Every day, the number of network blackouts has a distribution (probability mass function)
1 answer:
Expected Mean, E(X), is obtained by multiplying each pair of 
and its 
and add up the answers
E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4
The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²
E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16
Var(X) = 0.6 - 0.16 = 0.44
Translating these answers into the context we have
E(Y) = 0.4×500 = $200
Var(Y) = $110
E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4
The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²
E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16
Var(X) = 0.6 - 0.16 = 0.44
Translating these answers into the context we have
E(Y) = 0.4×500 = $200
Var(Y) = $110
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Answer:
Step-by-step explanation:
It is given that the regression equation
predicts an adult’s height (y) given the individual’s mother’s height (x1), his or her father’s height (x2), and whether the individual is male (x3 = 1) or female (x3 = 0).
The coefficient of x1 = 0.32 represents the increase in height of the child due to one inch increase in mother.
Similarly coefficient of x2 = 0.42 represents the increase in height of the child due to one inch increase in father.
and coefficient of x3 = 5.31 represents the increase in height of the child due to being a male than a female.
Answer: no
Step-by-step explanation: see attachment
Answer:
Answer is 0.25 to 0.5
Step-by-step explanation:
Let the linear function be
and exponential function be
x 0.25 0.5 0.75 1.0 1.25 1.5
y 0.25 0.5 0.75 1.0 1.25 1.5
Diff 0.25 0.25 0.25 0.25 0.25
e^x 1.29 1.65 2.12 2.72 3.49 4.48
Diff 0.36 0.47 0.60 0.77 0.99
Hence 0.36 to be nearer to 0.25, than other intervals
So answer is 0.25 to 0.5