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Babies Weights (Example 2) Some sources report that the weights of full-term newborn babies have a mean of 7 pounds and a standard deviation of \(0.6\) pound and are Normally distributed. a. What is the probability that one newborn baby will have a weight within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds, or within one standard deviation of the mean? b. What is the probability the average of four babies' weights will be within \(0.6\) pound of the mean - that is, between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and b.

Short Answer

Expert verified
a) The probability that one newborn baby will have a weight within 0.6 pound of the mean is approximately 68%. b) The probability that the average weight of four babies will also be within 0.6 pound of the mean is the same - about 68%. c) The difference between a and b is in the scope - a considers individual weights while b is about the average of four weights, yet the resulting probabilities are the same as we are still considering a range within one standard deviation of the mean for the respective cases.

Step by step solution

01

Understanding the problem

The problem informs that newborn babies' weights have a mean of 7 pounds, the standard deviation is 0.6 pound and follows a normal distribution. It asks for the probability of a baby's weight within 0.6 pound of the mean and the same for the average weight of four babies.
02

Calculate the probability for one baby

In normal distribution, about 68% of all the values lie within one standard deviation of the mean. So, the probability that a baby's weight is between 6.4 and 7.6 pounds is approximately 0.68 or 68%.
03

Calculate the probability for the mean weight of four babies

For a sample mean, the standard deviation (also known as standard error) is the original standard deviation divided by the square root of the sample size. Here, the sample size is 4. So, the standard error will be \(0.6/\sqrt{4} = 0.3\) pound. Since we still want the weights between 6.4 and 7.6 pounds, which is within one standard deviation of the mean for the sample mean distribution, the probability will be the same as in the previous step -- 68%.
04

Explain the difference between a and b

In both cases a and b, we are looking at weights within 0.6 pound of the mean. In a, it's the weight of a single baby, and in b, it's the average weights of four babies. The difference is that case b considers an aggregated metric (average) which has a smaller variability around the mean as compared to individual measurements in case a. Consequently, the standard deviation (or error, in case of average) is smaller in case b. Yet, for this specific scenario the probability is the same because we are still considering a range within one standard deviation.

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Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics and probability theory. It describes how data is distributed around a mean. Imagine a bell-shaped curve that is symmetrical on both sides. This curve is known as the "normal distribution curve." At its center is the mean, where most of the data points cluster. The shape of the curve implies that values closer to the mean are more frequent.

The tails of the curve show that as you move further from the mean, the occurrence of data points decreases. This pattern holds true for many naturally occurring datasets, such as heights, IQ scores, and the weights of newborn babies, as in our example. with full-term newborn babies' weights, we assume a standard bell-shaped curve with a mean of 7 pounds. Therefore, understanding the normal distribution is crucial for analyzing and predicting data behavior.
Standard Deviation
Standard deviation is a measure of how spread out numbers are in a dataset. It is denoted by the symbol \( \sigma \). The standard deviation tells us, on average, how far each data point is from the mean. It provides insight into the variability or consistency within a dataset.

In the example of newborn babies' weights, a standard deviation of 0.6 pounds means that most babies' weights are within 0.6 pounds of the average weight of 7 pounds. The smaller the standard deviation, the more concentrated the data is around the mean. A larger standard deviation indicates more variation and a wider spread of weights.

This consistency is essential for determining probabilities within the context of a normal distribution. Hence, knowing the standard deviation helps estimate the likelihood of specific outcomes occurring, such as the probability of a baby's weight falling between 6.4 and 7.6 pounds.
Sample Mean
The sample mean refers to the average value of a small group, or sample, taken from a larger population. It provides an estimate of the population mean but can be influenced by the sample's size and selection. To find the sample mean, sum up all the values in the sample and divide by the number of values.

In our problem, we are interested in the average weight of four babies. To compute this sample mean, you would add up the weights of these four babies and divide by four. The sample mean helps in assessing whether the smaller group behaves similarly to the entire population.

A close relationship between the sample mean and the population mean implies that our sample is a good representation of the broader dataset. This is crucial when predicting probabilities and understanding variability in an average context.
Standard Error
The standard error measures how much the sample mean of a group is expected to vary from the true population mean. It provides an estimate of the dispersion of sample means. Often confused with standard deviation, the standard error is the standard deviation of the sample mean.

In the exercise, the sample consists of four babies. The standard error is calculated by dividing the standard deviation (0.6 pounds) by the square root of the sample size (4). Thus, the standard error in this context is \(0.6/\sqrt{4} = 0.3\) pounds.

A smaller standard error indicates that the sample mean is likely to be close to the population mean. Understanding standard error is essential when making inferences about a population based on a sample. It allows you to construct confidence intervals and test hypotheses with greater accuracy, ensuring reliable conclusions about population characteristics.

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Most popular questions from this chapter

Driving (Example 1) Drivers in Wyoming drive more miles yearly than motorists in any other state. The annual number of miles driven per licensed driver in Wyoming is 22,306 miles. Assume the standard deviation is 5500 miles. A random sample of 200 licensed drivers in Wyoming is selected and the mean number of miles driven yearly for the sample is calculated. (Source: 2017 World Almanac and Book of Facts) a. What value would we expect for the sample mean? b. What is the standard error for the sample mean?

Why Is \(n-1\) in the Sample Standard Deviation? Why do we calculate \(s\) by dividing by \(n-1\), rather than just \(n\) ? $$ s^{2}=\frac{\sum(x-\bar{x})^{2}}{n-1} $$ The reason is that if we divide by \(n-1\), then \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), the population variance. We want to show that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. The mathematical proof that this is true is beyond the scope of an introductory statistics course, but we can use an example to demonstrate that it is. First we will use a very small population that consists only of these three numbers: 1,2, and 5 . You can determine that the population standard deviation, \(\sigma\), for this population is \(1.699673\) (or about \(1.70\) ), as shown in the \(\mathrm{TI}-84\) output. So the population variance, sigma squared, \(\sigma^{2}\), is \(2.888889\) (or about 2.89). Now take all possible samples, with replacement, of size 2 from the population, and find the sample variance, \(s^{2}\), for each sample. This process is started for you in the table. Average these sample variances \(\left(s^{2}\right)\), and you should get approximately \(2.88889 .\) If you do. then you have demonstrated that \(s^{2}\) is an unbiased estimator of \(\sigma^{2}\), sigma squared. Show your work by filling in the accompanying table and show the average of \(s^{2}\).

Men's Pulse Rates (Example 10) A random sample of 25 men's resting pulse rates shows a mean of 72 beats per minute and \(\mathrm{a}\) standard deviation of 13 . a. Find a \(95 \%\) confidence interval for the population mean pulse rate for men, and report it in a sentence. You may use the table given for Exercise \(9.25\) b. Find a \(99 \%\) confidence interval. c. Which interval is wider, and why?

Choose a test for each situation: one-sample \(t\) -test, two-sample \(t\) -test, paired \(t\) -test, and no \(t\) -test. a, A random sample of students who transfered to a 4 -year university from community colleges are asked their GPAs. Our goal is to determine whether the mean GPA for transfer students is significantly different from the population mean GPA for all students at the university. b. Students observe the number of office hours posted for a random sample of tenured and a random sample of untenured professors. c. A researcher goes to the parking lot at a large grocery chain and observes whether each person is male or female and whether they return the cart to the correct spot before leaving (yes or no).

Confidence Interval Changes State whether each of the following changes would make a confidence interval wider or narrower. (Assume that nothing else changes.) a. Changing from a \(95 \%\) level of confidence to a \(90 \%\) level of confidence b. Changing from a sample size of 30 to a sample size of 20 c. Changing from a standard deviation of 3 inches to a standard deviation of \(2.5\) inches

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