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Chapter 15: Problem 24

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. The probability that the average pregnancy length for six randomly chosen women exceeds 270 days is about a. \(0.40\) b. \(0.27\) c. \(0.07\).

Short Answer

Expert verified
The probability is approximately 0.27 (option b).

Step by step solution

01

Understand the Problem

We want to find the probability that the average pregnancy length for six women exceeds 270 days. The pregnancy length follows a normal distribution with a mean (\( \mu \)) of 266 days and a standard deviation (\( \sigma \)) of 16 days.
02

Find the Standard Error

First, we need to determine the standard error of the mean for the sample of six women. The formula for the standard error (SE) is given by \( SE = \frac{\sigma}{\sqrt{n}} \), where \( n \) is the sample size. In this case, \( \sigma = 16 \) and \( n = 6 \). Therefore, \( SE = \frac{16}{\sqrt{6}} \approx 6.53 \).
03

Compute the Z-Score

Next, we need to calculate the Z-score to find out how many standard errors 270 days is from the mean of 266 days. The formula for the Z-score is \( Z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} = 270 \), \( \mu = 266 \), and \( SE \approx 6.53 \). This gives us \( Z = \frac{270 - 266}{6.53} \approx 0.61 \).
04

Find the Probability

Now, we use the standard normal distribution table (or a calculator) to find the probability of a Z-score greater than 0.61. This corresponds to the area to the right of 0.61 in the standard normal distribution.
05

Interpret the Probability

Using the standard normal distribution table, we find the probability that \( Z \leq 0.61 \) is approximately 0.7291. Hence, the probability that \( Z > 0.61 \) is \( 1 - 0.7291 = 0.2709 \). This means there is about a 27% chance that the average pregnancy length for six randomly chosen women exceeds 270 days.

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Key Concepts

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

Standard Error
When dealing with samples instead of entire populations, the standard error (SE) becomes an essential concept. It provides a measure of how much variation we can expect in the sample mean compared to the true population mean.

To calculate the standard error, you use the formula: \[SE = \frac{\sigma}{\sqrt{n}} \]where:
  • \(\sigma\) is the standard deviation of the population.
  • \(n\) is the sample size.
In our exercise with six women, the standard deviation of pregnancy length is 16 days, and our sample size is 6. By plugging these values into the formula, the standard error is computed as approximately 6.53 days.

Understanding the standard error helps us determine how much the sample mean of these six pregnancies might differ if we draw different samples. It's smaller than the population standard deviation because we're looking at the average, not individual values.
Z-Score
The Z-score is a statistical measurement that tells us how many standard deviations a data point (or an average of a sample) is from the mean of a data set.

To compute the Z-score for our scenario, we use the formula:\[Z = \frac{\bar{x} - \mu}{SE}\]where:
  • \(\bar{x}\) is the sample mean we're interested in (270 days in this case).
  • \(\mu\) is the population mean (266 days).
  • \(SE\) is the standard error we calculated earlier (approximately 6.53).
By inserting these values, we find the Z-score to be around 0.61.

This result indicates that 270 days is 0.61 standard errors above the mean of 266 days. The calculation helps us translate a raw score into a standardized format, which is crucial for determining probabilities.
Probability
The concept of probability is key to interpreting statistical results. Probability enables us to understand the likelihood of a particular outcome, here the average pregnancy length for six women exceeding 270 days.

In our exercise, once we have the Z-score (0.61), we use it to find the corresponding probability from the standard normal distribution table. The table provides the likelihood that a value is less than our Z-score.

For a Z-score of 0.61, that probability is approximately 0.7291. This means there's about a 72.91% chance that a sampled mean is less than 270 days. Consequently, the probability that the average pregnancy length exceeds 270 days is the complement: \[1 - 0.7291 = 0.2709\]or about 27%.

Understanding how to turn a Z-score into a probability allows us to interpret what we see in our research context, making probability a powerful tool for statistical analysis.

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Most popular questions from this chapter

Runners. In a study of exercise, a large group of male runners walk on a treadmill for six minutes. After this exercise, their heart rates vary with mean \(8.8\) beats per five seconds and standard deviation \(1.0\) beats per five seconds. The researcher records the number of heartbeats per five seconds for each runner over a period of time. This distribution takes only whole-number values, so it is certainly not Normal. a. Let \(x\) be the mean number of beats per five seconds after measuring heart rate for 24 five-second intervals (two minutes). What is the approximate distribution of \(x\) according to the central limit theorem? b. What is the approximate probability that \(x\) is less than 8 ? c. What is the approximate probability that the heart rate of a runner is less than 100 beats per minute? (Hint: Restate this event in terms of \(x .)\)

Detecting the Emerald Ash Borer. The emerald ash borer is a serious threat to ash trees. A state agriculture department places traps throughout the state to detect the emerald ash borer. When traps are checked periodically, the mean number of ash borers trapped is only \(2.2\), but some traps have many ash borers. The distribution of ash borer counts is finite and strongly skewed, with standard deviation \(3.9 .\) a. What are the mean and standard deviation of the average number of ash borers \(x\) in 50 traps? b. Use the central limit theorem to find the probability that the average number of ash borers in 50 traps is greater than 3.0.

Testing Glass. How well materials conduct heat matters when designing houses. As a test of a new measurement process, 10 measurements are made on pieces of glass known to have conductivity 1 . The average of the 10 measurements is 1.07. For each of the boldface numbers, indicate whether it is a parameter or a statistic. Explain your answer.

A newborn baby has extremely low birth weight (ELBW) if it weighs less than 1000 grams. A study of the health of such children in later years examined a random sample of 219 children who had been born with ELBW. Their mean weight at birth was \(x=810\) grams. This sample mean is an unbiased estimator of the mean weight \(\mu\) in the population of all ELBW babies. This means that a. in many samples from this population, the mean of the many values of \(x\) will be equal to \(\mu\). b. as we take larger and larger samples from this population, \(x\) will get closer and closer to \(\mu\). c. in many samples from this population, the many values of \(x\) will have a distribution that is close to Normal.

Glucose Testing. Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood glucose levels during pregnancy). There is variation both in the actual glucose level and in the blood test that measures the level. In a test to screen for gestational diabetes, a patient is classified as needing further testing for gestational diabetes if the glucose level is above 130 milligrams per deciliter ( \(\mathrm{mg} / \mathrm{dL}\) ) one hour after having a sugary drink. Shelia's measured glucose level one hour after the sugary drink varies according to the Normal distribution with \(\mu=122 \mathrm{mg} / \mathrm{dL}\) and \(\sigma=12 \mathrm{mg} / \mathrm{dL}\). a. If a single glucose measurement is made, what is the probability of Shelia being diagnosed as needing further testing for gestational diabetes? b. If measurements are made on four separate days and the mean result is compared with the criterion 130 \(\mathrm{mg} / \mathrm{dL}\), what is the probability that Shelia is diagnosed as needing further testing for gestational diabetes?
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