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Chapter 1: Problem 27

A set of 340 examination scores exhibiting a bell-shaped relative frequency distribution has a mean of \(\bar{y}=72\) and a standard deviation of \(s=8\). Approximately how many of the scores would you expect to fall in the interval from 64 to \(80 ?\) The interval from 56 to \(88 ?\)

Short Answer

Expert verified
231 scores are expected between 64 and 80; 323 scores between 56 and 88.

Step by step solution

01

Understand the Empirical Rule

The Empirical Rule, or the 68-95-99.7 rule, tells us that for a bell-shaped distribution (normal distribution), approximately 68% of the data falls within one standard deviation (\([-s, +s]\)), 95% within two standard deviations, and 99.7% within three standard deviations of the mean.
02

Calculate the First Interval Limits

We are asked to find the number of scores within the interval from 64 to 80. Since the mean is 72 and the standard deviation is 8, calculate this interval from \(\bar{y} - s = 72 - 8 = 64\) to \(\bar{y} + s = 72 + 8 = 80\). This interval is within one standard deviation.
03

Calculate the Second Interval Limits

Determine the interval from 56 to 88. This involves calculating two standard deviations from the mean: \(\bar{y} - 2s = 72 - 16 = 56\) and \(\bar{y} + 2s = 72 + 16 = 88\). This is the range for two standard deviations from the mean.
04

Apply Empirical Rule to Estimate First Interval

According to the Empirical Rule, approximately 68% of the data falls within one standard deviation of the mean. Thus, 68% of the 340 scores is calculated as \(0.68 \times 340 = 231.2\). Approximately 231 scores are expected to fall within the range of 64 to 80.
05

Apply Empirical Rule to Estimate Second Interval

For the interval that spans two standard deviations from the mean (56 to 88), the Empirical Rule states that about 95% of data falls within this range. Calculate \(0.95 \times 340 = 323\). Hence, approximately 323 scores are expected to fall within the range of 56 to 88.

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Key Concepts

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

bell-shaped distribution
A bell-shaped distribution refers to a graphical representation of data that resembles the shape of a bell. In statistics, this typically indicates a normal distribution. A normal distribution is when most of the data points cluster around the mean, with fewer and fewer observations as you move away from the mean. This results in the characteristic 'bell' shape.
  • The peak of the bell represents where the mean, median, and mode have the same value.
  • The data is symmetrically distributed around the mean, creating the tallest point at the center.
  • As you move towards the edges of the bell, the number of observations decreases.
Given its shape, understanding a bell-shaped distribution helps in predicting where most data points are likely to fall. In practical terms, it also allows us to use tools like the Empirical Rule to forecast the likelihood of data points in a specific range.
standard deviation
Standard deviation is a key concept in statistics measuring the amount of variation or dispersion in a set of data. It tells us how much the individual data points tend to differ from the mean of the data set.
  • A low standard deviation means data points are close to the mean.
  • A high standard deviation indicates that the data points are spread out over a wider range of values.
Mathematically, standard deviation is represented as \( s \). It is computed by taking the square root of the variance, which is the average of squared differences from the mean.
For example, in the exercise, the standard deviation is 8. This means on average, the examination scores deviate by 8 points from the mean score of 72.
normal distribution
Normal distribution, often termed as the Gaussian distribution, is a probability distribution that is symmetric about its mean. It is one of the most important continuous probability distributions used in statistics and is defined by two parameters: the mean (average) and the standard deviation.
  • A perfectly normal distribution forms a perfectly symmetrical bell curve.
  • The total area under the curve is 1, which represents the whole population of data.
  • The mean divides the distribution into two equal halves.
Normal distribution is pivotal because many statistical tests and analyses assume or require data that is approximately normally distributed. This assumption allows for the use of the Empirical Rule to predict how data points distribute around the mean.
mean
The mean, or average, is a measure of central tendency that sums up a set of numbers and divides it by the count of those numbers. It gives us a central value that represents the entire set of data.
  • It's calculated as \( \frac{\text{sum of all data points}}{\text{number of data points}} \).
  • The mean is sensitive to extreme values, meaning it can be skewed by outliers in the data set.
  • In a normal distribution, the mean is located at the peak of the bell curve.
In the context of the exercise, the mean examination score is 72. This indicates that on average, students scored 72 on the examination. The mean helps in identifying central patterns and can be used, alongside standard deviation, to determine intervals where most data points lie according to the Empirical Rule.

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Most popular questions from this chapter

The discharge of suspended solids from a phosphate mine is normally distributed with mean daily discharge 27 milligrams per liter \((\mathrm{mg} / \mathrm{L})\) and standard deviation \(14 \mathrm{mg} / \mathrm{L}\). In what proportion of the days will the daily discharge be less than \(13 \mathrm{mg} / \mathrm{L} ?\)

The manufacturer of a new food additive for beef cattle claims that \(80 \%\) of the animals fed a diet including this additive should have monthly weight gains in excess of 20 pounds. A large sample of measurements on weight gains for cattle fed this diet exhibits an approximately normal distribution with mean 22 pounds and standard deviation 2 pounds. Do you think the sample information contradicts the manufacturer's claim? (Calculate the probability of a weight gain exceeding 20 pounds.)

Resting breathing rates for college-age students are approximately normally distributed with mean 12 and standard deviation 2.3 breaths per minute. What fraction of all college-age students have breathing rates in the following intervals? a. 9.7 to 14.3 breaths per minute b. 7.4 to 16.6 breaths per minute c. 9.7 to 16.6 breaths per minute d. Less than 5.1 or more than 18.9 breaths per minute

The following results on summations will help us in calculating the sample variance \(s^{2}\). For any constant \(c\), a. \(\sum_{i=1}^{n} c=n c\). b. \(\sum_{i=1}^{n} c y_{i}=c \sum_{i=1}^{n} y_{i}\). c. \(\sum_{i=1}^{n}\left(x_{i}+y_{i}\right)=\sum_{i=1}^{n} x_{i}+\sum_{i=1}^{n} y_{i}\). Use (a), (b), and (c) to show that $$s^{2}=\frac{1}{n-1} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}=\frac{1}{n-1}\left[\sum_{i=1}^{n} y_{i}^{2}-\frac{1}{n}\left(\sum_{i=1}^{n} y_{i}\right)^{2}\right].$$

Compared to their stay-at-home peers, women employed outside the home have higher levels of high-density lipoproteins (HDL), the "good" cholesterol associated with lower risk for heart attacks. A study of cholesterol levels in 2000 women, aged \(25-64\), living in Augsburg, Germany, was conducted by Ursula Haertel, Ulrigh Keil, and colleagues \(^{\star}\) at the GSF-Medis Institut in Munich. Of these 2000 women, the \(48 \%\) who worked outside the home had HDL levels that were between 2.5 and 3.6 milligrams per deciliter (mg/dL) higher than the HDL levels of their stay-at-home counterparts. Suppose that the difference in HDL levels is normally distributed, with mean 0 (indicating no difference between the two groups of women) and standard deviation \(1.2 \mathrm{mg} / \mathrm{dL}\). If you were to select an employed woman and a stay-at-home counterpart at random, what is the probability that the difference in their HDL levels would be between 1.2 and \(2.4 ?\)
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