Porisma

Worked solution

Fit a bounded growth model

A scientist is studying the growth of a bacteria colony in a petri dish. He places bacteria at one spot of the dish and he measures the area AA the bacteria cover on day tt. His results are recorded below:

Time, tt (days)01234567
Area, AA (cm2\mathrm{cm}^2)1.005.0027.1041.3946.9549.9551.0752.05

(i) Draw a scatter diagram to illustrate the data. mark

(ii) The scientist would like to predict the future growth of the bacteria. Explain why, in this context, a linear model is not likely to be appropriate. mark

It is proposed to fit a model of the form ln⁡(D−A)=a+bt\ln(D-A)=a+bt, where DD is a suitable constant. The product moment correlation coefficient between tt and ln⁡(D−A)\ln(D-A) is denoted by rr. The following table gives values of rr for some possible values of DD.

DD535455
rr−0.99132−0.98767

(iii) Calculate the value of rr for D=53D=53, giving your answer correct to 5 decimal places. Hence, explain which value of DD is the most appropriate. marks

(iv) With the value of DD in (iii), calculate the values of aa and bb correct to 5 decimal places, and use them to predict the area covered by the bacteria after 20 days. marks

(v) Give an interpretation, in context, of the value of DD. mark

[2013/CJC/II/11]

CJC · Review problems · Question 5

Answer

D=53D=53, r=−0.99349r=-0.99349; a=4.26272a=4.26272, b=−0.60899b=-0.60899; predicted area after 20 days 53.0 cm253.0\ \mathrm{cm}^2.

(i) Sketch the data

Plot time horizontally and area vertically using all eight pairs. The initially rapid increase levels off as the area approaches a limiting value; see the comparison sketch.

How did this part go?

(ii) Use the physical constraint

A straight line with a positive slope would predict unbounded area as time increases. The petri dish has finite area, so such a model is inappropriate for long-term predictions.

How did this part go?

(iii) Choose D

For D=53D=53, correlate tt and ln⁡(53−A)\ln(53-A). This gives r=−0.9934935112=−0.99349r=-0.9934935112=-0.99349 (5 d.p.). Of the three candidates, this is closest to −1-1, so D=53D=53 is preferred.

How did this part go?

(iv) Fit and predict

Regress ln⁡(53−A)\ln(53-A) on tt: a=4.262717499a=4.262717499, b=−0.6089913903b=-0.6089913903. To 5 d.p., a=4.26272a=4.26272, b=−0.60899b=-0.60899. Using these requested rounded coefficients, A=53−e4.26272−0.60899(20)=52.99963553≈53.0 cm2A=53-e^{4.26272-0.60899(20)}=52.99963553\approx53.0\ \mathrm{cm}^2.

How did this part go?

(v) Interpret the limiting value

Since b<0b<0, A=D−ea+bt→DA=D-e^{a+bt}\to D from below as t→∞t\to\infty. In the source context, D=53 cm2D=53\ \mathrm{cm}^2 represents the maximum available area of the petri dish.

How did this part go?

Compare your sketch

Expand diagram

Use the scrollbars or arrow keys to inspect details.

Scatter diagram for Question 5: Time t (days) horizontally, Area A (cm²) vertically. All 8 pairs from the question table are plotted.
Scatter diagram for Question 5: Time t (days) horizontally, Area A (cm²) vertically. All 8 pairs from the question table are plotted.
Scatter diagram for Question 5: Time t (days) horizontally, Area A (cm²) vertically. All 8 pairs from the question table are plotted.

Common mistakes

53.0 is rounded: the fitted area remains below D. The day 20 prediction also extrapolates beyond the observed 0–7 days.

Solution notes

The printed solution repeats label(i) for the response to (ii); student working uses the intended part. Its “r closest to 1” alternative is |r| closest to 1 in the visual source, not positive r.

Read the hint