Question

Let \(x\) denote the duration of a randomly selected pregnancy (the time elapsed between conception and birth). Accepted values for the mean value and standard deviation of \(x\) are 266 days and 16 days, respectively. Suppose that the probability distribution of \(x\) is (approximately) normal. a. What is the probability that the duration of pregnancy is between 250 and 300 days? b. What is the probability that the duration of pregnancy is at most 240 days? c. What is the probability that the duration of pregnancy is within 16 days of the mean duration? d. A "Dear Abby" column dated January 20,1973, contained a letter from a woman who stated that the duration of her pregnancy was exactly 310 days. (She wrote that the last visit with her husband, who was in the navy, occurred 310 days before birth.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you a bit skeptical of the claim? e. Some insurance companies will pay the medical expenses associated with childbirth only if the insurance has been in effect for more than 9 months ( 275 days). This restriction is designed to ensure that the insurance company pays benefits for only those pregnancies for which conception occurred during coverage. Suppose that conception occurred 2 weeks after coverage began. What is the probability that the insurance company will refuse to pay benefits because of the 275 -day insurance requirement?

Short answer

a. The probability that the duration of pregnancy is between 250 and 300 days is approximately 0.8186. b. The probability that the duration of pregnancy is at most 240 days is approximately 0.0520. c. The probability that the duration of pregnancy is within 16 days of the mean duration is approximately 0.6827. d. The probability that the duration of pregnancy is at least 310 days is approximately 0.0030. e. The probability that the insurance company will refuse to pay benefits because of the 275-day insurance requirement is approximately 0.2869.

Step-by-step solution

Understand the Problem and Identify the Given Elements

For this problem, the mean, \(\mu\), is 266 days and the standard deviation, \(\sigma\), is 16 days. The duration of pregnancy, \(x\), is assumed to be distributed normally.

Calculate Probability for 250 to 300 days

First, convert 250 days and 300 days to z-scores using formula \(z = (x - \mu) / \sigma\), which gives \(z_{250} = (250-266) / 16 = -1\) and \(z_{300} = (300-266) / 16 = 2.125\). The area between these two z-values on the standard normal distribution corresponds to the probability that \(x\) is between 250 and 300 days. Using statistical tables or software, this probability is approximately 0.8186.

Calculate Probability for at most 240 days

First, convert 240 days to a z-score, getting \(z_{240} = (240-266) / 16 = -1.625\). The area to left of this z-value corresponds to the probability that \(x\) is at most 240 days. Using statistical tables or software, this probability is approximately 0.0520.

Calculate Probability for within 16 days of mean

Here, \(x\) is within the range of 250 days (266 - 16) to 282 days (266 + 16). Converting these values to z-scores, we get \(z_{250} = -1\) and \(z_{282} = 1\). The probability that \(x\) is within this range can be found as the area between these two z-values, which is approximately 0.6827.

Calculate Probability for at least 310 days

First, convert 310 days to a z-score, getting \(z_{310} = (310-266) / 16 = 2.75\). The area to right of this z-value corresponds to the probability that \(x\) is at least 310 days. Using statistical tables or software, this probability is approximately 0.0030. This very small probability suggests that a pregnancy lasting exactly 310 days is quite rare, making the claim somewhat dubious.

Calculate Probability for more than 275 days

The converting 275 days to a z-score gives \(z_{275} = (275-266) / 16 = 0.5625\). The area to right of this z-value represents the probability that the insurance company will not pay, which is approximately 0.2869. Thus, if conception occurs two weeks after coverage begins, there is about a 28.69% chance the insurance company will refuse to pay due to the 275-day requirement.

Key concepts

Z-Score Calculations

Z-score calculations are pivotal in statistics to determine how far or how close a particular value is from the mean of a data set. They are expressed in terms of standard deviations. A z-score, also known as a standard score, provides insight into the position of a data point within a distribution and is calculated using the formula:

\[ z = \frac{{x - \mu}}{{\sigma}} \]
Here, \(x\) is the value of interest, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. For instance, in the context of pregnancy durations, to determine how unusual a 250-day pregnancy is, we would calculate its z-score by substituting the actual days for \(x\), the average duration of 266 days for \(\mu\), and 16 days for \(\sigma\). A negative z-score indicates that the value is below the mean, while a positive one suggests it is above the mean. Understanding z-score calculations is crucial to interpreting probability in a normal distribution context.

Standard Normal Distribution

When we talk about the standard normal distribution, we're referring to a special case of the normal distribution where the mean \(\mu\) is 0 and the standard deviation \(\sigma\) is 1. This standardized version is used to compare data across different scales by converting raw scores into z-scores.

The standard normal distribution is symmetrical and its total area under the curve equals 1, signifying the total probability. What's convenient about it is the universality of its tables or software, which enables us to quickly find the probability associated with any z-score. In practical terms, for the textbook exercise, after converting the days to z-scores, we can find the probability of pregnancy durations falling within certain ranges by looking up these z-scores in a standard normal distribution table or by using statistical software. It's a powerful tool for statistical analysis.

Probability Distribution

Probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. In the context of our exercise, the normal distribution, which is a type of probability distribution, is used to model the duration of pregnancies.

The characteristics of a normal distribution include its bell-shaped curve, mean equals median, and it is dictated by two parameters: the mean and the standard deviation. The area under the curve of a normal distribution signifies probability and is useful for calculating the likelihood of events, such as the chance that a pregnancy will last between a certain range of days. In the exercise, various probabilities are computed by finding the corresponding area under the normal curve, highlighting the practical application of probability distribution in real-world scenarios.

Statistical Significance

Statistical significance plays a central role in hypothesis testing and indicates whether the results from a study or an experiment are likely due to something other than random chance. It's determined by a p-value, which is the probability that the observed data would occur if the null hypothesis were true. In the exercise, assessing the statistical significance of the claim that a pregnancy lasted exactly 310 days involves computing a p-value and comparing it to a pre-determined significance level, often denoted as \(\alpha\).

Generally, if the p-value is less than \(\alpha\), which is commonly set at 0.05, the result is deemed statistically significant, indicating that the observed outcome is unlikely to have occurred by chance alone. As the probability associated with a 310-day pregnancy is very low, this would lead us to question the veracity of the claim, demonstrating the practical utility of understanding statistical significance in evaluating real-world data.