{200.9 - 65}{59.1} = \frac{135.9}{59.1} \approx 2.30$

Step 2: Use the z-score to find the probability

Now, we need to find $P(X > 200.9)$, which corresponds to $P(Z > 2.30)$.

From a z-score table or using a calculator, the probability for $Z \leq 2.30$ is approximately 0.9893. Since we want $P(Z > 2.30)$, we subtract this value from 1:

$P(X > 200.9) = 1 - 0.9893 = 0.0107$

Final Answer:

$P(X > 200.9) \approx 0.0107$

Would you like further explanation, or have any questions?

Here are 5 related questions to explore this topic further:

1. How do we interpret z-scores in the context of a normal distribution?
2. How can we find the probability for values less than a certain point?
3. What is the empirical rule, and how does it apply to normal distributions?
4. How do we calculate probabilities for values between two points in a normal distribution?
5. How does the standard deviation affect the shape of a normal distribution?

Tip: The z-score indicates how many standard deviations a data point is from the mean, which helps in determining its relative position in a normal distribution.