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A study of U.S. births published on the website Medscape from WebMD reported that the average birth length of babies was \(20.5\) inches and the standard deviation was about \(0.90\) inch. Assume the distribution is approximately Normal. Find the percentage of babies with birth lengths of 22 inches or less.

Short Answer

Expert verified
The percentage of babies with birth lengths of 22 inches or less is approximately \(94.85\%\).

Step by step solution

01

Calculation of Z-Score

First the Z-score is calculated using the formula:\[ Z = \frac{{X - \mu}}{{\sigma}} \]where: - \(X\) is the value we're interested in understanding (in this case, 22 inches), - \(\mu\) is the mean of the population (20.5 inches in this case), - \(\sigma\) is the standard deviation of the population (0.90 inches in this case).Plugging in the values, one gets:\[ Z = \frac{{22 - 20.5}}{{0.90}} \]
02

Calculate the Z-Score

After executing the previous operation, the result for the Z-score is approximately 1.67. This result tells us that the value of 22 inches is about 1.67 standard deviations above the mean of the population.
03

Use the Z-Score to Find Probability

Next, the standard normal distribution table (Z-table) is used to find the probability that corresponds to a Z-score of 1.67. In most Z-tables, the value found for Z=1.67 will be 0.9485. This value represents the probability that a baby's birth length, selected from the population, is less than 22 inches.

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Key Concepts

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

Z-score Calculation
Understanding the concept of Z-score calculation is crucial for analyzing how individual data points relate to a group, especially in a normally distributed set. In the context of the given exercise, the Z-score measures the number of standard deviations a particular measurement, like the birth length of a baby, is from the mean or average of the population.

In mathematical terms, the Z-score is calculated using the formula:
\begin{align*} Z = \frac{{X - \(mu\)}}{{\(sigma\)}} \end{align*}
where \(X\) represents the individual data point we're evaluating, \(\mu\) is the mean of the population, and \(\sigma\) is the population's standard deviation. By substituting the values from our study—\(X = 22\) inches, \(\mu = 20.5\) inches, and \(\sigma = 0.90\) inches—we find the Z-score that indicates how far the value 22 inches is from the mean in units of standard deviation.

To gain a clearer understanding, let’s consider a student's grade in a class. If the class has an average grade of 70 with a standard deviation of 10, a student scoring 90 on a test would have a Z-score of 2. This means the student's score is two standard deviations above the average grade. Similarly, in our birth length example, a Z-score of 1.67 means that a baby's length of 22 inches falls 1.67 standard deviations above the average birth length.
Standard Deviation
The concept of standard deviation is a measure that indicates the amount of variation or dispersion of a set of values. A low standard deviation means that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

In probability and statistics, standard deviation plays a vital role because it quantifies uncertainty. For instance, when considering the average birth length reported by Medscape from WebMD, knowing the standard deviation of \(0.90\) inch adds context. It tells us that most birth lengths cluster within a range of \(0.90\) inch above or below the average, allowing for predictions about how common or rare certain measurements are.

Practical Importance of Standard Deviation

In everyday terms, standard deviation helps us to understand predictability. If you are a pediatrician, you could use the standard deviation of birth lengths to reassure parents that their baby's length is within the expected range. If it's significantly higher or lower, that might prompt additional assessments. Beyond healthcare, standard deviation is important in finance, quality control, and many other fields as a key metric for variability and risk assessment.
Probability and Statistics
Probability and statistics are branches of mathematics concerned with collecting, analyzing, interpreting, and presenting data. Probability deals with the likelihood of events occurring, while statistics involves summarizing and making inferences from data.

The Z-score obtained in our earlier calculation translates to probability using a standard normal distribution table, also known as a Z-table. This tool provides the probability of a value being at or below a certain Z-score in a normally distributed dataset. In the context of our exercise, the Z-score of 1.67 is found in the Z-table correlating to a probability of 0.9485, or 94.85%. This means there's a 94.85% chance that a randomly picked baby will have a birth length of 22 inches or less.

Interpreting Results in Real Situations

Grasping these concepts allows you to interpret data and make informed decisions. For example, if you're an obstetrician analyzing birth length data, understanding probabilities can guide your expectations for patient outcomes. In a broader sense, learning to interpret statistical data equips us for critical thinking about information we encounter daily, from health metrics to socio-economic studies, thus enhancing overall decision-making proficiency.

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Most popular questions from this chapter

According to data from the U.S. State Department, the percentage of Americans who have a passport has risen dramatically. In 2007 , only \(27 \%\) of Americans had a passport; in 2017 that percentage had risen to \(42 \%\). Assume that currently \(42 \%\) of Americans have a passport. Suppose 50 Americans are selected at random. a. Find the probability that fewer than 20 have a passport. b. Find the probability that at most 24 have a passport. c. Find the probability that at least 25 have a passport.

The Empirical Rule applies rough approximations to probabilities for any unimodal, symmetric distribution. But for the Normal distribution we can be more precise. Use the figure and the fact that the Normal curve is symmetric to answer the questions. Do not use a Normal table or technology. According to the Empirical Rule, a. Roughly what percentage of \(z\) -scores are between \(-2\) and 2 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) b. Roughly what percentage of \(z\) -scores are between \(-3\) and 3 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) c. Roughly what percentage of \(z\) -scores are between \(-1\) and 1 . i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) d. Roughly what percentage of \(z\) -scores are greater than 0 ? i. almost all iii. \(68 \%\) ii. \(95 \%\) iv. \(50 \%\) e. Roughly what percentage of \(z\) -scores are between 1 and 2 ? i. almost all iii. \(50 \%\) ii. \(13.5 \%\) iv. \(2 \%\)

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