Question

Consider babies born in the "normal" range of \(37-\) 43 weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth wcight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\) ("Are Babies Normal?" The American Statistician [1999]: \(298-302\) ). a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .)\) d. How would you characterize the most extreme \(0.1 \%\) of all birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

Short answer

a) The probability of a birth weight exceeding \(4000g\) is \(0.119\) and between \(3000 - 4000g\) is \(0.815\). b) The probability is \(0.003\) for the birth weight to be less than \(2000g\) or greater than \(5000g\). c) The probability a birth weight exceeds \(7lbs\) is \(0.701\). d) The most extreme \(0.1\%\) of all birth weights are below \(1915.44g\) or above \(4948.56g\). e) Birth weights in pounds also follows a normal distribution with mean \(7.57lbs\) and standard deviation \(1.06lbs\). Recalculating the probability for weight exceeding \(7lbs\) gives the same result as previously calculated.

Step-by-step solution

Determine Z-scores

The Z-score is a measure of how many standard deviations an element is from the mean. To calculate the Z-score for a value \(x\), the formula is \(Z = \frac{x - μ}{σ}\), where \(μ\) is the mean and \(σ\) is the standard deviation. We calculate the Z-scores for the weights given in the exercise. So, for \(4000g\), \(Z_1 = \frac{4000 - 3432}{482} = 1.18\), for \(3000g\), \(Z_2 = \frac{3000 - 3432}{482} = -0.9\), for \(2000g\), \(Z_3 = \frac{2000 - 3432}{482} = -2.97\), for \(5000g\), \(Z_4 = \frac{5000 - 3432}{482} = 3.25\), and for \(7lbs\) (converted to grams = \(3175.13g\)), \(Z_5 = \frac{3175.13 - 3432}{482} = -0.53\).

Determine probabilities

We then lookup these Z-scores in the standard normal distribution table (or use a common function from a software package) to determine the corresponding probabilities. That yields \(P(Z > 1.18) = 0.119, P( -0.9 < Z < 1.18) = 0.815, P(Z < -2.97 or Z > 3.25) = 0.003, P(Z > -0.53) = 0.701\). Use of 0.1% to find Z value in table leads to \(Z = ± 3.32\), so weight boundaries are \(3432 - 3.32×482=1915.44 g\) and \(3432 + 3.32×482=4948.56 g\).

Convert grams into pounds

To find the values in pounds for part (e), convert \(3432 g\) and \(482 g\) into pounds by dividing by \(453.59 g/lb\), which yields \(3432 / 453.59 = 7.57 lbs\) and \(482 / 453.59 = 1.06 lbs\). After the conversion, the z score for \(7lbs\) is \(Z = \frac{7 - 7.57}{1.06} = -0.53\); comparing with previous step, the result equals.

Key concepts

Z-score Calculation

The Z-score is an incredibly useful statistic that translates an individual score into a standardized form, making comparison across different datasets or populations possible. In the realm of birth weight statistics, for example, Z-scores allow us to determine how a particular baby’s weight compares to a 'normal' population average.

To calculate a Z-score, we use the formula: \[ Z = \frac{x - \mu}{\sigma} \] where \(x\) is the value we’re examining, \(\mu\) is the mean of the population, and \(\sigma\) is the standard deviation. Essentially, the Z-score tells us the number of standard deviations a value is from the mean. A Z-score of 0 indicates the value is exactly at the mean, while a positive Z-score indicates a value above the mean and a negative one indicates a value below it.

For instance, if we take the mean birth weight of 3432 grams and a standard deviation of 482 grams, the Z-score for a baby weighing 4000 grams would be calculated as \[ Z_1 = \frac{4000 - 3432}{482} = 1.18 \]. This tells us that a baby weighing 4000 grams is 1.18 standard deviations heavier than the average birth weight.

Understanding Z-scores is vital for many statistical applications, including assessing probabilities and anomalies within a given dataset.

Birth Weight Statistics

Birth weight serves as an important indicator of a newborn’s health and development. Statistics on birth weights are comprehensive, especially for infants born within the full-term gestational window of 37 to 43 weeks. In such cases, the birth weights follow a normal distribution.

A normal distribution is a bell-shaped curve where most of the observations cluster around the central peak and the probabilities for values further away from the mean taper off equally in both directions. For full-term US babies, the mean weight is typically around 3432 grams, and the spread of these weights is measured by the standard deviation of 482 grams.

Understanding these statistics helps healthcare professionals to identify atypical birth weights for full-term babies. For example, a weight below the 3rd or above the 97th percentile might be flagged for further medical evaluation. These statistics are also instrumental in health policy-making and in ensuring that the appropriate support and interventions are available for babies born outside of the 'normal' range.

Incorporating the use of standard deviation and mean in Z-score calculations, as seen in the exercise solution, allows for a precise understanding of where an individual baby's weight stands in relation to the statistical norm.

Probability Determination

Determining probabilities in a normal distribution is a crucial aspect of statistical analysis. Once we have the Z-scores, we determine the likelihood of an event occurring within the distribution. For example, we can calculate the probability of a randomly selected baby's birth weight falling within a certain range.

To find these probabilities, we use Z-score tables or statistical software. These tools give us the probability that corresponds to any given Z-score under the standard normal curve. For instance, if a baby’s Z-score for weight is 1.18, the related probability would indicate how common it is to find a baby's weight above this Z-score value in our normal distribution.

In our exercise example, with Z-scores of 1.18, -0.9, -2.97, and 3.25, we find corresponding probabilities of 0.119, 0.815, 0.003, and 0.701. These values help us understand a wide range of outcomes, such as the rarity of extremely low or high birth weights (part b) or how common it is to exceed a weight of 7 pounds (part c).

Importance of Clear Probability Interpretation

Effective communication of probability results is essential. An educator must express these figures in a manner that conveys the practical implications, such as the likelihood of certain weights and their health implications, without relying on technical jargon that may confuse learners.