91Ó°ÊÓ

Among the many letters sent to a popular adviceto-the-lovelorn columnist, was one involving a paternity issue that raised an interesting statistical question. The distraught writer-call her "San Diego Reader" - said her husband is in the military and that she got pregnant the last day before he left for an extended tour of duty. Ten months and four days later the baby was born - usually a happy occasion - but her husband, accustomed to pregnancies being nine months long, became obsessed with the possibility that he might not be the child's father. DNA testing was not yet available. The only relevant information known at the time was that pregnancy durations are normally distributed with a mean ( \(\mu\) ) of 266 days and a standard deviation \((\sigma)\) of 16 days. For the benefit of San Diego Reader's husband, how would you associate a probability with a pregnancy lasting 10 months and 4 days? Do you think San Diego Reader is telling the truth?

Step by step solution

01

Convert the given time into days

We start by converting 10 months and 4 days into days. As a month is anthropically assumed to bear 30.44 days, this sums up to 10*30.44 days + 4 days = 304.4 days ≈ 304 days. Hence, the duration of interest is 304 days.

02

Calculate the Z-Score

Next, we calculate the z-score, which gives us the number of standard deviations the given duration is from the mean. The Z-score is calculated by the formula \( Z = \frac {X - \mu} {\sigma} \), where X is the observation (304 days), \( \mu \) is the mean (266 days) and \( \sigma \) is the standard deviation (16 days). Hence, \( Z = \frac {304 - 266} {16} ≈ 2.375 \). This indicates the given duration is 2.375 standard deviations above the mean.

03

Find the Probability

We find the probability corresponding to the Z-score from the standard normal distribution table or using computational software. The Z-score 2.375 corresponds to a probability (to the left of the z-value) of approximately 0.991. However, we are interested in the probability to the right of the z-value (because the question asks for the likelihood of a pregnancy duration of 304 days or more), which is \( 1 - 0.991 ≈ 0.009 \) or 0.9%.

04

Conclusion

Considering a probability of 0.9%, it can be concluded that there's a small but existent chance for pregnancies to last 304 days or longer. This does not directly imply dishonesty from the San Diego Reader, though it is a rare situation. Our analysis implies that the Reader's claim is far from impossible, though improbable.

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Key Concepts

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

Normal Distribution

The concept of normal distribution is core to understanding various phenomena in nature, including the duration of pregnancy. It represents a bell-shaped curve where the bulk of the data points are clustered around the mean, with fewer and fewer occurrences as you move away from the mean on either side.

Mathematically, a normal distribution is symmetric about the mean, ensuring that the mean, median, and mode of the distribution are equal. It is completely described by its mean \(\mu\) and standard deviation \(\sigma\). In the context of pregnancy durations, the mean represents the average length of a pregnancy, and the standard deviation indicates the variability of pregnancy lengths around this average.

The normal distribution helps us calculate the probability of observing a specific range of values within a dataset. For example, in our exercise, we use it to assess if a pregnancy duration significantly deviates from the norm.

Z-score Calculation

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It is a useful way to find out how unusual or usual a data point is within a data set.

To calculate a Z-score, the formula \( Z = \frac {X - \mu} {\sigma} \) is used. In this formula, \(X\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. If a Z-score is 0, it indicates that the data point's score is identical to the mean score. A positive Z-score says the data point is above the mean, and a negative Z-score indicates it is below the mean.

In the pregnancy duration problem, the Z-score tells us how far, in standard deviation units, the pregnancy duration is from the average pregnancy duration.

Standard Deviation

Standard deviation is a widely used measure of variability or diversity used in statistics and probability theory. It shows how much variation or dispersion from the average exists.

The formula for standard deviation is \(\sigma = \sqrt{\frac{1}{N}\sum_{i=1}^{N}(x_{i} - \mu)^2}\), where \(N\) is the number of observations, \(x_{i}\) are the observations themselves, and \(\mu\) is the mean of the observations. In simpler terms, it determines on average how much individual measurements differ from the mean value of the set.

In our exercise, the standard deviation helps us understand the distribution of pregnancy durations around the average and is key to interpreting the Z-score and the overall probability.

Probability

Probability is a statistical measure that quantifies the chance of a certain event occurring. It ranges from 0 to 1, where 0 indicates impossibility and 1 indicates certainty.

In our exercise, we used probability to determine how likely it is for a pregnancy to last 304 days or more. The standard normal distribution table or computational software can provide us with this probability once we have the Z-score. Typically, a high Z-score translates to a low probability of the event occurring by chance, which in the San Diego Reader's case suggests that a pregnancy lasting 304 days is quite unusual, but not impossible, as the calculated probability is 0.9%.