Question

if the density function for a uniform distribution is nonzero on the interval (0,16), then the mean and variance of the distribution are
Select one:
0
O a. 12,6
O b. 6,12
O c. 4/3,16
O d. 8,4/3​

Answer 1

Answer:

THE ANSWER IS:-

(d) 4/3,16

I HOPE YOU UNDERSTOOD THE QUESTION!!!

Answer 2

Answer:

option (D) 8, 64 /3

Step-by-step explanation:

General density function:

f(x) = 1 / b ; for x in [0, b], and it is 0 elsewhere.  

(since the full area under the curve = 1 for a probability density function)

Since it's uniform distribution:

Mean = b / 2 (since the distribution will be symmetric).

If X is a random variable with this distribution, then:

E(X) = b/2 (the expected value of X is b/2).

For b = 16:

E(X)= 16 / 2 = 8

For the variance:

Var (X) = E( X²) - ( E( X) )².

To get E (X²):

E (X²) = ∫ x² (1 / b) dx  (integrating from 0 to b)

= (1 / b) ∫ x² dx  (integrating from 0 to b)

= (1 / b) [ x³ / 3 ]   (evaluating from 0 to b)

= (1 / b) (b³/3) = b²/3.

Now we plug the values:

Var (X) = E(X²) - ( E(X) )²

= b²/3 - ( b/2 )²

= b²/3 - b²/4

= b² / 12.

For b = 16:

Var (X) = (16)² / 12.

Var (X) = 64 / 3.

Therefore, option (D) 8, 64 /3 is the correct option.

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