Question

After scoring a touchdown, a football team may elect to attempt a two-point conversion, by running or passing the ball into the end zone. If successful, the team scores two points. For a certain football team, the probability that this play is successful is 0.40.
a.â Let X =1 if successful, X= 0 if not. Find the mean and variance of X.

b.â If the conversion is successful, the team scores 2 points; if not the team scores 0 points. Let Y be the number of points scored. Does Y have a Bernoulli distribution? If so, find the success probability. If not, explain why not.

c.â Find the mean and variance of Y.

Answer 1

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!!!