In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.)

(a)    Φ(c) = 0.9850

(b)    P(0 ≤ Z ≤ c) = 0.2939

(c)    P(c ≤ Z) = 0.1292

(d)    P(−c ≤ Z ≤ c) = 0.6528

(e)    P(c ≤ |Z|) = 0.0128

We need to determine the value of the constant \( c \) that makes each probability statement correct. For these types of problems, we are dealing with the standard normal distribution \( Z \), which has a mean of 0 and a standard deviation of 1. We can use the cumulative distribution function (CDF) \( \Phi(c) \), the properties of the standard normal distribution, and tables or technology to find the values of \( c \).

### (a) \( \Phi(c) = 0.9850 \)

\( \Phi(c) \) represents the cumulative distribution function for a standard normal distribution, meaning \( \Phi(c) \) gives the probability that \( Z \) is less than or equal to \( c \).

We need to find \( c \) such that:
\[
\Phi(c) = 0.9850
\]
Using a standard normal table or calculator, we find that:
\[
c \approx 2.17
\]

### (b) \( P(0 \leq Z \leq c) = 0.2939 \)

This represents the probability that \( Z \) lies between 0 and \( c \), so we need to find \( c \) such that:
\[
P(0 \leq Z \leq c) = \Phi(c) - \Phi(0) = 0.2939
\]
Since \( \Phi(0) = 0.5 \), we have:
\[
\Phi(c) - 0.5 = 0.2939 \quad \Rightarrow \quad \Phi(c) = 0.7939
\]
Using the standard normal table:
\[
c \approx 0.82
\]

### (c) \( P(c \leq Z) = 0.1292 \)

This represents the probability that \( Z \) is greater than or equal to \( c \). This is the complementary probability of the cumulative distribution:
\[
P(Z \geq c) = 1 - \Phi(c) = 0.1292
\]
So, we have:
\[
\Phi(c) = 1 - 0.1292 = 0.8708
\]
Using the standard normal table:
\[
c \approx 1.13
\]

### (d) \( P(-c \leq Z \leq c) = 0.6528 \)

This represents the probability that \( Z \) lies between \( -c \) and \( c \), which is:
\[
P(-c \leq Z \leq c) = 2\Phi(c) - 1 = 0.6528
\]
Solving for \( \Phi(c) \):
\[
2\Phi(c) - 1 = 0.6528 \quad \Rightarrow \quad 2\Phi(c) = 1.6528 \quad \Rightarrow \quad \Phi(c) = 0.8264
\]
Using the standard normal table:
\[
c \approx 0.94
\]

### (e) \( P(c \leq |Z|) = 0.0128 \)

This represents the probability that \( Z \) is greater than or equal to \( c \) in absolute value, meaning:
\[
P(|Z| \geq c) = 2P(Z \geq c) = 0.0128
\]
Thus:
\[
2[1 - \Phi(c)] = 0.0128 \quad \Rightarrow \quad 1 - \Phi(c) = 0.0064 \quad \Rightarrow \quad \Phi(c) = 0.9936
\]
Using the standard normal table:
\[
c \approx 2.48
\]

### Final answers:
- (a) \( c \approx 2.17 \)
- (b) \( c \approx 0.82 \)
- (c) \( c \approx 1.13 \)
- (d) \( c \approx 0.94 \)
- (e) \( c \approx 2.48 \)

Would you like more details on any of these solutions? Here are 5 related questions to expand the topic:

1. How can you calculate probabilities using the standard normal distribution without a table?
2. What is the relationship between the CDF and the probability for a normal distribution?
3. How can we interpret the value of \( \Phi(c) \) in real-world applications?
4. How does the symmetry of the normal distribution affect these types of problems?
5. What is the meaning of \( P(c \leq |Z|) \) and when is it useful?

**Tip:** Always remember that the cumulative distribution function \( \Phi(c) \) gives the area under the curve from \( -\infty \) to \( c \), representing the probability that a standard normal variable \( Z \) is less than or equal to \( c \).

In each case, determine the value of the constant c that makes the probability statement correct. (Round your answers to two decimal places.)

(a) Φ(c) = 0.9850

(b) P(0 ≤ Z ≤ c) = 0.2939

(c) P(c ≤ Z) = 0.1292

(d) P(−c ≤ Z ≤ c) = 0.6528

(e) P(c ≤ |Z|) = 0.0128

Determine the constant c in probability statements involving the standard normal distribution, including calculations with Φ(c) and using probability properties. This problem covers important concepts like the cumulative distribution function (CDF) and probability in normal distribution, suitable for advanced students.

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