The problem is asking for the value of $c$ such that $P(c \leq Z \leq 1.04) = 0.8215$, where $Z$ is a standard normal variable (mean 0, standard deviation 1).

Let's break down the steps to find $c$:

1. Identify the probability for $Z \leq 1.04$: Using the standard normal table or a calculator, we find the cumulative probability for $Z = 1.04$.

   $P(Z \leq 1.04) \approx 0.8508$.

2. Set up the equation using the probability difference: Since $P(c \leq Z \leq 1.04) = 0.8215$, we can write: $P(Z \leq 1.04) - P(Z \leq c) = 0.8215$ Substituting $P(Z \leq 1.04) = 0.8508$: $0.8508 - P(Z \leq c) = 0.8215$

3. Solve for $P(Z \leq c)$: Rearranging the equation gives: $P(Z \leq c) = 0.8508 - 0.8215 = 0.0293$

4. Find the corresponding $c$-value: Now, we need to find the z-score that corresponds to a cumulative probability of 0.0293 in the standard normal table. This is approximately $c \approx -1.88$.

Thus, the value of $c$ is approximately -1.88.

Would you like further details or have any questions on this?

Related Questions:

1. How can we find probabilities for ranges that do not end at common z-scores?
2. Why is it necessary to look up cumulative probabilities for problems involving the standard normal distribution?
3. How would the solution change if a different standard deviation were used?
4. What steps are involved in converting between z-scores and raw scores in a normal distribution?
5. How can we verify answers when using a standard normal table versus a calculator?

Tip:

When solving normal distribution problems, always ensure that cumulative probabilities are used consistently to avoid errors in probability ranges.