The problem asks to determine the value of $c$ such that the probability $P(c \leq Z \leq -0.69) = 0.2157$, where $Z$ follows a standard normal distribution.

Step-by-Step Solution:

1. Understand the probability range: We need to find $c$ such that the cumulative probability between $c$ and $-0.69$ is 0.2157.

2. Use the cumulative distribution function (CDF) of the standard normal distribution: The probability $P(c \leq Z \leq -0.69) = 0.2157$ can be written as: $P(Z \leq -0.69) - P(Z \leq c) = 0.2157$ We can break this down into two parts:

   • The cumulative probability $P(Z \leq -0.69)$ can be found using a standard normal table or calculator.

3. Find $P(Z \leq -0.69)$: Using the standard normal distribution table, we find: $P(Z \leq -0.69) \approx 0.2451$

4. Set up the equation: Now, using the known probability: $0.2451 - P(Z \leq c) = 0.2157$ Solving for $P(Z \leq c)$: $P(Z \leq c) = 0.2451 - 0.2157 = 0.0294$

5. Determine $c$: From the standard normal distribution table, we find the z-score corresponding to a cumulative probability of 0.0294, which is approximately $c = -1.89$.

Answer:

The value of $c$ is approximately -1.89.

Would you like any further details or clarifications?

5 Related Questions:

1. How do we calculate the cumulative probability for any z-value using the standard normal distribution?
2. What is the significance of the standard normal distribution in statistics?
3. Can you explain the process of finding probabilities for ranges of z-values?
4. How does the value of $c$ affect the cumulative probability?
5. How do we use z-scores to determine the probability for any given range?

Tip:

Always ensure you're using at least four decimal places when dealing with probabilities in the standard normal distribution for precision, especially in exams or technical problems.