Question

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Answer 1

Using a discrete probability distribution, it is found that:

a) There is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.

b) There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.

c) The expected value is of 1.3 lawns mowed on a randomly selected day.

What is the discrete probability distribution?

Researching the problem on the internet, it is found that the distribution for the number of lawns mowed on a randomly selected dayis given by:

• P(X = 0) = 0.2.

• P(X = 1) = 0.4.

• P(X = 2) = 0.3.

• P(X = 3) = 0.1.

Item a:

P(X = 2) = 0.3, hence, there is a 0.3 = 30% probability that he will mow exactly 2 lawns on a randomly selected day.

Item b:

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2 = 0.8$

There is a 0.8 = 80% probability that he will mow at least 1 lawn on a randomly selected day.

Item c:

The expected value of a discrete distribution is given by the sum of each value multiplied by it's respective probability, hence:

E(X) = 0(0.2) + 1(0.4) + 2(0.3) + 3(0.1) = 1.3.

The expected value is of 1.3 lawns mowed on a randomly selected day.

More can be learned about discrete probability distributions at brainly.com/question/24855677