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According to National Vital Statistics, the average length of a newborn baby is \(19.5\) inches with a standard deviation of \(0.9\) inch. The distribution of lengths is approximately Normal. Use a table or technology for each question. Include an appropriately labeled and shaded Normal curve for each part. There should be three separate curves. an What is the probability that a baby will have a length of \(20.4\) inches or more? b. What is the probability that a baby will have a length of \(21.4\) inches or more? c. What is the probability that a baby will be between 18 and 21 inches in length?

Short Answer

Expert verified
The probabilities of a baby being 20.4 inches or more and 21.4 inches or more need to be calculated by finding the Z-scores and corresponding probabilities. For the range between 18 and 21 inches, two Z-scores need to be found and the the probability between them calculated.

Step by step solution

01

Calculate Z-Score for Part a

A Z-Score can be calculated using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(X\) is the value from our dataset, \(\mu\) is the mean of the data and \(\sigma\) is the standard deviation. For part a, \(X = 20.4, \mu = 19.5, \text{and } \sigma = 0.9\). So, the Z-Score will be \(Z_a = \frac{20.4 - 19.5}{0.9}.\)
02

Calculate Probability for Part a

After calculating the Z-Score, the corresponding probability can be found using a standard normal table or a calculator. The probability that a baby will be 20.4 inches or more is the probability that Z is more than \(Z_a\), which can be calculated by \(P(Z > Z_a) = 1 - P(Z ≤ Z_a)\). The value of \(P(Z ≤ Z_a)\) can be obtained from the standard normal table.
03

Repeat Steps for Part b and c

Repeat steps 1 and 2 for part b and c. Remember that for part c, we want the probability that the length of the baby is between 18 and 21 inches, or \(P(18 < X < 21)\). This means we have to find two Z-Scores (one for X = 18 and one for X = 21) and calculate the probability that Z is between these two Z-Scores, which can be found using the formula \(P(Z_b < Z < Z_c) = P(Z < Z_c) - P(Z < Z_b)\), where \(Z_b\) and \(Z_c\) are the calculated Z-Scores.

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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 Z-score is crucial when working with normal distributions. A Z-score indicates how many standard deviations an element is from the mean. In essence, it converts the measurements onto a standard scale with zero mean. For example, in the provided exercise, calculating the Z-score for a newborn's length allows us to determine how unusual a measurement is in the context of the given data set.

The formula for calculating a Z-score is \( Z = \frac{X - \mu}{\sigma} \) where \(X\) is the value we are examining, \(\mu\) is the mean of the data, and \(\sigma\) represents the standard deviation. Applying this formula to the exercise, for a baby length of 20.4 inches (value \(X\)), the Z-score is calculated using the provided mean length of 19.5 inches (value \(\mu\)) and the standard deviation of 0.9 inches (value \(\sigma\)).

The Z-score is then used to find the probability that a randomly selected baby will be a certain length, in this case, 20.4 inches or more. The simplicity of this process is what makes the Z-score a valuable tool for statisticians and researchers alike.
Standard Deviation
Standard deviation, symbolized as \(\sigma\), is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that the values tend to be close to the mean (the expected value) of the set, while a high standard deviation means that the values are spread out over a wider range. It is a key concept in statistics, especially when dealing with the normal distribution.

In the context of the exercise, the standard deviation of 0.9 inches informs us about the variation in the length of newborn babies around the mean length of 19.5 inches. To put it simply, it helps us understand how concentrated or spread out the lengths are. If a baby's length is exactly 19.5 inches, it's equivalent to the mean; if a baby's length differs from 19.5 inches, the standard deviation helps to quantify that difference. In practice, calculating the standard deviation is a foundational step to further statistical analysis, such as Z-score calculations and probability estimations.
Normal Curve Representation
The normal curve is a graph that represents the distribution of values in a data set that is symmetrically distributed, and it has a distinct bell shape. This bell curve is a visual representation of a normal distribution, with the highest point over the mean and the curve tapering off equally on both sides.

An important property of the normal curve is that it's defined by its mean and standard deviation. The mean dictates where the peak of the curve is centered, and the standard deviation indicates the width of the curve. When looking at the normal curve representation in the exercise, it's imperative to accurately label and shade the relevant sections to visually illustrate the probabilities of different ranges of baby lengths.

For instance, to determine the probability of a baby being between 18 and 21 inches, you would look for the area under the curve between the Z-scores associated with those lengths. Essentially, the normal curve helps us to visualize probabilities, and the area under the curve is key in determining these probabilities. In practice, the normal curve reinforces the relationship between a probability and the Z-scores calculated from a data set.

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Most popular questions from this chapter

The distribution of the math portion of SAT scores has a mean of 500 and a standard deviation of 100 , and the scores are approximately Normally distributed. a. What is the probability that one randomly selected person will have an SAT score of 550 or more? b. What is the probability that four randomly selected people will all have SAT scores of 550 or more? c. For 800 randomly selected people, what is the probability that 250 or more will have scores of 550 or more? d. For 800 randomly selected people, on average how many should have scores of 550 or more? Round to the nearest whole number. e. Find the standard deviation for part d. Round to the nearest whole number. f. Report the range of people out of 800 who should have scores of 550 or more from two standard deviations below the mean to two standard deviations above the mean. Use your rounded answers to part \(\mathrm{d}\) and \(\mathrm{e}\). g. If 400 out of 800 randomly selected people had scores of 550 or more, would you be surprised? Explain.

The average birth weight of domestic cats is about 3 ounces. Assume that the distribution of birth weights is Normal with a standard deviation of \(0.4\) ounce. a. Find the birth weight of cats at the 90 th percentile. b. Find the birth weight of cats at the 10 th percentile.

The Normal model \(N(65,2.5)\) describes the distribution of heights of college women (inches). Which of the following questions asks for a probability and which asks for a measurement (and is thus an inverse Normal question)? a. What is the probability that a random college woman has a height of 68 inches or more? b. To be in the Tall Club, a woman must have a height such that only \(2 \%\) of women are taller. What is this height?

A study of human body temperatures using healthy men showed a mean of \(98.1^{\circ} \mathrm{F}\) and a standard deviation of \(0.70^{\circ} \mathrm{F}\). Assume the temperatures are approximately Normally distributed. a. Find the percentage of healthy men with temperatures below \(98.6^{\circ} \mathrm{F}\) (that temperature was considered typical for many decades). b. What temperature does a healthy man have if his temperature is at the 76 th percentile?

Suppose college women's heights are approximately Normally distributed with a mean of 65 inches and ? population standard deviation of \(2.5\) inches. What height is at the 20 th percentile? Include an appropriately labeled sketch of the Normal curve to support your answer.

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