Let $$ Z \sim N(0,1) $$. Accurate to 4 decimal places, evaluate $$ P(-0.37 \leq Z \leq-0.07) $$ 0.3557 0.1164 0.8836 0.4721 None of the above.

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We are to evaluate 
$$
P(-0.37 \leq Z \leq -0.07)
$$
for $$ Z \sim N(0,1) $$.

**Step 1. Express the probability in terms of the CDF**

The probability can be written as
$$
P(-0.37 \leq Z \leq -0.07) = \Phi(-0.07) - \Phi(-0.37)
$$
where $$\Phi(z)$$ is the cumulative distribution function of the standard normal distribution.

**Step 2. Use the symmetry property of the standard normal distribution**

Recall the property
$$
\Phi(-z) = 1 - \Phi(z).
$$
Using this:
$$
\Phi(-0.07) = 1 - \Phi(0.07)
$$
$$
\Phi(-0.37) = 1 - \Phi(0.37).
$$

**Step 3. Substitute and simplify**

Substitute in the original expression:
$$
P(-0.37 \leq Z \leq -0.07) = \bigl[1 - \Phi(0.07)\bigr] - \bigl[1 - \Phi(0.37)\bigr] = \Phi(0.37) - \Phi(0.07).
$$

**Step 4. Look up the standard normal values**

Using a standard normal table or calculator:
$$
\Phi(0.07) \approx 0.5279,
$$
$$
\Phi(0.37) \approx 0.6443.
$$

**Step 5. Compute the difference**

Now subtract:
$$
P(-0.37 \leq Z \leq -0.07) \approx 0.6443 - 0.5279 = 0.1164.
$$

Thus, accurate to 4 decimal places, the probability is $$ \boxed{0.1164} $$.
The probability $$ P(-0.37 \leq Z \leq -0.07) $$ is approximately 0.1164.

Let \( Z \sim N(0,1) \). Accurate to 4 decimal places, evaluate \( P(-0.37 \leq Z \leq-0.07) \) 0.3557 0.1164 0.8836 0.4721 None of the above.

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