| Problem | Part | Solution |
|---|---|---|
| 1 | 1 | The parent population is normally distributed, so the sample mean is automatically normally distributed. |
| 1 | 2 | The sample size is large, and the Central Limit Theorem implies that the sample mean is normally distributed. |
| 2 | - | z = (value - mean)/standard deviation. \(Z = \frac{\bar{x}-\mu}{\sigma/\sqrt{n}}\) |
| 3 | A | About 68% using the 68, 95, 99.7 rule or 0.6827 ‘exact’ |
| 3 | B | 40\(^{th}\)percentile = 8.7333 |
| 3 | C | \(Z = 1.7\) |
| 3 | D | \(Z = 0.4\) so probability = 0.3446 |
| 3 | E | \(z(\bar{x}<8)= -0.4\text{ probability} = 0.3446\) \(z(\bar{x}>12)=0.4\text{ probability} = 0.3446\) \(0.3446+0.3446=0.6892\) |
| 4 | - | Normal |
| 5 | - | Normal |
| 6 | - | About 16% using the 68, 95, 99.7 rule or 0.1587 ‘exact’ |
| 7 | - | About 95% using the 68, 95, 99.7 rule or 0.9545 ‘exact’ |
| 8 | - | About 95% using the 68, 95, 99.7 rule or 0.9545 ‘exact’ |
| 9 | - | \(\text{Probability}(\bar{x}>50)= 0.00621\) |
| 10 | - | Right Skewed |
| 11 | - | Approximately Normal |
| 12 | - | Central Limit Theorem |
| 13 | - | No, the distribution is not normal, and the normal probability applet is only for normal distribution. |
| 14 | - | \(\text{Probability} = 0.9599\) |
| 15 | - | \(\text{ probability}(\bar{x}<37.5) = 0.1056\) \(\text{ probability}(\bar{x}>42.5) = 0.1056\) \(1-0.1056-0.1056=0.7887\) |