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 x2/4.
c. What percentage of the time is the distance of the center of the first spot to appear from the center of the petri dish

1. at most 1 inch?
2. between 0.25 inch and 1.5 inches?
3. more than 0.75 inch?

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batolisis batolisis · Answer 1 · 2021-03-20T08:51:34+01:00

Answer:

a) attached below

b) Area = 1/2 ( base ) (height )

= 1/2 ( x ) ( x/2)

= x^2 / 4

C) i)25% ii) 54.69% iii) 85.94%

Step-by-step explanation:

Given density curve y = x/2 for 0 < x < 2 and y = 0

a) Graph the density curve of this variable

attached below

b) show that the area under the density curve to the left of any number x between 0 and 2 equals x^2/4

base of triangle = x

height = x/2

Hence to show the area under the density curve is equal to x^2/4 we will calculate the area to the left

Area = 1/2 ( base ) (height )

= 1/2 ( x ) ( x/2)

= x^2 / 4

c) Determine the percentage of the time of the distance of the center of the first spot to appear from the center of the petri dish

i) At most 1 inch

(area to the left of 1 ) = 1^2 / 4

= 1/4 = 0.25 ≈ 25%

ii) between 0.25 inch and 1.5 inches?

( area between 0.25 and 1.5 ) = ( area to the left of 1.5 ) - ( Area to the left of 0.25 )

= (1.5^2 / 4 )- ( 0.25^2 / 4 )

= 0.5469 = 54.69%

iii) more than 0.75 inch

( Area to the right of 0.75 ) = 1 - (0.75^ / 4)

= 0.8594 = 85.94 %