91Ó°ÊÓ

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 a randomly selected pregnancy is between 250 and 300 days? b. What is the probability that the duration is at most 240 days? c. What is the probability that the duration is within 16 days of the mean duration? d. A "Dear Abby" newspaper 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 the birth of her child.) What is the probability that the duration of pregnancy is at least 310 days? Does this probability make you 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 benefits are only paid if 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 requirement?

Step by step solution

01

Identify the given information

The duration of pregnancy follows a normal distribution with a mean of 266 days and a standard deviation of 16 days. Mean (μ) = 266 days Standard deviation (σ) = 16 days

02

Find the probability of a pregnancy duration between 250 and 300 days

We need to find P(250 < x < 300). To do this, we'll first convert the values into z-scores. z1 = \( \frac{250 - 266}{16} = -1 \) z2 = \( \frac{300 - 266}{16} = 2.125 \) Now, we can use a Z-table or calculator to find the probability between -1 and 2.125. P(-1 < z < 2.125) = P(z < 2.125) - P(z < -1) Using a Z-table or calculator: P(z < 2.125) ≈ 0.9832 P(z < -1) ≈ 0.1587 P(250 < x < 300) ≈ 0.9832 - 0.1587 ≈ 0.8245

03

Find the probability of a pregnancy duration at most 240 days

We need to find P(x ≤ 240). First, convert the value into a z-score. z = \( \frac{240 - 266}{16} = -1.625 \) Now, we can use a Z-table or calculator to find the probability. P(z ≤ -1.625) ≈ 0.0521

04

Find the probability of a pregnancy duration within 16 days of the mean duration

We need to find P(250 < x < 282). First, convert the values into z-scores. z1 = \( \frac{250 - 266}{16} = -1 \) z2 = \( \frac{282 - 266}{16} = 1 \) Now, we can use a Z-table or calculator to find the probability between -1 and 1. P(-1 < z < 1) = P(z < 1) - P(z < -1) Using a Z-table or calculator: P(z < 1) ≈ 0.8413 P(z < -1) ≈ 0.1587 P(250 < x < 282) ≈ 0.8413 - 0.1587 ≈ 0.6826

05

Find the probability of a pregnancy duration at least 310 days

We need to find P(x ≥ 310). First, convert the value into a z-score. z = \( \frac{310 - 266}{16} = 2.75 \) Now, we can use a Z-table or calculator to find the probability. P(z ≥ 2.75) ≈ 1 - P(z < 2.75) ≈ 1 - 0.9930 ≈ 0.0070

06

Evaluate the insurance claim

Suppose conception occurred 2 weeks (14 days) after coverage began, x = 275 - 14 = 261 days. We need to find P(x ≤ 261). First, convert the value into a z-score. z = \( \frac{261 - 266}{16} = -0.3125 \) Now, we can use a Z-table or calculator to find the probability. P(z ≤ -0.3125) ≈ 0.3771 The insurance company will refuse to pay benefits in about 37.71% of the cases where conception occurred 2 weeks after coverage began.

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Key Concepts

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

Probability Calculations

In the world of statistics, probability calculations help us understand the likelihood of different outcomes. When we talk about the duration of pregnancy, we're interested in how often certain ranges or values occur. This is where probability comes in handy. The probability of an event tells us how likely that event is to happen.
This is especially useful when dealing with normally distributed data, like pregnancy durations, where we have a bell-shaped curve. To find probabilities of certain events or outcomes, we rely on tools such as Z-scores and Z-tables. These allow us to convert everyday data points into standardized values, making it easier to compute and understand probabilities.
In practical terms, by using probability calculations, we can answer questions like: "What are the chances of a pregnancy lasting between 250 and 300 days?" These calculations empower us to make informed decisions based on statistical evidence.

Z-scores

Z-scores are a crucial tool in statistics for understanding where a specific data point falls within a normal distribution. A Z-score tells us how many standard deviations a particular value is from the mean. This is extremely useful because it standardizes different values, allowing us to compare them directly on a one-to-one basis, even if they originally had different scales.
Calculating a Z-score involves subtracting the mean from your value of interest and then dividing by the standard deviation. The formula looks like this:

• Z = \( \frac{x - \mu}{\sigma} \)

Where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

• Z < 0 means the data point is below the mean.
• Z = 0 means it's exactly at the mean.
• Z > 0 means it's above the mean.

This standardization helps in determining the probability of a particular data point or range within a distribution, making Z-scores a powerful tool in statistical analysis.

Pregnancy Duration Statistics

Pregnancy duration is a well-researched area of statistics, as it offers insights into typical gestational periods and their variations. Statistically, the duration of a pregnancy is often modeled as a normal distribution, with an average (mean) duration and a certain variability. For many populations, the average is roughly 266 days, and the typical variability, measured by standard deviation, is around 16 days.
These statistics are important for healthcare professionals and researchers to understand natural variations and make informed decisions. Moreover, using normal distribution in this context helps deal with questions about uncommon pregnancy durations, like those under 240 days, or extraordinarily long ones, such as 310 days.
By analyzing pregnancy durations with a normal distribution lens, we can evaluate probabilities related to specific duration ranges, assess potentially suspicious claims, or make policy decisions, as exemplified by insurance scenarios. This standardized approach ensures that such assessments are consistent and based on robust statistical methods.

Standard Deviation Analysis

Standard deviation is a measure of how spread out the values in a dataset are around the mean. In simplified terms, it gives us an idea of how "wide" or "narrow" the distribution is. A small standard deviation means the values tend to be close to the mean, whereas a large standard deviation indicates the values are spread out.
For instance, in pregnancy duration statistics, a standard deviation of 16 days tells us that most pregnancy durations are within this range from the mean of 266 days. This allows us to predict that about two-thirds of pregnancies will last between 250 and 282 days, assuming a normal distribution.
Standard deviation is critical in forming the basis for more complex statistical calculations. It helps in computing Z-scores and probability calculations, enabling us to draw meaningful conclusions about datasets, such as determining how unusual a specific pregnancy duration is. Understanding how data varies informs decisions in real-world applications, like healthcare policies and insurance claims.