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To study how recognition memory decreases with time, the following experiment was conducted. The subject read a list of 20 words slowly aloud, and later, at different time intervals, was shown a list of 40 words containing the 20 words that he or she had read. The percentage, [tex]P[/tex], of words recognized was recorded as a function of [tex]t[/tex], the time elapsed in minutes. The table below shows the averages for 5 different subjects. This is modeled by [tex]P = a \ln t + b[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
t, \text{min} & 5 & 15 & 30 & 60 & 120 & 240 \\
\hline
P \% & 73 & 61.7 & 58.3 & 55.7 & 50.3 & 46.7 \\
\hline
\end{tabular}
\][/tex]

[tex]\[
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
t , \text{min} & 480 & 720 & 1440 & 2880 & 5760 & 10080 \\
\hline
P \% & 40.3 & 38.3 & 29 & 24 & 18.7 & 10.3 \\
\hline
\end{tabular}
\][/tex]

(a) Find [tex]\ln t[/tex] for each value of [tex]t[/tex] in the table above, and then use linear regression on a calculator to estimate [tex]a[/tex] and [tex]b[/tex] in the linear fit [tex]P = a \ln t + b[/tex]. In the blanks below, enter the corresponding values for [tex]a[/tex] and [tex]b[/tex]:

[tex]a = \square[/tex] (round values to 4 decimal places)

[tex]b = \square[/tex] (round values to 4 decimal places)

(b) When does this model (in part a) predict that the subjects will recognize no words?

In [tex]\square[/tex] DAYS (1440 minutes = 1 day)

(c) When does this model (in part a) predict that the subjects will recognize all words?

In [tex]\square[/tex] SECONDS.

Sagot :

GinnyAnsweridn
### Part (a)

First, let's calculate the natural logarithm ([tex]\(\ln t\)[/tex]) for each value of [tex]\(t\)[/tex] using the given time intervals in minutes.

[tex]\[ \begin{aligned} &\ln 5 \approx 1.6094 \\ &\ln 15 \approx 2.7081 \\ &\ln 30 \approx 3.4012 \\ &\ln 60 \approx 4.0943 \\ &\ln 120 \approx 4.7875 \\ &\ln 240 \approx 5.4806 \\ &\ln 480 \approx 6.1738 \\ &\ln 720 \approx 6.5793 \\ &\ln 1440 \approx 7.2724 \\ &\ln 2880 \approx 7.9655 \\ &\ln 5760 \approx 8.6587 \\ &\ln 10080 \approx 9.2183 \\ \end{aligned} \][/tex]

After converting all [tex]\(t\)[/tex] values to [tex]\(\ln t\)[/tex] values, we proceed with linear regression to estimate the coefficients [tex]\(a\)[/tex] and [tex]\(b\)[/tex] for the linear model [tex]\(P = a \ln t + b\)[/tex].

From the linear regression, we find:
[tex]\[ a \approx -7.7867 \][/tex]
[tex]\[ b \approx 86.2831 \][/tex]

So, the values are:
[tex]\[ a = -7.7867 \][/tex]
[tex]\[ b = 86.2831 \][/tex]

### Part (b)

We need to find the time [tex]\(t\)[/tex] when [tex]\(P = 0\)[/tex]. We solve the equation [tex]\(0 = a \ln t + b\)[/tex]:

[tex]\[ 0 = -7.7867 \ln t + 86.2831 \][/tex]

Solving for [tex]\(\ln t\)[/tex]:

[tex]\[ \ln t = \frac{-b}{a} = \frac{-86.2831}{-7.7867} \approx 11.0796 \][/tex]

Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:

[tex]\[ t = e^{11.0796} \approx 64318.3 \text{ minutes} \][/tex]

To convert this time into days:

[tex]\[ \text{Days} = \frac{64318.3}{1440} \approx 45.0822 \][/tex]

Therefore, the model predicts that the subjects will recognize no words in about [tex]\(45.0822\)[/tex] days.

### Part (c)

To find the time [tex]\(t\)[/tex] when [tex]\(P = 100\)[/tex], solve the equation [tex]\(100 = a \ln t + b\)[/tex]:

[tex]\[ 100 = -7.7867 \ln t + 86.2831 \][/tex]

Solving for [tex]\(\ln t\)[/tex]:

[tex]\[ \ln t = \frac{100 - b}{a} = \frac{100 - 86.2831}{-7.7867} \approx -1.7645 \][/tex]

Now, converting [tex]\(\ln t\)[/tex] back to [tex]\(t\)[/tex]:

[tex]\[ t = e^{-1.7645} \approx 0.1718 \text{ minutes} \][/tex]

To convert this time into seconds:

[tex]\[ \text{Seconds} = 0.1718 \times 60 \approx 10.3062 \][/tex]

Therefore, the model predicts that the subjects will recognize all words in about [tex]\(10.3062\)[/tex] seconds.
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