Let the random variable X be equally likely to assume any of the values 1/14, 1/7, or 3/14. Determine the mean and variance of X. The mean is Enter your answer; The mean is 1/7 [Give exact answer in form of fraction.] The variance is Enter your answer; The variance is 0.1190 [Round your answer to four decimal places (e. g. 98.7654).]

1/7 ;

0.0034

Step-by-step explanation:

Given that:

X = 1/14, 1/7, 3/14

n = number of observations = 3

Since, probability of X is equally likely ;

P(X) = Σp(x) = 1 ; p(x) = 1/n = 1/3

The mean, m ;

Σx*p(x) = (1/3 * 1/14) + (1/3 * 1/7) + (1/3 * 3/14)

= 1/42 + 1/21 + 3/42

= (1 + 2 + 3) / 42

= 6 / 42

= 1/7

The variance :

Σx²*p(x) - m²

((1/3) * (1/14)^2) + ((1/3) * (1/7)^2) + ((1/3) * (3/14)^2) - (1/7)^2

= 0.0238095 - 0.0204081

= 0.0034014

= 0.0034

the answer is a

step-by-step explanation:

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: