Question

determine the area under the standard normal curve that lies to the right of (a) \\(z = -0.77\\), (b) \\(z = 0.79\\), (c) \\(z = -1.87\\), and (d) \\(z = -1.46\\).

Explanation:

⚡ Using what you learned: normal distribution

Step 1: Area to the right of \( Z = -0.77 \)

Using a standard normal distribution table, the cumulative area to the left of \( Z = -0.77 \) is:

$$ P(Z < -0.77) = 0.2206 $$

The area to the right is the complement:

$$ P(Z > -0.77) = 1 - P(Z < -0.77) = 1 - 0.2206 = 0.7794 $$

Step 2: Area to the right of \( Z = 0.79 \)

The cumulative area to the left of \( Z = 0.79 \) is:

$$ P(Z < 0.79) = 0.7852 $$

The area to the right is:

$$ P(Z > 0.79) = 1 - P(Z < 0.79) = 1 - 0.7852 = 0.2148 $$

Step 3: Area to the right of \( Z = -1.87 \)

The cumulative area to the left of \( Z = -1.87 \) is:

$$ P(Z < -1.87) = 0.0307 $$

The area to the right is:

$$ P(Z > -1.87) = 1 - P(Z < -1.87) = 1 - 0.0307 = 0.9693 $$

Step 4: Area to the right of \( Z = -1.46 \)

The cumulative area to the left of \( Z = -1.46 \) is:

$$ P(Z < -1.46) = 0.0721 $$

The area to the right is:

$$ P(Z > -1.46) = 1 - P(Z < -1.46) = 1 - 0.0721 = 0.9279 $$

Answer:

(a) 0.7794
(b) 0.2148
(c) 0.9693
(d) 0.9279