Question

The probabilities that a player will get 4-9 questions right on a trivia quiz are shown belovw PLEASE FIND MEAN, VARIANCE, AND STANDARD DEVIATION X 4 56 789 P(X) 0.06 0.2 0.4 0.1 0.14 0.1

Answer 1

Mean of the probability distribution is given as

$\dpi{100} \small E(X)=\sum xP(X=x)$

Mean of the probability distribution that a player will get 4-9 questions right on a trivia quiz is

$\dpi{100} \small E(X)=4(0.06)+5(0.2)+6(0.4)+7(0.1)+8(0.14)+9(0.1)=6.36$

Variance of a probability distribution is given as

$\dpi{100} \small Var(X)=E(X^{2})-\left ( E(X) \right )^{2}$

$\dpi{100} \small E(X^{2})=\sum x^{2}P(X=x)$

$\dpi{100} \small E(X^{2})=4^{2}(0.06)+5^{2}(0.2)+6^{2}(0.4)+7^{2}(0.1)+8^{2}(0.14)+9^{2}(0.1)=42.32$

$\dpi{100} \small Var(X)=42.32-6.36^{2}=1.8704$

Variance that a player will get 4-9 questions right on a trivia quiz is 1.87.

Standard deviation is the square root of the variance. So the standard deviation that a player will get 4-9 questions right on a trivia quiz is $\dpi{100} \small \sqrt{1.8704}=1.37$

Mean = 6.36

Variance = 1.87

Standard deviation = 1.37