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Chapter 15: Problem 23

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
(b) 0.27

Step by step solution

01

Understand the Problem

We need to find the probability that the average pregnancy length for six women exceeds 270 days. Given is a normal distribution with mean 266 days and standard deviation 16 days.
02

Calculate the Standard Error

The standard error (SE) is used here since we are dealing with a sample mean of six women. The formula for SE is \( \text{SE} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = 16 \) and \( n = 6 \). Thus, \( \text{SE} = \frac{16}{\sqrt{6}} \approx 6.53 \).
03

Calculate the Z-score

The Z-score is calculated by using the formula \( Z = \frac{\bar{x} - \mu}{\text{SE}} \), where \( \bar{x} = 270 \), \( \mu = 266 \), and \( \text{SE} \approx 6.53 \). Therefore, \( Z \approx \frac{270 - 266}{6.53} \approx 0.61 \).
04

Find the Probability

Using the Z-table, find the probability of a Z-score greater than 0.61. The table gives a probability of 0.7291 for Z being less than 0.61. Thus, the complement probability is \( 1 - 0.7291 = 0.2709 \).
05

Choose the Nearest Option

From the calculated probability of 0.2709, we match this closest to the options given. Thus, the correct choice is (b) 0.27.

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Key Concepts

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

Standard Error
The standard error (SE) is a crucial concept when dealing with the means of samples rather than individual data points. It provides a measure of how much the sample mean is expected to vary from the population mean. In the context of our exercise, we are interested in understanding how the average pregnancy length for six women might differ from the overall average pregnancy length.
The SE is calculated using the formula:\[\text{SE} = \frac{\sigma}{\sqrt{n}}\]
  • \( \sigma \) (sigma) is the standard deviation of the population.
  • \( n \) is the sample size.
In our case, \( \sigma = 16 \) days and \( n = 6 \). By plugging the values into the formula, we find:\[\text{SE} = \frac{16}{\sqrt{6}} \approx 6.53\]This SE tells us how much the average length of pregnancies for a sample of six women is expected to fluctuate around the population mean of 266 days.
Z-score
The Z-score is a statistical measure that describes the position of a raw score in terms of its distance from the mean, measured in standard deviations. In simpler terms, it tells us how far away a data point is from the average, and whether it is above or below the average.
In our exercise, we are looking at whether the average pregnancy length for six women is longer than the population average. To calculate the Z-score, use the formula:\[Z = \frac{\bar{x} - \mu}{\text{SE}}\]
  • \( \bar{x} \) is the sample mean we are assessing, which in this case is 270 days.
  • \( \mu \) is the population mean (266 days).
  • \( \text{SE} \) is the standard error (\( \approx 6.53 \) days, as we calculated previously).
Substituting in these values:\[Z = \frac{270 - 266}{6.53} \approx 0.61\]A Z-score of 0.61 means the sample mean is 0.61 standard deviations above the population mean. This is a moderate distance, suggesting a probability calculation is needed for precise interpretation.
Probability Calculation
Once we have calculated the Z-score, we can move on to finding the probability that a sample mean lies beyond a certain value. A Z-score translates to a probability through the use of a Z-table, which reflects the cumulative probability of a Z-score being less than a specified value in a standard normal distribution.
For a Z-score of 0.61, the Z-table tells us that the cumulative probability of a score being less than 0.61 is approximately 0.7291. However, our interest lies in the probability that the mean pregnancy length exceeds 270 days (i.e., a Z-score greater than 0.61).
To find this probability, we calculate the complement:
  • The complement probability is found by subtracting the cumulative probability from 1.
Thus, we calculate:\[1 - 0.7291 = 0.2709\]This value, 0.2709, tells us that there is about a 27% chance that the average pregnancy length for six randomly selected women will exceed 270 days. This probability helps in deciding if a sample's mean is unusually high, given the variation in the data.

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Most popular questions from this chapter

Florida Voters. Florida played a key role in recent presidential elections. Voter registration records in August 2016 show that \(38 \%\) of Florida voters are registered as Democrats and \(36 \%\) as Republicans. (Most of the others did not choose a party.) In September 2016 you wsed a random digit dialing device to poll voters for the 2016 presidential elections. You used it to call 250 randomly chosen residential telephones in Florida. Of the registered voters contacted, \(35 \%\) are registered Democrats. Is each of the boldface numbers a parameter or a statistic?

Glucose testing. Shelia's doctor is concerned that she may suffer from gestational diabetes (high blood ghucose 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 (mg/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 that Shelia is 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?

Insurance. The idea of insurance is that we all face risks that are unlikely but carry high cost. Think of a fire or flood destroying your apartment. Insurance spreads the risk: we all pay a small amount, and the insurance policy pays a large amount to those few of us whose apartments are damaged. An insurance company looks at the records for millions of apartment owners and sees that the mean loss from apartment damage in a year is \(\mu=130\) per person. (Most of us have no loss, but a few lose most of their possessions. The \(\$ 130\) is the average loss.) The company plans to sell renter's insurance for \(\$ 130\) plus enough to cover its costs and profit. Explain clearly why it would be unwise to sell only 10 policies. Then explain why selling thousands of such policies is a safe business.

Sampling students, continued. To estimate the mean score \(\mu\) of those who took the Medical College Admission Test on your campus, you will obtain the scores of an SRS of students. From published information, you know that the scores are approximately Normal with standard deviation about 10.4. You want your sample mean \(x^{-} \nu\) to estimate \(\mu\) with an error of no more than one point in either direction. (a) What standard deviation must \(x^{-} \bar{x}\) have so that \(99.7 \%\) of all samples give an \(x^{-} \bar{x}\) within one point of \(\mu\) ? (Use the 68-95-99.7 rule.) (b) How large an SRS do you need in order to reduce the standard deviation of \(x^{-} \bar{x}\) to the value you found in part (a)?

Larger Sample, More Accurate Estimate. Suppose that, in fact, the total cholesterol level of all men aged 20-34 follows the Normal distribution with mean \(\mu=182\) milligrams per deciliter (mg/dL) and standard deviation \(\sigma=37\) \(\mathrm{mg} / \mathrm{dL}\). (a) Choose an SRS of 100 men from this population. What is the sampling distribution of \(\mathrm{x}^{-} \vec{x}\) ? What is the probability that \(\mathrm{x}^{-} \vec{x}\) takes a value between 180 and \(184 \mathrm{mg} / \mathrm{dL}\) ? This is the probability that \(\mathrm{x}^{-} \bar{x}\) estimates \(\mu\) within \(\pm 2\) \(\mathrm{mg} / \mathrm{dL}\). (b) Choose an SRS of 1000 men from this population. Now what is the probability that \(\mathrm{x}^{-}=\)falls within \(\pm 2 \mathrm{mg} / \mathrm{dL}\) of \(\mu\) ? The larger sample is much more likely to give an accurate estimate of \(\mu\).
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