Chapter 1.2, Problem 87E

a.

To determine

Verify the Population Statistics.

Answer to Problem 87E

The model fit the data well.

Explanation of Solution

Given information:

The table shows the life expectancies of a child (at birth) in the United States for selected years from $1930$ through $2000$ .

  $\begin{array}{cc}Year& \begin{array}{l}Life\\ Expec\tan cy\left(y\right)\end{array}\\ 1930& 59.7\\ 1940& 62.9\\ 1950& 68.2\\ 1960& 69.7\\ 1970& 70.8\\ 1980& 73.7\\ 1990& 75.4\\ 2000& 76.8\end{array}$

A model for the life expectancy during this period is

  $\begin{array}{l}y=-0.002t^2+0.50t+46.6\\ 30\le t\le 100\end{array}$

Where $y$ represents the life expectancy and $t$ is the time in years, with $t=30$ corresponding to $1930$ .

Use a graphing utility to graph the data from the table and the model in the same viewing window. How well does the model fit the data? Explain.

Calculation:

The graph of data from the given table,

  

From the graph it is clear that model fit the data well.

Hence the model fit the data well.

b.

To determine

Verify the Population Statistics.

Answer to Problem 87E

The life expectancy in $1990$ is $75.4$ years.

Explanation of Solution

Given information:

The table shows the life expectancies of a child (at birth) in the United States for selected years from $1930$ through $2000$ .

  $\begin{array}{cc}Year& \begin{array}{l}Life\\ Expec\tan cy\left(y\right)\end{array}\\ 1930& 59.7\\ 1940& 62.9\\ 1950& 68.2\\ 1960& 69.7\\ 1970& 70.8\\ 1980& 73.7\\ 1990& 75.4\\ 2000& 76.8\end{array}$

A model for the life expectancy during this period is

  $\begin{array}{l}y=-0.002t^2+0.50t+46.6\\ 30\le t\le 100\end{array}$

Where $y$ represents the life expectancy and $t$ is the time in years, with $t=30$ corresponding to $1930$ .

Determine the life expectancy in $1990$ both graphically and algebraically.

Calculation:

The graph shows that line parallel to $y-axis$ intersect the graph on point $75.4$ which relate to $1990$ of data from the given table,

  

Now verify algebraically put $t=90$ in the equation,

  $\begin{array}{l}y=-0.002{\left(90\right)}^2+0.50\left(90\right)+46.6\\ =-0.002\times 8100+45+46.6\\ =-16.2+45+46.6\\ =75.4years\end{array}$

Hence the life expectancy in $1990$ is $75.4$ years.

c.

To determine

Verify the Population Statistics.

Answer to Problem 87E

The year when life expectancy $76.0$ is $1994$ .

Explanation of Solution

Given information:

The table shows the life expectancies of a child (at birth) in the United States for selected years from $1930$ through $2000$ .

  $\begin{array}{cc}Year& \begin{array}{l}Life\\ Expec\tan cy\left(y\right)\end{array}\\ 1930& 59.7\\ 1940& 62.9\\ 1950& 68.2\\ 1960& 69.7\\ 1970& 70.8\\ 1980& 73.7\\ 1990& 75.4\\ 2000& 76.8\end{array}$

A model for the life expectancy during this period is

  $\begin{array}{l}y=-0.002t^2+0.50t+46.6\\ 30\le t\le 100\end{array}$

Where $y$ represents the life expectancy and $t$ is the time in years, with $t=30$ corresponding to $1930$ .

Use the graph to determine the year when life expectancy was approximately $76.0$ . Verify your answer algebraically.

Calculation:

The graph shows the year when life expectancy was approximately $76.0$ which is $1994$ .

  

Hence the year when life expectancy $76.0$ is $1994$ .

d.

To determine

Verify the Population Statistics.

Answer to Problem 87E

The projection given by model $77.65$ is less than $78.9$ .

Explanation of Solution

Given information:

The table shows the life expectancies of a child (at birth) in the United States for selected years from $1930$ through $2000$ .

  $\begin{array}{cc}Year& \begin{array}{l}Life\\ Expec\tan cy\left(y\right)\end{array}\\ 1930& 59.7\\ 1940& 62.9\\ 1950& 68.2\\ 1960& 69.7\\ 1970& 70.8\\ 1980& 73.7\\ 1990& 75.4\\ 2000& 76.8\end{array}$

A model for the life expectancy during this period is

  $\begin{array}{l}y=-0.002t^2+0.50t+46.6\\ 30\le t\le 100\end{array}$

Where $y$ represents the life expectancy and $t$ is the time in years, with $t=30$ corresponding to $1930$ .

One projection for the life expectancy of a child born in $2015$ is $78.9$ . How does this compare with the projection given by the model?

Calculation:

For the life expectancy of a child born in $2015$ put $t=115$

  $\begin{array}{l}y=-0.002{\left(115\right)}^2+0.50\left(115\right)+46.6\\ =-0.002\times 13225+57.5+46.6\\ =-26.45+57.5+46.6\\ =77.65years\end{array}$

Hence the projection given by model $77.65$ is less than $78.9$ .

e.

To determine

Verify the Population Statistics.

Answer to Problem 87E

Yes this model can be used to predict the life expectancy of a child $50$ years from now.

Explanation of Solution

Given information:

The table shows the life expectancies of a child (at birth) in the United States for selected years from $1930$ through $2000$ .

  $\begin{array}{cc}Year& \begin{array}{l}Life\\ Expec\tan cy\left(y\right)\end{array}\\ 1930& 59.7\\ 1940& 62.9\\ 1950& 68.2\\ 1960& 69.7\\ 1970& 70.8\\ 1980& 73.7\\ 1990& 75.4\\ 2000& 76.8\end{array}$

A model for the life expectancy during this period is

  $\begin{array}{l}y=-0.002t^2+0.50t+46.6\\ 30\le t\le 100\end{array}$

Where $y$ represents the life expectancy and $t$ is the time in years, with $t=30$ corresponding to $1930$ .

Do you think this model can be used to predict the life expectancy of a child $50$ years from now? Explain.

Calculation:

Yes this model can be used to predict the life expectancy of a child $50$ years from now.

Hence this model can be used to predict the life expectancy of a child $50$ years from now.