(a)
0.309

\( X \sim N(1000, 100^2) \)

Seek: \( P(X < 950) \)

Z-score: \( z = \frac{950 – 1000}{100} = -0.5 \)

\( P(Z < -0.5) \approx 0.3085 \)

Round to 3 decimal places: 0.309

Result: 0.309 [3]

(b)
0.868

Method 1
Let \( \overline{X} \) be the mean weight of 5 bags

\( \overline{X} \sim N\left(1000, \frac{100^2}{5}\right) = N(1000, 2000) \)

Standard deviation: \( \sigma_{\overline{X}} = \frac{100}{\sqrt{5}} \approx 44.721 \)

Seek: \( P(\overline{X} > 950) \)

Z-score: \( z = \frac{950 – 1000}{44.721} \approx -1.118 \)

\( P(Z > -1.118) = 1 – P(Z < -1.118) \approx 1 – 0.1318 = 0.8682 \)

Round to 3 decimal places: 0.868

Method 2
Let \( T \) be the total weight of 5 bags

\( T \sim N(5 \times 1000, 5 \times 100^2) = N(5000, 50000) \)

Standard deviation: \( \sigma_T = \sqrt{5 \times 100^2} \approx 223.607 \)

\( \overline{X} > 950 \implies T > 5 \times 950 = 4750 \)

Z-score: \( z = \frac{4750 – 5000}{223.607} \approx -1.118 \)

\( P(Z > -1.118) \approx 0.8682 \approx 0.868 \)

Result: 0.868 [4]