Questions: The birth weights for twins are normally distributed with a mean of 2353 grams and a standard deviation of 647 grams. Use z-scores to determine which birth weight could be considered unusual. A. 3647 g B. 1200 g C. 2000 g D. 2353 g

Solution Steps

To determine which birth weights are considered unusual, we calculate the z-scores for each weight using the formula:

\[ z = \frac{X - \mu}{\sigma} \]

where:

• \( X \) is the birth weight,
• \( \mu = 2353 \) grams (mean),
• \( \sigma = 647 \) grams (standard deviation).

1. For \( X = 3647 \) g: \[ z = \frac{3647 - 2353}{647} = \frac{1294}{647} \approx 2.0 \] The birth weight \( 3647 \) g has a z-score of \( 2.0 \), which is not considered unusual.

2. For \( X = 1200 \) g: \[ z = \frac{1200 - 2353}{647} = \frac{-1153}{647} \approx -1.7821 \] The birth weight \( 1200 \) g has a z-score of \( -1.7821 \), which is not considered unusual.

3. For \( X = 2000 \) g: \[ z = \frac{2000 - 2353}{647} = \frac{-353}{647} \approx -0.5456 \] The birth weight \( 2000 \) g has a z-score of \( -0.5456 \), which is not considered unusual.

4. For \( X = 2353 \) g: \[ z = \frac{2353 - 2353}{647} = \frac{0}{647} = 0.0 \] The birth weight \( 2353 \) g has a z-score of \( 0.0 \), which is not considered unusual.

Final Answer