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Chapter 3: Problem 33

The dotplot shows heights of college women; the mean is 64 inches \((5\) feet 4 inches), and the standard deviation is 3 inches. a. What is the \(z\) -score for a height of 58 inches ( 4 feet 10 inches)? b. What is the height of a woman with a z-score of \(1 ?\)

Short Answer

Expert verified
a. The z-score for a height of 58 inches is -2. b. The height of a woman with a z-score of 1 is 67 inches.

Step by step solution

01

Calculate Z-Score for 58 Inches

In order to calculate the z-score for a height of 58 inches, subtract the mean height (64 inches) from 58, then divide the result by the standard deviation (3 inches). The formula we use is \[Z = \frac{X - \mu}{\sigma}\] where \(X\) is the value for which we want the z-score, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Substituting in our known values, we get \[Z = \frac{58 - 64}{3} = -2\]
02

Calculate the Height for a Z-score of 1

To find the height that corresponds to a z-score of 1, we need to multiply the z-score by the standard deviation and then add the mean height. Using the formula \(X = Z \cdot \sigma + \mu\), where \(Z\) is the z-score, \(\sigma\) is the standard deviation, and \(\mu\) is the mean, substituting in the given values, we get $X = 1 \cdot 3 + 64 = 67$ inches.

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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 key concept in statistics. It's often visualized as a bell-shaped curve, representing data that tends to cluster around a central value. This curve is symmetric, with most occurrences taking place near the mean or average of the data.

Some essential features of a normal distribution include:
  • Symmetry around the mean
  • The mean, median, and mode are all located at the center
  • The curve approaches the horizontal axis but never actually touches it
When the heights of college women were analyzed, they were assumed to follow such a distribution. This means that most women would have heights close to the mean, or average. Very tall or very short women are fewer, making them reside on either tail of the bell curve.
Mean and Standard Deviation
Understanding the mean and standard deviation is critical when working with normally distributed data. The mean is simply the average of all data points. It's calculated by summing all values and dividing by the number of values.

In the exercise, the mean height of the women is given as 64 inches. This means that, on average, the women are about 5 feet 4 inches tall.

The standard deviation measures the amount of variation or dispersion from the mean. A smaller standard deviation indicates that the data points tend to be close to the mean. Conversely, a larger standard deviation means the data points are spread out over a wider range of values.

The standard deviation for the height data given is 3 inches. This suggests that most women's heights are within 3 inches of the mean height, either above or below.
Height Data Analysis
Height data analysis often involves calculating the z-score, which gives you an idea of how an individual data point compares to the mean.

The z-score formula is:\[ Z = \frac{X - \mu}{\sigma} \]
  • \(X\) represents the data point you're interested in
  • \(\mu\) is the mean
  • \(\sigma\) is the standard deviation
In the exercise, calculating the z-score for a height of 58 inches involved using this formula. Subtracting the mean height (64 inches) from 58 inches, and dividing by the standard deviation (3 inches) gave a z-score of \(-2\).

To find the height that corresponds to a z-score of 1, you do the reverse. Multiply the z-score by the standard deviation and add the mean. So, for a z-score of 1, the height would be 67 inches. This kind of analysis helps in understanding where specific heights lie relative to the overall distribution.

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Most popular questions from this chapter

In 2017 a pollution index was calculated for a sample of cities in the western states using data on air and water pollution. Assume the distribution of pollution indices is unimodal and symmetric. The mean of the distribution was \(43.0\) points with a standard deviation of \(11.3\) points. a. What percentage of western cities would you expect to have a pollution index between \(31.7\) and \(54.3\) points? b. What percentage of western cities would you expect to have a pollution index between \(20.4\) and \(65.6\) ? c. The pollution index for San Jose in 2017 was \(51.9\) points. Based on this distribution, was this unusually high? Explain.

Wechsler IQ tests have a mean of 100 and a standard deviation of \(15 .\) Which is more unusual: an IQ above 110 or an IQ below 80 ?

In 2017 a pollution index was calculated for a sample of cities in the eastern states using data on air and water pollution. Assume the distribution of pollution indices is unimodal and symmetric. The mean of the distribution was \(35.9\) points with a standard deviation of \(11.6\) points. (Source: numbeo. com) see Guidance page \(142 .\) a. What percentage of eastern cities would you expect to have a pollution index between \(12.7\) and \(59.1\) points? b. What percentage of castern cities would you expect to have a pollution index between \(24.3\) and \(47.5\) points? c. The pollution index for New York, in 2017 was \(58.7\) points. Based on this distribution, was this unusually high? Explain.

The mean weight gain for women during a full-term pregnancy is \(30.2\) pounds. The standard deviation of weight gain for this group is \(9.9\) pounds, and the shape of the distribution of weight gains is symmetric and unimodal. a. State the weight gain for women one standard deviation below the mean and for one standard deviation above the mean. b. Is a weight gain of 35 pounds more or less than one standard deviation from the mean?

The tuition costs (in dollars) for a sample of four-year state colleges in California and Texas are shown below. Compare the means and the standard deviations of the data and compare the state tuition costs of the two states in a sentence or two. CA: \(7040,6423,6313,6802,7048,7460\) TX: \(7155,7504,7328,8230,7344,5760\)
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