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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; will be between \(6.4\) and \(7.6\) pounds? c. Explain the difference between a and \(\mathrm{b}\).

Short Answer

Expert verified
a. The probability that one newborn weight falls within 0.6 pound of the mean is approximately 0.68 or 68%. b. The probability that the average of four babies' weights will fall within 0.6 pound of the mean is approximately 0.95 or 95%. c. For a single baby, the probability of its weight being within 0.6 pound of the mean is less than the probability of the average weight of four babies being within 0.6 pound of the mean. This is because the average of four weights is more likely to be close to the true mean due to the Central Limit Theorem.

Step by step solution

01

Establish the Distribution

The data is normally distributed, which means it follows the properties of a normal distribution. In a normal distribution, 68% of data falls within one standard deviation of the mean.
02

Probability Calculation for Part a

To solve part a, which asks for the probability that one newborn baby will have a weight within 0.6 pound of the mean, we use the fact that, for a normal distribution, about 68% of observations are within one standard deviation of the mean. So the answer here is about 0.68 or 68%.
03

Establish the Average for Multiple Observations

For part b, we need to find the probability that the average weights of four babies will fall within 0.6 pound of the mean. However, when considering the average weights of four babies, the distribution's standard deviation decreases by the square root of the number of babies (n). Therefore the new standard deviation is \(0.6 / \sqrt{4} = 0.3\)
04

Probability Calculation for Part b

Since the range 6.4 to 7.6 pounds is two standard deviations away from the mean (7 pounds), we can use the 95% rule of the normal distribution (which states that about 95% of the observations lie within two standard deviations of the mean). The probability is therefore approximately 0.95 or 95%.
05

Differentiate Between a and b

In step c, we're asked to explain the difference between a and b. The key difference lies in the fact that in question a, we're dealing with the weight of a single baby, while in question b, we're dealing with the average weight of four babies. Due to the Central Limit Theorem, the distribution of averages tends to have a smaller standard deviation compared to the distribution of individual observations, thus making the collected data closer to the true mean.

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Key Concepts

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

Central Limit Theorem
The Central Limit Theorem (CLT) is a fundamental principle in statistics that provides great power in predicting outcomes from sample data. This theorem states that if you have a group of independent random variables, their averages will form a normal distribution as long as the sample size is large enough.
Even if the original set of data isn't perfectly normal, the average of a series of samples from that data will approximate a normal distribution. It forms the basis for understanding how sample means behave and is crucial for making inferences about population parameters from sample statistics.
  • For example, when calculating the average weight of four babies, the CLT tells us that the distribution of these averages is normal even if baby weights themselves were not normally distributed.
  • This allows us to calculate probabilities about sample means more easily, such as in part b of the exercise where we look at the average weight of a group of babies.
The application of the CLT helps us reduce the complexity of data analysis and make predictions about larger populations.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It is a crucial concept when dealing with any kind of data analysis, especially with normal distributions. The standard deviation tells us how much the individual data points typically deviate from the mean.
In our exercise, the standard deviation is given as 0.6 pounds for the weights of newborn babies. This means most baby weights will fall within 0.6 pounds above or below the mean value of 7 pounds.
  • This aligns with the empirical rule for normal distributions, where roughly 68% of values lie within one standard deviation of the mean.
  • When looking at the average weights of multiple babies, we use a "standard deviation of the mean" which is calculated by dividing the original standard deviation by the square root of the sample size. This results in reduced variation among averages, making them less spread out and closer to the population mean.
This concept is essential in comparing individual occurrences to a wider trend in data.
Probability Calculation
The calculations of probability, especially within normal distributions, provide insights into how likely an event is to occur. In the context of normally distributed data, probabilities can be deduced using the mean and standard deviation.
In part a of our exercise, we calculate the probability that a single baby's weight will lie within one standard deviation from the mean. According to the properties of a normal distribution, this probability is approximately 68%. This is derived from the empirical rule which applies to normal distributions, stating that 68% of data falls within one standard deviation of the mean.
For part b, we calculate the probability that the average weight of four babies falls within one altered standard deviation (0.3 pounds in this case) from the mean. Here the probability increases to about 95% because the reduced deviation means the data points are more concentrated around the mean.
Understanding probability in this way helps to interpret real-world data and predict outcomes more accurately.

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Most popular questions from this chapter

Tomatoes Use the data from Exercise \(9.36\). a. Using the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 5 pounds using a significance level of \(0.05\) ? Why or why not? The confidence interval is reported here: \(\mathrm{I}\) am \(95 \%\) confident the population mean is between \(4.9\) and \(5.3\) pounds. b. Now test the hypothesis that the population mean is not 5 pounds using the four step procedure. Use a significance level of \(0.05\) and number your steps.

Potatoes Use the data from Exercise \(9.35\). a. If you use the four-step procedure with a two-sided alternative hypothesis, should you be able to reject the hypothesis that the population mean is 20 pounds using a significance level of \(0.05 ?\) Why or why not? The confidence interval is reported here: I am \(95 \%\) confident that the population mean is between \(20.4\) and \(21.7\) pounds. b. Now test the hypothesis that the population mean is not 20 pounds using the four-step procedure. Use a significance level of \(0.05\). c. Choose one of the following conclusions: i. We cannot reject a population mean of 20 pounds. ii. We can reject a population mean of 20 pounds. iii. The population mean is \(21.05\) pounds.

Carrots The weights of four randomly chosen bags of horse carrots, each bag labeled 20 pounds, were \(20.5,19.8,20.8\), and \(20.0\) pounds. Assume that the distribution of weights is Normal. Find a \(95 \%\) confidence interval for the mean weight of all bags of horse carrots. Use technology for your calculations. a. Decide whether each of the following three statements is a correctly worded interpretation of the confidence interval, and fill in the blanks for the correct option(s). i. \(95 \%\) of all sample means based on samples of the same size will be between and ii. I am \(95 \%\) confident that the population mean is between and iii. We are \(95 \%\) confident that the boundaries are and b. Can you reject a population mean of 20 pounds? Explain.

Vegetarians' Weights The mean weight of all 20-yearold women is 128 pounds (http://www.kidsgrowth.com). A random sample of 40 vegetarian women who are 20 years old showed a sample mean of 122 pounds with a standard deviation of 15 pounds. The women's measurements were independent of each other. a. Determine whether the mean weight for 20 -year old vegetarian women is significantly less than 128 , using a significance level of \(0.05\). b. Now suppose the sample consists of 100 vegetarian women who are 20 years old, and repeat the test. c. Explain what causes the difference between the \(\mathrm{p}\) -values for parts a and \(\mathrm{b}\).

Exam Scores The distribution of the scores on a certain exam is \(N(70,10)\), which means that the exam scores are Normally distributed with a mean of 70 and standard deviation of \(10 .\) a, Sketch the curve and label, on the \(x\) -axis, the position of the mean, the mean plus or minus one standard deviation, the mean plus or minus two standard deviations, and the mean plus or minus three standard deviations. b. Find the probability that a randomly selected score will be between 50 and 90 . Shade the region under the Normal curve whose area corresponds to this probability.

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