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Chapter 5: Problem 18

John Beale of Stanford, CA, recorded the speeds of cars driving past his house, where the speed limit was \(20 \mathrm{mph}\). The mean of 100 readings was \(23.84 \mathrm{mph},\) with a standard deviation of \(3.56 \mathrm{mph} .\) (He actually recorded every car for a two-month period. These are 100 representative readings.) a) How many standard deviations from the mean would a car going under the speed limit be? b) Which would be more unusual, a car traveling \(34 \mathrm{mph}\) or one going \(10 \mathrm{mph} ?\)

Short Answer

Expert verified
Cars going under 20 mph are 1.08 standard deviations below the mean. A car going 10 mph is more unusual than 34 mph.

Step by step solution

01

Define the Parameters

First, let's define what we are dealing with. The mean speed of the cars is \( \mu = 23.84 \, \text{mph} \) and the standard deviation is \( \sigma = 3.56 \, \text{mph}\). The speed limit is given as \( 20 \, \text{mph}\).
02

Calculate Z-score for Speed Under Limit

To calculate how many standard deviations a speed of \(20 \, \text{mph}\) is from the mean, we can use the z-score formula: \[ z = \frac{X - \mu}{\sigma} \] Where \(X\) is the value of interest (here, \(20\) mph). Substitute \( X = 20, \mu = 23.84, \sigma = 3.56 \) into the formula: \[ z = \frac{20 - 23.84}{3.56} \approx -1.08 \] This means a car traveling at \(20\) mph is approximately \(-1.08\) standard deviations from the mean.
03

Calculate Z-score for 34 mph

Next, we calculate how far a speed of \(34 \, \text{mph}\) is from the mean using the formula: \[ z_{34} = \frac{34 - 23.84}{3.56} \approx 2.85 \] So, a car traveling at \(34\) mph is approximately \(2.85\) standard deviations above the mean.
04

Calculate Z-score for 10 mph

Similarly, calculate the z-score for a speed of \(10 \, \text{mph}\) using the formula: \[ z_{10} = \frac{10 - 23.84}{3.56} \approx -3.91 \] Thus, a car traveling at \(10\) mph is approximately \(-3.91\) standard deviations below the mean.
05

Compare Unusual Speeds

To determine which speed is more unusual, compare the absolute values of the z-scores. The z-score for \(10\) mph is \(-3.91\), while for \(34\) mph it is \(2.85\). Since \(|-3.91| > |2.85|\), a car traveling at \(10\) mph is more unusual than one traveling at \(34\) mph.

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Key Concepts

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

Understanding Z-score
The z-score is a statistical measurement that helps us understand how far away a specific data point is from the mean of a dataset. In simpler terms, it tells us how unusual or typical a value is compared to the average. A z-score is calculated by taking the difference between a data point and the mean (average), and then dividing by the standard deviation. The formula for calculating a z-score is:
  • \( z = \frac{X - \mu}{\sigma} \)
Where:
  • \( X \) is the data point of interest.
  • \( \mu \) is the mean of the dataset.
  • \( \sigma \) is the standard deviation.
A z-score of 0 indicates the value is exactly on the mean. Positive z-scores show the data point is above the mean, while negative z-scores indicate it's below the mean. Large absolute values of z-scores signal unusual data points.
Mean Speed and Its Significance
Mean speed is simply the average speed recorded from a set of data. To find the mean, sum up all the individual speeds, then divide by the number of readings. For example, if John collected 100 speed readings, he would add them together, and divide by 100 to find the mean.In this context:
  • The mean speed noted was \( 23.84 \) mph.
  • This value represents the typical, or expected, speed of the cars passing by John's house.
Mean is crucial because it provides a baseline from which to measure variation, or deviation, of individual speeds.
Calculating Standard Deviation
Standard deviation measures the amount of variation or spread in a set of data. A low standard deviation indicates the data points are close to the mean, while a high standard deviation means they are spread out over a wider range.The formula for standard deviation is:
  • \( \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^N (X_i - \mu)^2} \)
Where:
  • \( N \) is the number of data points.
  • \( X_i \) is each individual data point.
  • \( \mu \) is the mean of the dataset.
In John's data, the standard deviation was \( 3.56 \) mph. This helps to contextualize how much the speeds varied around the mean of \( 23.84 \) mph.
Spotting Unusual Data Points
Unusual data points, or outliers, are those that significantly deviate from the mean. To identify these, we can use the z-score. In John's example:
  • A car going at the speed limit of \( 20 \) mph had a z-score of approximately \(-1.08\), showing it is slightly below the mean, but not very unusual.
  • A car traveling at \( 34 \) mph had a z-score of \( 2.85 \), indicating it's well above the average speed, so it's quite unusual.
  • A car moving at \( 10 \) mph had a z-score of \(-3.91\), marking it as extremely below average and therefore, highly unusual.
Z-scores help highlight which speeds stand out from the rest of the driving population. Comparatively, a speed of \( 10 \) mph is more unusual than \( 34 \) mph, due to its greater deviation from the mean, as indicated by z-score absolutes.

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Most popular questions from this chapter

Recall that the beef cattle described in Exercise 17 had a mean weight of 1152 pounds, with a standard deviation of 84 pounds. a) Cattle buyers hope that yearling Angus steers will weigh at least 1000 pounds. To see how much over (or under) that goal the cattle are, we could subtract 1000 pounds from all the weights. What would the new mean and standard deviation be? b) Suppose such cattle sell at auction for 40 cents a pound. Find the mean and standard deviation of the sale prices for all the steers.

A friend tells you about a recent study dealing with the number of years of teaching experience among current college professors. He remembers the mean but can't recall whether the standard deviation was 6 months, 6 years, or 16 years. Tell him which one it must have been, and why.

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The winning scores of all college men's basketball games for the \(2011-12\) season were approximately normally distributed with mean 77.5 points and standard deviation 12.5 points. a) Draw the Normal model for winning scores. b) What interval of winning scores would be the central \(95 \%\) of all winning scores for the \(2011-12\) season? c) About what percent of the winning scores should be less than 65 points? d) About what percent of the winning scores should be between 65 and 102 points? e) About what percent of the winning scores should be over 102 points?

Consider the Angus weights model \(N(1152,84)\) one last time. a) What weight represents the 40th percentile? b) What weight represents the 99th percentile? c) What's the IQR of the weights of these Angus steers?
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