Question

A petri dish is a small, shallow dish of thin glass or plastic, used especially for cultures in bacteriology. A 2-inch-radius petri dish, containing nutrients upon which bacteria can multiply, is smeared with a uniform suspension of bacteria. Subsequently, spots indicating colonies of bacteria appear. The distance (in inches) of the center of the first spot to appear from the center of the petri dish is a variable with density curve $y=\frac{x}{2}$ for 0<x<2, and y=0 otherwise

(a) Graph the density curve of this variable

(b) Why is this a density curve?

what percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish?

(c) at most 1 inch?

(d) between 0.25 inch and 1.5 inches?

(e) more than 0.75 inch?

Short answer

(a) The graph of the density function is:

(b) Thus, the area under this density curve to the left of any number x between 0 and 2equals $\frac{x^2}{4}$.

(c) $25\%$ percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish is at most 1 inch.

(d)$54.69\%$percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish is between 0.25 and 1.5inches.

(e)85.94% of the time is the distance of the center of the first spot to appear from the center of the petri dish is more than 0.75inches.

Step-by-step solution

Part (a) Step 1. Given Information.  

A variable has a density curve whose equation is $y=\frac{x}{2},0<x<2,y=0$.

Part (a) Step 2. Graph the density function.  

Graph the density function $y=\frac{x}{2},0<x<2,y=0$.

Part (b) Step 1. The area between 0,2

From part (a),

Base,$b=x$

Height,$h=\frac{x}{2}$

Thus the area of the rectangle is given by,

$$\begin{gathered} A=\frac{1}{2}\times x\times \left(\frac{x}{2}\right) \\ =\frac{x^2}{4} \end{gathered}$$

Part (c) Step 1. Find the percentage.  

Area to the left of $1=\frac{1^2}{4}$

role="math" localid="1652602718242" $$\begin{gathered} =\frac{1}{4} \\ =0.25 \end{gathered}$$

$25\%$percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish is at most1inch.

Part (d) Step 1. Find the percentage.

Area between $0.25,1.5=$[Area to the left of $1.5]-$ [Area to the left of $0.25]$

$$\begin{gathered} =\frac{1.5^2}{4}-\frac{0.{25}^2}{4} \\ =0.5469 \end{gathered}$$

$54.69\%$ percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish is between 0.25 and 1.5inches.

Part (e) Step 1. Find the percentage.  

[Area to the right of $0.75]=1-$[Area to the left of $0.75]$

role="math" $$\begin{gathered} =1-\frac{0.{75}^2}{4} \\ =0.8594 \end{gathered}$$

85.94% of the time is the distance of the center of the first spot to appear from the center of the petri dish is more than 0.75inches.