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

Short Answer

Expert verified
The height at the 20th percentile is approximately 62.9 inches.

Step by step solution

01

Find the z-score for the 20th percentile

Look up the z-score that corresponds to the 20th percentile in a standard normal distribution table or use a calculator capable of inverse normal distribution calculations. The z-score for the 20th percentile is approximately -0.84.
02

Use the z-score to find the corresponding height

The formula for converting a z-score to an original score (in this case, height) is \(X = μ + zσ\), where \(X\) is the original score, \(μ\) is the mean, \(z\) is the z-score, and \(σ\) is the standard deviation. Apply the values: \(X = 65 + (-0.84) * 2.5\).
03

Compute the height

Perform the calculation to find the height that corresponds to the 20th percentile: \(X = 65 - 2.1 = 62.9\) inches.

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Key Concepts

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

Percentile
Understanding percentiles is crucial in comparing a particular value to a broader dataset. A percentile indicates the relative standing of a value within a dataset. Hence, a score in the 20th percentile means that it is as high or higher than 20% of the other data points in the distribution. In the context of our exercise, we're tasked with finding the height at the 20th percentile. This translates to finding a height where only 20% of women are shorter or equal in height, while the remaining 80% are taller. Percentiles help in assessing benchmarks such as test scores, weights, and heights, providing a sense of where an individual falls in comparison to a larger population.

Percentiles do not divide data into equal blocks; instead, they offer a ranking system that can be very illuminating in understanding how data is spread across a distribution.
Z-score
Z-scores are a statistical measure that provides insight into how far away a particular data point is from the mean of a dataset, measured in units of standard deviations. Considering a normal distribution, where the mean is the center point, a z-score tells us how many standard deviations away the data point lies. A negative z-score implies the value is below the mean, while a positive one, above.

In our problem, we used a z-score to determine the height corresponding to the 20th percentile. By finding a z-score of -0.84, it indicated that a height at the 20th percentile is 0.84 standard deviations below the mean. Calculating or finding z-scores from a statistical table allows us to convert a percentile into a data point with context.

Understanding z-scores is fundamental in statistics for transforming raw data into a normalized form.
Standard Deviation
Standard deviation is a key concept in understanding how data points spread out around the mean in a dataset. In a normal distribution, it gives a concrete measure of the variability or dispersion. A small standard deviation indicates data points are clustered closely around the mean, while a large one signifies more spread out data. In terms of units, standard deviation allows us to express data variability in a comparable way.

In our exercise, the standard deviation of 2.5 inches is applied to measure the dispersion of heights around the mean height of 65 inches. Using this measure, we were able to calculate the height at the 20th percentile, which turns out to be approximately 62.9 inches. Understanding the role of standard deviation helps in appreciating the concept of z-scores, as those are expressed in terms of this unit.
Normal Curve
The normal curve, often referred to as the bell curve, represents the distribution of data where most values cluster around a central point (mean), and probabilities gradually decrease as one moves away from the mean in either direction. It is characterized by its symmetrical shape, with values evenly distributed around the mean. This concept is fundamental in statistics given the prevalence of normal distributions in various datasets.

In our problem scenario, the heights of women can be understood as a normal distribution, allowing us to leverage the properties of this distribution to find specific percentiles. This includes using a standard normal distribution table when seeking z-scores and calculating specific data points such as height.

Visualizing data with a normal curve helps in understanding how values distribute around the mean, making it easier to grasp concepts like percentiles and standard deviations.

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Most popular questions from this chapter

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.

Babies weighing \(5.5\) pounds or less at birth are said to have low birth weights, which can be dangerous. Full-term birth weights for single babies (not twins or triplets or other multiple births) are Normally distributed with a mean of \(7.5\) pounds and a standard deviation of 1.1 pounds. a. For one randomly selected full-term single-birth baby, what is the probability that the birth weight is \(5.5\) pounds or less? b. For two randomly selected full-term, single-birth babies, what is the probability that both have birth weights of \(5.5\) pounds or less? c. For 200 random full-term single births, what is the approximate probability that 7 or fewer have low birth weights? d. If 200 independent full-term single-birth babies are born at a hospital, how many would you expect to have birth weights of \(5.5\) pounds or less? Round to the nearest whole number. e. What is the standard deviation for the number of babies out of 200 who weigh \(5.5\) pounds or less? Retain two decimal digits for use in part f. f. Report the birth weight for full-term single babies (with 200 births) for two standard deviations below the mean and for two standard deviations above the mean. Round both numbers to the nearest whole number. g. If there were 45 low-birth-weight full-term babies out of 200 , would you be surprised?

Professional basketball player Draymond Green has a free-throw success rate of \(70 \%\). Suppose Green takes as many free throws as he can in one minute. Why would it be inappropriate to use the binomial model to find the probability that he makes at least 5 shots in one minute? What condition or conditions for use of the binomial model is or are not met?

Support for the legalization of marijuana has continued to grow among Americans. A 2017 Gallup poll found that \(64 \%\) of Americans now say that marijuana use should be legal. Suppose a random sample of 150 Americans is selected. a. Find the probability that at most 110 people support marijuana legalization. b. Find the probability that between 90 and 110 support marijuana legalization. c. Complete this sentence: In a group of 150 , we would expect _____ support marijuana legalization, give or take ____.

The distribution of white blood cell count per cubic millimeter of whole blood is approximately Normal with mean 7500 and standard deviation 1750 for healthy patients. Use technology or a table to answer these questions. For each include an appropriately labeled and shaded Normal curve. a. What is the probability that a randomly selected person will have a white blood cell count between 6000 and \(10,000 ?\) b. An elevated white blood cell count can be a sign of infection somewhere in the body. A white blood cell count can be considered elevated if it is over 10,500 . What percentage of people have white blood cell counts in this elevated range? c. A white blood cell count below 4500 is considered low. People in this range may be referred for additional medical testing. What is the probability that a randomly selected person has a white blood cell count below \(4500 ?\)

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