math question

Uniform Probabilities

The uniform distribution is a continuous probability distribution and is concerned

with events that are equally likely to occur. When working out problems that have

a uniform distribution, be careful to note if the data is inclusive or exclusive.

Formula Review

x = is a real number between a and b

In some instances, x can take on the values a and b, if that happens then

a = smallest X; b = largest X

X ~ U (a, b), where a ≤ x ≤ b

The mean is 𝜇 = 𝑎+𝑏

2

The standard deviation is 𝜎 = √ (𝑏−𝑎)2

12

Probability density function: 𝑓(𝑥) = 1

𝑏−𝑎 where a ≤ x ≤ b

Cumulative density function: P(X ≤ x) = 𝑥−𝑎

𝑏−𝑎

Area to the Left of x: P(X < x) = (𝑥 − 𝑎) ( 1

𝑏−𝑎 ) =

𝑥−𝑎

𝑏−𝑎

Area to the Right of x: P(X > x) = (𝑏 − 𝑥) ( 1

𝑏−𝑎 ) =

𝑏−𝑥

𝑏−𝑎

Area Between c and d: P(c < x < d) = (base)(height) = (𝑑 − 𝑐) ( 1

𝑏−𝑎 ) =

𝑑−𝑐

𝑏−𝑎

Note: (d – c) is the base and ( 1

𝑏−𝑎 ) is the height

Example:

Researchers have developed a safe method for rapidly detecting anthrax spores in

powders and on surfaces. The method has been found to work well even when

there are very few anthrax sports in a powered specimen. Consider a powder

specimen has exactly 30 anthrax spores. Supposed that the number of anthrax

spores in the sample detected by the new method follows a uniform distribution

from 10 to 30. Find the following probabilities.

First, we see that a = 10 and b = 30.

1) Find f(x)

𝑓(𝑥) = 1

𝑏 − 𝑎

𝑓(𝑥) = 1

30 − 10 =

1

20 = .05

2) Find the mean and standard deviation

𝜇 = 𝑎 + 𝑏

2

𝜇 = 10 + 30

2 = 20

𝜎 = √ (𝑏 − 𝑎)2

12

𝜎 = √ (30 − 10)2

12 = √

400

12 = 5.7735

3) Find the probability that 22 or fewer anthrax spores are detected in the

powdered specimen.

P(X ≤ x) = 𝑥−𝑎

𝑏−𝑎

P(X ≤ 22) = 22−10

30−10 =

12

20 = .6

4) Find the probability that between 10 and 25 anthrax spores are detected in the

powdered specimen.

P(c < x < d) = (base)(height) = (𝑑 − 𝑐) ( 1

𝑏−𝑎 )

P(10 < x < 25) = (base)(height) = (25 − 10) ( 1

30−10 ) = 15 ∗ (

1

20 ) = .75

5) Find the probability that fewer than 13 anthrax spores are detected in the

powdered specimen.

P(X < x) = (𝑥 − 𝑎)( 1

𝑏−𝑎 )

P(X < 13) = (13 − 10) ( 1

30−10 ) = 3 (

1

20 ) = .15

6) Find the probability that more than 26 anthrax spores are detected in the

powdered specimen.

P(x > x) = (𝑏 − 𝑥)( 1

𝑏−𝑎 )

P(x > 26) = (30 − 26) ( 1

30−10 ) = 4 (

1

20 ) = .20

7) Find the probability that more than 19 anthrax spores are detected given that

12 anthrax spores have already been detected in the powdered specimen.

P(x > 19 | x > 12) = 𝑃(𝑥>19)

𝑃(𝑥>12)

From here, we will use this probability twice and then divide the two answers.

P(x > x) = (𝑏 − 𝑥)( 1

𝑏−𝑎 )

P(x > 19) = (30 − 19) ( 1

30−10 ) = 11 (

1

20 ) = .55

P(x > x) = (𝑏 − 𝑥)( 1

𝑏−𝑎 )

P(x > 12) = (30 − 12) ( 1

30−10 ) = 18 (

1

20 ) = .9

P(x > 19 | x > 12) = 𝑃(𝑥>19)

𝑃(𝑥>12) =

.55

.9 = .6111

8) 78% of all anthrax spores that are detected in the powdered specimen fall

below the 78th percentile. Find x.

P(X < x) = .78

Using this equation,

P(X < x) = 𝑥−𝑎

𝑏−𝑎

.78 = 𝑥−𝑎

𝑏−𝑎

.78 = 𝑥−10

30−10

.78 = 𝑥−10

20

.78*20 = x – 10

15.6 = x – 10

15.6 + 10 = x

25.6 = x

The 78th percentile of all anthrax spores detected by the powdered specimen is

25.6.

9) 33% of all anthrax spores that are detected in the powdered specimen fall

above the 33rd percentile. Find x.

P(X > x ) = .33

P(X > x) = 𝑏−𝑥

𝑏−𝑎

.33 = 𝑏−𝑥

𝑏−𝑎

.33 = 30−𝑥

30−10

.33 = 30−𝑥

20

.33*20 = 30 – x

6.6 – 30 = -x

-23.4 = - x

23.4 = x

The upper 33rd percentile of all anthrax spores detected by the powdered

specimen is 23.4.