Question

6.32 Bacteria on a Petri Dish. 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 of the center of the first spot to appear from the center of the petri dish is a variable with density curve $y=x/2$for $0<x<2$, and $y=0$ otherwise.
a. Graph the density curve of this variable.
b. Show that the area under this density curve to the left of any number $x$between 0 and 2 equals $x^2/4$.
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 the density curve of this variable as:

(b) Shown that the area under this density curve to the left of any number $x$between 0 and 2 equals $x^2/4$.

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

(d) The percentage of the time the distance of the center of the first spot to appear from the center of the petri dish between $0.25$ inch and $1.5$ inches is $54.6875\%$.

(e) The percentage of the time the distance of the center of the first spot to appear from the center of the petri dish more than $0.75$inch is $85.9375\%$.

Step-by-step solution

Part (a) Step 1: Given information

To graph the density curve of this variable.

Part (a) Step 2: Explanation

With the density curve, the distance between the center of the first spot to form and the center of the petri dish is a variable$y=\frac{x}{2}$for $0<x<2$, and $y=0$.
The probability distribution curve of a normal random variable is called a normal curve. A normal distribution is represented graphically in this graph. If $X$is a continuous random variable with a mean of $\mu$and a standard deviation of $\sigma$, then a normal curve with random variable $X$ has the following equation:
$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$
Furthermore, a normal curve with random variable $Z$ has the following equation.
$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}},$
where $Z$ has a mean of $0$ and a standard deviation of $1$accordingly.

Part (a) Step 3: Explanation

A normal curve typically includes two population parameters: population mean $\mu$and population standard deviation $\sigma$.
$f\left(y\right)=\left\{\begin{array}{l}\frac{x}{2},0<x<2\\ 0,\text{otherwise}\end{array}\right.$
The curve is calculated as follows:

Part (b) Step 1: Explanation

To show that the area under this density curve to the left of any number $x$between 0 and 2 equals $x^2/4$.

Part (b) Step 2: Explanation

Between $0$and $x$, the area under the density curve is a triangle with a base of $x$and a height of $\frac{x}{2}$. As a result, a triangle's area equals half of the product of its base and height.
$A=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times x\times \frac{x}{2}$

$=\frac{x^2}{4}$

The curve is calculated as follows:

Part (c) Step 1: Given information

To find the percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish at most 1 inch.

Part (c) Step 2: Explanation

According from part (b):

The area to the left of $x$is $\frac{x^2}{4}$
So, the area to the left of 1 inch is:
$=\frac{(1{)}^2}{4}$

$=\frac{1}{4}$

$=0.25$

$=25\%$

As a result, the percentage of the time the distance of the center of the first spot to appear from the center of the petri dish at most 1 inch is $25\%$.

Part (d) Step 1: Given information

To find the percentage of the time the distance of the center of the first spot to appear from the center of the petri dish between 0.25 inch and 1.5 inches.

Part (d) Step 2: Explanation

According to part (b):

The area to the left of $x$is $\frac{x^2}{4}$
So, the area to the left of $0.25$ inch is:
$=\frac{(0.25{)}^2}{4}$

$=\frac{0.0625}{4}$

$=0.015625$

Then the area to the left of $1.5$inches is determined as:

$=\frac{(1.5{)}^2}{4}$

$=\frac{2.25}{4}$

$=0.5625$

Hence, the area between $0.25$inch and $1.5$inches is calculated as follows:

$=0.5625-0.015625$

$=0.546875$

$=54.6875\%$

As a result, the percentage of the time the distance of the center of the first spot to appear from the center of the petri dish between $0.25$inch and $1.5$inches is $54.6875\%$.

Part (e) Step 1: Given information

To find the percentage of the time the distance of the center of the first spot to appear from the center of the petri dish more than $0.75$inch.

Part (e) Step 2: Explanation

According to part (b):

The area to the left of $x$is $\frac{x^2}{4}$
So, the area to the left of $0.75$ inch is calculated as:
$=\frac{(0.75{)}^2}{4}$

$=\frac{0.5625}{4}$

$=0.140625$

The probability to the left reduces the area to the right of $0.75$inch by 1.

Determine the area to the right of $0.75$inch as:

$=1-(0.140625)$

$=0.859375$

$=85.9375\%$

As a result, the percentage of the time the distance of the center of the first spot to appear from the center of the petri dish more than $0.75$inch is $85.9375\%$.