To solve the problem, we need to use the properties of the standard normal distribution and a standard normal distribution table (also known as the Z-table). The standard normal distribution is a normal distribution with a mean ($\mu$) of 0 and a standard deviation ($\sigma$) of 1. The Z-table provides the cumulative probabilities for values of $Z$ up to a given value.
Step-by-Step Solution:
#### (a) $ P(Z < 2.16) $
- This is the probability that a standard normal random variable $Z$ is less than 2.16.
- Using the Z-table, look up the value 2.16.
- The Z-table gives the cumulative probability up to 2.16, which is approximately 0.9846.
- Therefore, $ P(Z < 2.16) = 0.9846 $.
#### (b) $ P(Z > 1.13) $
- This is the probability that a standard normal random variable $Z$ is greater than 1.13.
- Using the Z-table, look up the value 1.13.
- The Z-table gives the cumulative probability up to 1.13, which is approximately 0.8708.
- Since $ P(Z > 1.13) = 1 - P(Z \leq 1.13) $, we have:
$$
P(Z > 1.13) = 1 - 0.8708 = 0.1292
$$
#### (c) $ P(Z < -1.65) $
- This is the probability that a standard normal random variable $Z$ is less than -1.65.
- Using the Z-table, look up the value 1.65 (since the table typically only provides positive values, we use the symmetry of the normal distribution).
- The Z-table gives the cumulative probability up to 1.65, which is approximately 0.9505.
- Since the normal distribution is symmetric, $ P(Z < -1.65) = 1 - P(Z \leq 1.65) $, we have:
$$
P(Z < -1.65) = 1 - 0.9505 = 0.0495
$$
#### (d) $ P(Z > 1.65) $
- This is the probability that a standard normal random variable $Z$ is greater than 1.65.
- Using the Z-table, look up the value 1.65.
- The Z-table gives the cumulative probability up to 1.65, which is approximately 0.9505.
- Since $ P(Z > 1.65) = 1 - P(Z \leq 1.65) $, we have:
$$
P(Z > 1.65) = 1 - 0.9505 = 0.0495
$$
#### (e) $ P(|Z| > 1.65) $
- This is the probability that the absolute value of $Z$ is greater than 1.65.
- This can be split into two parts: $ P(Z > 1.65) + P(Z < -1.65) $.
- From parts (c) and (d), we know:
$$
P(Z > 1.65) = 0.0495 \quad \text{and} \quad P(Z < -1.65) = 0.0495
$$
- Therefore:
$$
P(|Z| > 1.65) = P(Z > 1.65) + P(Z < -1.65) = 0.0495 + 0.0495 = 0.0990
$$
#### (f) $ P(|Z| < 1.65) $
- This is the probability that the absolute value of $Z$ is less than 1.65.
- This is the complement of $ P(|Z| > 1.65) $.
- From part (e), we know:
$$
P(|Z| > 1.65) = 0.0990
$$
- Therefore:
$$
P(|Z| < 1.65) = 1 - P(|Z| > 1.65) = 1 - 0.0990 = 0.9010
$$
#### (g) $ P(Z > 4) $
- This is the probability that a standard normal random variable $Z$ is greater than 4.
- Using the Z-table, look up the value 4.00.
- The Z-table gives the cumulative probability up to 4.00, which is approximately 0.99997.
- Since $ P(Z > 4) = 1 - P(Z \leq 4) $, we have:
$$
P(Z > 4) = 1 - 0.99997 = 0.00003
$$
Final Answers:
\[
\boxed{
\begin{array}{ll}
(a) & 0.9846 \\
(b) & 0.1292 \\
(c) & 0.0495 \\
(d) & 0.0495 \\
(e) & 0.0990 \\
(f) & 0.9010 \\
(g) & 0.00003 \\
\end{array}
}
\]
Parent Tip: Review the logic above to help your child master the concept of normal curve worksheet.
