Question

determine the area under the standard normal curve that lies between (a) \\(z = -0.88\\) and \\(z = 0.88\\), (b) \\(z = -2.39\\) and \\(z = 0\\), and (c) \\(z = 1.28\\) and \\(z = 1.54\\).

Explanation:

⚡ Using what you learned: normal distribution

Step 1: Part (a) Area between \( Z = -0.88 \) and \( Z = 0.88 \)

$$ P(-0.88 < Z < 0.88) = P(Z < 0.88) - P(Z < -0.88) $$

Using standard normal distribution table values:

• \( P(Z < 0.88) \approx 0.8106 \)
• \( P(Z < -0.88) \approx 0.1894 \)

$$ 0.8106 - 0.1894 = 0.6212 $$

Step 2: Part (b) Area between \( Z = -2.39 \) and \( Z = 0 \)

$$ P(-2.39 < Z < 0) = P(Z < 0) - P(Z < -2.39) $$

Using standard normal distribution table values:

• \( P(Z < 0) = 0.5000 \)
• \( P(Z < -2.39) \approx 0.0084 \)

$$ 0.5000 - 0.0084 = 0.4916 $$

Step 3: Part (c) Area between \( Z = 1.28 \) and \( Z = 1.54 \)

$$ P(1.28 < Z < 1.54) = P(Z < 1.54) - P(Z < 1.28) $$

Using standard normal distribution table values:

• \( P(Z < 1.54) \approx 0.9382 \)
• \( P(Z < 1.28) \approx 0.8997 \)

$$ 0.9382 - 0.8997 = 0.0385 $$

Answer:

(a) 0.6212
(b) 0.4916
(c) 0.0385