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Chapter 6: Problem 31

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days. a. In a letter to "Dear Abby," a wife claimed to have given birth 308 days after a brief visit from her husband, who was working in another country. Find the probability of a pregnancy lasting 308 days or longer. What does the result suggest? b. If we stipulate that a baby is premature if the duration of pregnancy is in the lowest \(3 \%\) find the duration that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care.

Short Answer

Expert verified
a. Probability: 0.37%. This result suggests the claim is highly unlikely.b. Premature if < 240 days.

Step by step solution

01

Understand the problem

The problem is about finding probabilities and boundaries in a normal distribution. The given distribution has a mean (\(\text{\mu}\)) of 268 days and a standard deviation (\(\text{\sigma}\)) of 15 days.
02

Standardize the value for part (a)

Convert the pregnancy length of 308 days to a z-score using the formula \( z = \frac{x - \mu}{\sigma} \). Here, x = 308 days, \( \mu = 268 \) days, and \( \sigma = 15 \) days. Therefore, \( z = \frac{308 - 268}{15} = \frac{40}{15} \approx 2.67 \).
03

Find the probability for part (a)

Using the standard normal distribution table or a calculator, find the probability corresponding to z = 2.67. The cumulative probability for z = 2.67 is approximately 0.9963. The probability of a pregnancy lasting 308 days or longer is \(1 - 0.9963 = 0.0037 \) (or 0.37%).
04

Interpret the result for part (a)

Since the probability of a pregnancy lasting 308 days or longer is extremely low (0.37%), it suggests that the claim is highly unlikely and should be questioned.
05

Find the z-score for the lowest 3% for part (b)

To find the duration separating premature babies, locate the z-score corresponding to the lowest 3% in the standard normal distribution table. The z-score for the 3rd percentile is approximately -1.88.
06

Convert the z-score back to the original value for part (b)

Use the z-score formula in reverse to find the threshold duration: \( x = \mu + z \cdot \sigma \). Substituting the values, we get \( x = 268 + (-1.88) \cdot 15 = 268 - 28.2 \approx 240 \) days.
07

Conclusion for part (b)

A baby is considered premature if the pregnancy lasts less than approximately 240 days. Hospital administrators can use this to plan special care for such babies.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Standard Deviation
In statistics, the mean (often symbolized as \(\text{\textmu}\)) represents the average of a data set. For pregnancy durations, the mean is the typical length of pregnancy. For our exercise, the mean is 268 days. The standard deviation (\(\text{\textsigma}\)) measures how spread out the numbers in the data set are from the mean. Here, the standard deviation is 15 days. This tells us that most pregnancy durations fall within 15 days of the mean (either above or below). Understanding these two values is crucial because they help us gauge the distribution's center and spread, key to making probability calculations in normal distributions.
Z-score Calculation
A z-score helps us understand how far a specific value is from the mean, in terms of standard deviations. It can be calculated using the formula: \(z = \frac{x - \text{\textmu}}{\text{\textsigma}}\). For instance, in our problem, we want to find the z-score for a pregnancy lasting 308 days. Plugging in our values: \(z = \frac{308 - 268}{15}\), we get a z-score of approximately 2.67. This score tells us that 308 days is 2.67 standard deviations above the mean.
Probability in Normal Distribution
Normal distributions are symmetric and bell-shaped, showing that most values cluster around the mean. Using z-scores and standard normal distribution tables, we can determine probabilities. In our exercise, we found a z-score of 2.67 for a 308-day pregnancy. By looking it up, we see the cumulative probability is about 0.9963. This means there is a 99.63% chance that a pregnancy length is less than 308 days. Thus, the probability of a pregnancy lasting 308 days or longer is \(1 - 0.9963 \) or 0.37%. This low probability suggests it's unlikely for a pregnancy to last this long.
Percentiles in Normal Distribution
Percentiles indicate the value below which a certain percentage of data falls. In our exercise, we were asked to find the pregnancy duration that separates the lowest 3% of durations (premature babies) from the rest. By looking up the 3rd percentile in the standard normal distribution table, we find a z-score of about -1.88. Converting this z-score back to a pregnancy duration using \(x = \text{\textmu} + z \cdot \text{\textsigma}\), we get \(x = 268 + (-1.88) \cdot \ 15\) which equals approximately 240 days. So, pregnancies lasting fewer than 240 days are considered premature.
Interpretation of Statistical Results
Understanding statistical results is vital. For part (a) of our exercise, a 308-day pregnancy has a probability of only 0.37%, suggesting it's highly unusual, and the claim should be questioned. For part (b), knowing that pregnancies under 240 days are in the lowest 3% (premature) helps administrators prepare adequately. These findings support decision-making and improved readiness in healthcare settings. Emphasizing the low probability for part (a) helps understand rare occurrences, while part (b) highlights the importance of early identification for special medical care.

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