To solve the probability problems using the standard normal distribution (Z) and rounding to at least three decimal places, we will use standard normal distribution properties and cumulative probabilities:

(a) $P(Z > -1.93)$

The probability that $Z$ is greater than a certain value is equivalent to:

$P(Z > -1.93) = 1 - P(Z \leq -1.93)$

Looking up $P(Z \leq -1.93)$ from the cumulative distribution table, we get approximately:

$P(Z \leq -1.93) \approx 0.0268$

So:

$P(Z > -1.93) = 1 - 0.0268 = 0.9732$

(b) $P(Z \leq -1.14)$

Looking up the cumulative probability for $Z \leq -1.14$:

$P(Z \leq -1.14) \approx 0.1271$

(c) $P(0.87 < Z < 2.05)$

This is the probability that $Z$ lies between two values, and can be calculated as:

$P(0.87 < Z < 2.05) = P(Z < 2.05) - P(Z \leq 0.87)$

First, look up the cumulative probabilities:

$P(Z \leq 2.05) \approx 0.9798$ $P(Z \leq 0.87) \approx 0.8078$

Now subtract the two:

$P(0.87 < Z < 2.05) = 0.9798 - 0.8078 = 0.1720$

Final answers:

(a) $P(Z > -1.93) = 0.9732$

(b) $P(Z \leq -1.14) = 0.1271$

(c) $P(0.87 < Z < 2.05) = 0.1720$

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Do you want more details on these calculations or have any questions?

Here are five related questions to further explore the topic:

1. What is the probability of $Z \leq 0$ in the standard normal distribution?
2. How do you compute $P(Z > 1.5)$ in a standard normal distribution?
3. What is the symmetry property of the normal distribution that helps simplify these calculations?
4. How would you calculate $P(-2 < Z < 1)$?
5. Can you explain how the standard normal distribution is related to z-scores in real-world data?

Tip: Always remember that probabilities for $P(Z > x)$ can be computed by subtracting the cumulative probability from 1.