Answer:
d
Step-by-step explanation:
The expectation, E(3y +2) and variance, Var(3y+2) of the random variable are 13.4 and 19.44 respectively
<h3>How to determine the expectation and variance of a random variable?</h3>
The expectations or expected value E(y) of a random variable can be thought of as the “average” value of the random variable. It is also called its mean
By definition:
if y = ax + b
then E(y) = aE(x) + b
where a,b = constant
The variance V(y) of a random variable is the measure of spread for the distribution of a random variable that determines the degree to which the values of a random variable differ from the expected value
By definition
if y = ax + b
V(b) = 0
V(y) = V(ax) + V(b)
= a²V(x) + 0
where a,b = constant
Given: E(y)= 3.8 and Var(y)= 2.16
Calculate E( 3y +2) and Var( 3y+ 2)
E(3y +2) = 3E(y) + 2 since E(y) = 3.8
= 3×3.8 + 2
= 11.4+2
= 13.4
Var(3y+2) = 3²Var(y) + 0
= 9×2.16
= 19.44
Therefore, E(3y +2) is 13.4 and Var(3y+2) is 19.44
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Answer:
1 gallon = 4 quarts
20 gallon = (4×20) quarts=<u>80quarts</u>
<h3>C. is the right answer.</h3>
Answer:
Step-by-step explanation:
Given that a data set includes 109 body temperatures of healthy adult humans for which sample mean =98.2°F and s equals 0.62 degrees
Std error of mean =
df = 108
Critical value for 99% t =2.581
Margin of error = 2.581*std error = 0.1533
99% confidence interval = Mean ±0.1533
=
a) Since the confidence interval does not contain 98.6 degrees we reject the null hypothesis that mean = 98.6 degrees
b) confidence interval for population mean = 98.6±0.1533
=