The birth weight of full-term babies is known to be normally distributed (a bell shape) with a mean of 3400 grams and a standard deviation of 505 grams. Use the empirical rule to answer the following: Approximately 95% of the full-term babies will fall between what weights? Approximately what percent of the full-term babies will weigh more than 4410 grams (about 9 pounds 11 1/2 oz)?
The Correct Answer and Explanation is:
Correct Answers:
- Approximately 95% of full-term babies will weigh between 2390 grams and 4410 grams.
- Approximately 2.5% of full-term babies will weigh more than 4410 grams.
Explanation (without using the word “we”):
The empirical rule, also called the 68-95-99.7 rule, provides a quick estimate of the spread of data in a normal distribution. According to this rule:
- About 68% of the data falls within 1 standard deviation of the mean.
- About 95% of the data falls within 2 standard deviations of the mean.
- About 99.7% of the data falls within 3 standard deviations of the mean.
The mean birth weight of full-term babies is 3400 grams, and the standard deviation is 505 grams.
To find the range where approximately 95% of the birth weights fall:
- Calculate 2 standard deviations below and above the mean.
- Lower bound: 3400 − (2 × 505) = 3400 − 1010 = 2390 grams
- Upper bound: 3400 + (2 × 505) = 3400 + 1010 = 4410 grams
So, approximately 95% of full-term babies will have birth weights between 2390 grams and 4410 grams.
Next, to determine the percentage of babies who weigh more than 4410 grams, consider that 4410 grams is exactly 2 standard deviations above the mean. In a normal distribution, about 95% of the data lies between −2σ and +2σ, which leaves 5% in the two tails of the distribution (outside that range). These two tails are split evenly:
- About 2.5% lie above the +2σ mark.
- About 2.5% lie below the −2σ mark.
Therefore, approximately 2.5% of full-term babies will weigh more than 4410 grams, which is equivalent to about 9 pounds 11½ ounces.
