Let $$ Z \sim N(0,1) $$. Accurate to 4 decimal places, evaluate $$ P(Z \geq-0.65) $$ 0.2422 0.2578 0.4843 0.7422 None of the above.

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We start with the probability we want to compute:

$$
P(Z \geq -0.65)
$$

Since $$Z$$ is a standard normal random variable, the standard normal distribution is symmetric about 0. Using the symmetry property of the normal distribution, we have:

$$
P(Z \geq -0.65) = P(Z \leq 0.65)
$$

Next, we look up the cumulative distribution function (CDF) of the standard normal at $$0.65$$, which is:

$$
\Phi(0.65) \approx 0.7422
$$

Thus, we conclude:

$$
P(Z \geq -0.65) \approx 0.7422
$$
$$ P(Z \geq -0.65) \approx 0.7422 $$

Let \( Z \sim N(0,1) \). Accurate to 4 decimal places, evaluate \( P(Z \geq-0.65) \) 0.2422 0.2578 0.4843 0.7422 None of the above.

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