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The mean of the 100 car speeds in Exercise 20 was \(23.84 \mathrm{mph}\), with a standard deviation of \(3.56 \mathrm{mph}\). a) Using a Normal model, what values should border the middle \(95 \%\) of all car speeds? b) Here are some summary statistics. $$ \begin{array}{lll} \hline \text { Percentile } & & \text { Speed } \\ \hline 100 \% & \text { Max } & 34.060 \\ 97.5 \% & & 30.976 \\ 90.0 \% & & 28.978 \\ 75.0 \% & \text { Q3 } & 25.785 \\ 50.0 \% & \text { Median } & 23.525 \\ 25.0 \% & \text { Q1 } & 21.547 \\ 10.0 \% & & 19.163 \\ 2.5 \% & & 16.638 \\ 0.0 \% & \text { Min } & 16.270 \\ \hline \end{array} $$

Short Answer

Expert verified
The middle 95% of car speeds range from approximately 16.862 mph to 30.818 mph.

Step by step solution

01

Understanding the Normal Distribution

To find the values that border the middle 95% of car speeds, we use the properties of the normal distribution. In a normal distribution, the middle 95% is captured within approximately two standard deviations from the mean.
02

Calculating Z-scores for 95% Middle Range

For the middle 95% of a normal distribution, the z-scores corresponding to the lower and upper bounds are approximately -1.96 and 1.96, based on standard normal distribution tables.
03

Calculate Lower Bound Value

To find the lower boundary value, use the formula: \( \text{Lower Bound} = \mu + (z \times \sigma) \), where \( \mu = 23.84 \) and \( z = -1.96 \) and \( \sigma = 3.56 \). Substitute these into the equation:\[ \text{Lower Bound} = 23.84 + (-1.96 \times 3.56) = 23.84 - 6.978 = 16.862 \]
04

Calculate Upper Bound Value

To find the upper boundary value, similarly use: \( \text{Upper Bound} = \mu + (z \times \sigma) \), where \( \mu = 23.84 \), \( z = 1.96 \) and \( \sigma = 3.56 \). Substitute these into the equation:\[ \text{Upper Bound} = 23.84 + (1.96 \times 3.56) = 23.84 + 6.978 = 30.818 \]
05

Confirm with Given Summary Statistics

The summary statistics provided show that the 2.5th percentile is at a speed of 16.638, and the 97.5th percentile is at a speed of 30.976. These match our calculated bounds of the middle 95%, confirming the calculations.

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Key Concepts

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

Z-scores
The concept of Z-scores is crucial in understanding how data points relate to the mean of a dataset. A Z-score measures how many standard deviations an element is from the mean. For example, in a normal distribution, a Z-score of 0 means the element is exactly at the mean. A positive Z-score indicates the data point is above the mean, while a negative Z-score signifies it is below the mean.
To find Z-scores, you use the formula: \[Z = \frac{(X - \mu)}{\sigma}\]where \(X\) is a data value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
This measure is particularly useful when comparing different datasets or when trying to find the probability of a result occurring within a certain part of a normal distribution. For instance, when determining the range of the middle 95% of any dataset, the Z-scores are usually -1.96 and 1.96.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation indicates the data points tend to be close to the mean of the set, while a high standard deviation indicates the data points are spread out over a wider range of values.
In mathematical terms, the formula for standard deviation \(\sigma\) is:\[\sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2}\]where \(N\) is the number of data points, \(X_i\) represents each data point, and \(\mu\) is the mean of the dataset.
Standard deviation is a critical component of the normal distribution. It helps to determine the confidence intervals and probability of values occurring within that range. It is frequently used in various fields like finance, physics, and social sciences to compare different datasets by providing a scale that measures the variability of data.
Mean
The mean, also known as the average, is the sum of all numbers in a dataset divided by the number of numbers in that dataset. This is one of the most straightforward and commonly used statistical measures.
The formula to calculate mean \(\mu\) is:\[\mu = \frac{1}{N} \sum_{i=1}^{N} X_i\]where \(N\) is the total number of values, and \(X_i\) represents each value in the dataset.
In the context of a normal distribution, the mean is the central point around which data is distributed, forming the peak of the bell curve. This is why the mean is also regarded as a measure of central tendency. It's important to note that the mean can be influenced by outliers or extreme values. Therefore, in datasets with significant skewness, other measures like the median might provide better insight.

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Most popular questions from this chapter

. Agricultural scientists are working on developing an improved variety of Roma tomatoes. Marketing research indicates that customers are likely to bypass Romas that weigh less than 70 grams. The current variety of Roma plants produces fruit that averages 74 grams, but \(11 \%\) of the tomatoes are too small. It is reasonable to assume that a Normal model applies. a) What is the standard deviation of the weights of Romas now being grown? b) Scientists hope to reduce the frequency of undersized tomatoes to no more than \(4 \%\). One way to accomplish this is to raise the average size of the fruit. If the standard deviation remains the same, what target mean should they have as a goal? c) The researchers produce a new variety with a mean weight of 75 grams, which meets the \(4 \%\) goal. What is the standard deviation of the weights of these new Romas? d) Based on their standard deviations, compare the tomatoes produced by the two varieties.

The first Stats exam had a mean of 65 and a standard deviation of 10 points; the second had a mean of 80 and a standard deviation of 5 points. Derrick scored an 80 on both tests. Julie scored a 70 on the first test and a 90 on the second. They both totaled 160 points on the two exams, but Julie claims that her total is better. Explain.

One of the authors has an adopted grandson whose birth family members are very short. After examining him at his 2 -year checkup, the boy's pediatrician said that the \(z\) -score for his height relative to American 2-year-olds was \(-1.88\). Write a sentence explaining what that means.

Environmental Protection Agency (EPA) fuel economy estimates for automobile models tested recently predicted a mean of \(24.8 \mathrm{mpg}\) and a standard deviation of \(6.2 \mathrm{mpg}\) for highway driving. Assume that a Normal model can be applied. a) Draw the model for auto fuel economy. Clearly label it, showing what the \(68-95-99.7\) Rule predicts. b) In what interval would you expect the central \(68 \%\) of autos to be found? c) About what percent of autos should get more than \(31 \mathrm{mpg}\) ? d) About what percent of cars should get between 31 and \(37.2 \mathrm{mpg}\) ? e) Describe the gas mileage of the worst \(2.5 \%\) of all cars.

Based on the model \(N(1152,84)\) describing Angus steer weights, what are the cutoff values for a) the highest \(10 \%\) of the weights? b) the lowest \(20 \%\) of the weights? c) the middle \(40 \%\) of the weights?

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