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The average age of U.S. astronaut candidates in the past has been 34, but candidates have ranged in age from 26 to 46. Use the range rule of thumb to estimate the standard deviation of the applicants’ ages.

Short Answer

Expert verified
The estimated standard deviation of the applicants' ages is 5.

Step by step solution

01

Understand the Range Rule of Thumb

The range rule of thumb states that to estimate the standard deviation of a data set, we divide the range by 4. This approximation assumes a roughly normal distribution.
02

Determine the Range of Ages

The range is the difference between the maximum and minimum values. Here, the age range is from 46 (maximum) to 26 (minimum). Thus, the range of the ages is calculated as follows: \( \text{Range} = 46 - 26 \).
03

Calculate the Range

Subtract the minimum value (26) from the maximum value (46): \( \text{Range} = 46 - 26 = 20 \). The range of the ages is 20.
04

Estimate the Standard Deviation

According to the range rule of thumb, we estimate the standard deviation by dividing the range by 4: \( \text{Estimated Standard Deviation} = \frac{20}{4} \).
05

Calculate the Estimated Standard Deviation

Divide the range (20) by 4: \( \text{Estimated Standard Deviation} = \frac{20}{4} = 5 \). The estimated standard deviation of the applicants' ages is 5.

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Key Concepts

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

Range Rule of Thumb
The range rule of thumb is a handy shortcut used to estimate the standard deviation of a dataset. This method is particularly useful when dealing with a normal distribution or any distribution that approximates it closely. Instead of computing the standard deviation from scratch, which can be complex and require detailed data points, the range rule simplifies the process by using the range of the dataset. The concept is straightforward:
  • Identify the range, which is the difference between the maximum and minimum values of the dataset.
  • Divide this range by 4 to estimate the standard deviation.
This technique works well because, in a perfect normal distribution, most data falls within a certain spread from the mean. Although it's an approximation, this method provides a reasonable standard deviation estimate when full calculation isn't feasible or necessary. The usefulness of the range rule lies in its simplicity and quick application, making it a helpful tool in preliminary data analysis.
Standard Deviation Estimation
Estimating the standard deviation is crucial in statistics as it gives us insight into the variability of a dataset. The standard deviation tells us how much the data deviates from the mean, providing a measure of spread or dispersion. When utilizing the range rule of thumb, you calculate the standard deviation by taking the range of your data and dividing it by four. This assumes a roughly normal distribution, where data tend to cluster around a central value with defined spread:
  • The smaller the range, the less variation there is among data points.
  • Conversely, a large range suggests greater variability.
By estimating the standard deviation quickly, you can determine if your dataset has wide spread or is closely packed around the mean. This estimation method is efficient in settings where a quick, rough measure of variability is needed without detailed calculations.
Normal Distribution Approximation
Normal distribution approximation plays a vital role in statistical analysis, as many datasets naturally exhibit this pattern. A normal distribution, commonly known as the bell curve, shows that most observations cluster around the mean, with fewer instances occurring as you move away from the center.
  • The range rule of thumb assumes that the data is approximately normally distributed.
  • In a normal distribution, about 68% of data falls within one standard deviation of the mean, 95% within two, and 99.7% within three.
This characteristic allows us to estimate various statistics, like the standard deviation, using simple techniques like the range rule. When data approximates a normal distribution, using these methods can be reliable. Understanding normal distribution helps in making predictions and informed decisions based on statistical data, ensuring that we comprehend variability and its implications.

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Most popular questions from this chapter

The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2. The average is found as shown. Since $$\text { Time }=\text { distance } \div \text { rate }$$ then $$\begin{array}{l}{\text { Time } 1=\frac{100}{40}=2.5 \text { hours to make the trip }} \\ {\text { Time } 2=\frac{100}{50}=2 \text { hours to return }}\end{array}$$ Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

The number of people killed in each state from passenger vehicle crashes for a specific year is shown. Find the range, variance, and standard deviation for the data. \(\begin{array}{ccccc}{778} & {309} & {1110} & {324} & {705} \\ {1067} & {826} & {76} & {205} & {152} \\ {218} & {492} & {65} & {186} & {712} \\ {193} & {262} & {452} & {875} & {82} \\ {730} & {1185} & {2707} & {1279} & {390} \\\ {305} & {123} & {948} & {343} & {602} \\ {69} & {450} & {2080} & {565} & {875} \\ {414} & {981} & {2786} & {82} & {793} \\ {214} & {130} & {396} & {620} & {797}\end{array}\)

The average of the number of trials it took a sample of mice to learn to traverse a maze was 12. The standard deviation was 3. Using Chebyshev’s theorem, find the minimum percentage of data values that will fall in the range of 4–20 trials.

An employment evaluation exam has a variance of 250. Two particular exams with raw scores of 142 and 165 have z scores of ?0.5 and 0.955, respectively. Find the mean of the distribution.

Using the weighted mean, find the average number of grams of fat per ounce of meat or fish that a person would consume over a 5-day period if he ate these: \(\begin{array}{ll}{\text { Meat or fish }} & {\text { Fat }(\mathrm{g} / \mathrm{oz})} \\ \hline 3 \text { oz fried shrimp } & {3.33} \\ {3 \text { oz veal culet (broiled) }} & {3.00} \\ {2 \mathrm{oz} \text { roast beef (lean) }} & {2.50} \\ {2.5 \text { oz fried chicken drumstick }} & {4.40} \\\ {4 \text { oz tuna (canned in oil) }} & {1.75}\end{array}\)

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