Answer:
a) Mean of X = 0.40
Variance of X = 0.24
b) Y is a Bernoulli's distribution. Check Explanation for reasons.
c) Mean of Y = 0.80 points
Variance of Y = 0.96
Step-by-step explanation:
a) The probability that play is successful is 0.40. Hence, the probability that play isn't successful is then 1 - 0.40 = 0.60.
Random variable X represents when play is successful or not, X = 1 when play is successful and X = 0 when play isn't successful.
The probability mass function of X is then
X | Probability of X
0 | 0.60
1 | 0.40
The mean is given in terms of the expected value, which is expressed as
E(X) = Σ xᵢpᵢ
xᵢ = each variable
pᵢ = probability of each variable
Mean = E(X) = (0 × 0.60) + (1 × 0.40) = 0.40
Variance = Var(X) = Σx²p − μ²
μ = mean = E(X) = 0.40
Σx²p = (0² × 0.60) + (1² × 0.40) = 0.40
Variance = Var(X) = 0.40 - 0.40² = 0.24
b) If the conversion is successful, the team scores 2 points; if not the team scores 0 points. If Y ia the number of points that team scores.Y can take on values of 2 and 0 only.
A Bernoulli distribution is a discrete distribution with only two possible outcomes in which success occurs with probability of p and failure occurs with probability of (1 - p).
Since the probability of a successful conversion and subsequent 2 points is 0.40 and the probability of failure and subsequent 0 point is 0.60, it is evident that Y is a Bernoulli's distribution.
The probability mass function for Y is then
Y | Probability of Y
0 | 0.60
2 | 0.40
c) Mean and Variance of Y
Mean = E(Y)
E(Y) = Σ yᵢpᵢ
yᵢ = each variable
pᵢ = probability of each variable
E(Y) = (0 × 0.60) + (2 × 0.40) = 0.80 points
Variance = Var(Y) = Σy²p − μ²
μ = mean = E(Y) = 0.80
Σy²p = (0² × 0.60) + (2² × 0.40) = 1.60
Variance = Var(Y) = 1.60 - 0.80² = 0.96
Hope this Helps!!!
