Auniform distribution is defined over the interval from 6 to 10. a. what are the values for a and b? b. what is the mean of this uniform distribution? c. what is the standard deviation? d. show that the total area is 1.00 (use the area formula) e. find the probability of a value more than 7 f. find the probability of a value between 7 and 9

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

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb