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Chapter 12: Problem 55

It is expected that the female population in a city will double in two decades. a. Explain why this is possible for a growth rate of \(3.6 \%\) a year. (Hint: What does \((1.036)^{20}\) equal?) b. You might think that a growth rate of \(5 \%\) a year would result in \(100 \%\) growth (i.e. the female population doubles) over two decades. Explain why a growth rate of \(5 \%\) a year would actually cause the female population to multiply by 2.65 over two decades.

Short Answer

Expert verified
3.6% growth doubles population in 20 years, as \((1.036)^{20} \approx 2.03\). 5% growth results in multiplying by 2.65, \((1.05)^{20} \approx 2.653\), due to exponential growth.

Step by step solution

01

Understanding the Doubling Formula

The doubling of a population over time can be calculated using the formula for compound interest. If something grows by a fixed percentage every year, the final amount becomes the initial amount multiplied by \[(1 + r)^t\]where \(r\) is the growth rate expressed as a decimal and \(t\) is the time in years.
02

Calculating Doubling with 3.6% Growth Rate

For part a, we need to show that the female population doubles in 20 years at a growth rate of 3.6% per year. This means \(r = 0.036\) and \(t = 20\). Therefore, the calculation becomes \[(1.036)^{20}\].
03

Performing the Calculation for 3.6%

Calculate \[(1.036)^{20}\] which is approximately \[2.03\]. This shows that the population will double, since \[2.03\] is approximately equal to doubling (which is 2). Thus, a 3.6% growth rate is sufficient for doubling over 20 years.
04

Understanding the Exponential Growth

For part b, rather than growth effects accumulating linearly, they compound, i.e., each year's 5% growth is based on the previous year's total. This compounding effect significantly increases the population.
05

Calculating Population with 5% Growth Rate

With a growth rate of 5%, \(r = 0.05\) and \(t = 20\). The equation becomes \[(1.05)^{20}\].
06

Performing the Calculation for 5%

Calculate \[(1.05)^{20}\], which equals approximately \[2.653\]. This means the population increases by this factor, not just doubling (which would be 2). Hence, it doesn't just double but increases by about 165% (as 2.65 times the original population).

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Key Concepts

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

Compound Interest Formula
The concept of exponential growth in population is closely tied to the compound interest formula. This formula is essential to understand when calculating how populations change over time due to consistent growth rates.
At its core, the formula is expressed as \((1 + r)^t\), where:
  • \(r\) is the growth rate, expressed as a decimal.
  • \(t\) represents the time in years.
This formula shows how, with each passing period, the interest is calculated not just on the initial principal but also on the accumulated interest from previous periods. In population terms, this means every new individual added in a year contributes to further growth in subsequent years. Thus, populations experience exponential growth鈥� the rate of increase itself increases over time due to accumulated growth.
Population Doubling Time
Population doubling time refers to the period it takes for a given population to double in size. When dealing with consistent growth rates, the compound interest formula helps in determining this doubling time.
For example, with a growth rate of 3.6%, we use the formula \((1 + 0.036)^{20}\). Calculating \((1.036)^{20}\) yields approximately 2.03, which means the population has doubled (since doubling equals a factor of 2). The formula allows us to see that with a 3.6% growth rate, a population will double in about 20 years. This is an insightful calculation, especially when making projections for future population sizes.
Growth Rate Calculation
Understanding growth rates is essential for predicting how quickly populations expand over time. When you have a higher annual growth rate, the compounding effect becomes more pronounced.
Given a growth rate of 5% per year, expressed as \(r = 0.05\), we analyze the compounding over 20 years using \((1.05)^{20}\). Calculating \((1.05)^{20}\) gives us approximately 2.653, far exceeding a simple doubling.
This showcases the effect of compounding, where each year's growth builds upon the last. Therefore, a population with a 5% growth rate over two decades doesn't merely double; it multiplies by more than 2.5 times, demonstrating the powerful effect of exponential growth.

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Most popular questions from this chapter

Higher income with experience Suppose the regression line \(\mu_{y}=-10,000+9500 x\) models the relationship for the population of working adults in a country between \(x=\) experience (in years) and the mean of \(y=\) annual income (in US dollars). The conditional distribution of \(y\) at each value of \(x\) is modeled as normal with \(\sigma=6500 .\) Use this regression model to describe the mean and the variability around the mean for the conditional distribution at an experience of (a) 5 years and (b) 10 years.

For the Georgia Student Survey data file on the book's website, the correlation between daily time spent watching TV and college GPA is -0.35 . a. Interpret \(r\) and \(r^{2}\). Use the interpretation of \(r^{2}\) that (i) refers to the prediction error and (ii) the percent of variability explained. b. One student is 2 standard deviations above the mean on time watching TV. (i) Would you expect that student to be above or below the mean on college GPA? Explain (ii) How many standard deviations would you expect that student to be away from the mean on college GPA? Use your answer to explain "regression toward the mean."

In the first round of a golf tournament, five players tied for the lowest round, at \(65 .\) The mean score of all players was \(75 .\) If the mean score of all players is also 75 in the second round, what does regression toward the mean suggest about how well we can expect the five leaders to do, on the average, in the second round?

Assumption violated For prediction intervals, an important inference assumption is a constant standard deviation \(\sigma\) of \(y\) values at different \(x\) values. In practice, the standard deviation often tends to be larger when \(\mu_{y}\) is larger. a. Sketch a hypothetical scatterplot for which this happens, using observations on \(x=\) family income and \(y=\) amount donated to charity. b. Explain why a \(95 \%\) prediction interval would not work well at very small or at very large \(x\) values.

A study of 375 women who lived in pre-industrial Finland (by S. Helle et al., Science, vol. \(296,\) p. 1085,2002 ), using Finnish church records from 1640 to \(1870,\) found that there was roughly a linear relationship between \(y=\) life length (in years) and \(x=\) number of sons the woman had, with a slope estimate of \(-0.65(s e=0.29)\) a. Interpret the sign of the slope. Is the effect of having more boys good or bad? b. Show all steps of the test of the hypothesis that life length is independent of number of sons for the twosided alternative hypothesis. Interpret the P-value. c. Construct a \(95 \%\) confidence interval for the true slope. Interpret. Is it plausible that the effect is relatively weak, with true slope near \(0 ?\)
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