Question

A uniform distribution is defined over the interval from 6 to 10.

Answer 1

Answer:

Refer below

Step-by-step explanation:

a) a and b are the lower and higher values of the interval for which uniform distribution is defined.

Here a= 6 and b =10

b) Mean of the uniform distribution= (a+b)/2 = (6+10)/2 =8

Or int x (1/4) dx = x^2/8 = 8

c) Variance of the uniform distribution = (b^2-a^2)/12 = (100-64)/12

= 36/12 =3

Std dev = sq rt of 3 = 1.732

d) To find total area

PDF of the distribution = 1/(b-a) = 1/4, 6<x<10

Area = \int 6 to 10 of 1/4 dx

= x/4

Subtitute limits

= (10-6)/4 =1

So total area = 1

d)P(X>7) = int 7 to 10 of 1/4 dx = 3/4

e) P(7<x<9) = Int 7 to 9 of 1/4 dx = 2/4 = 1/2