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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.
<h3>What is the discrete probability distribution?</h3>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:
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 <u>sum of each value multiplied by it's respective probability</u>, 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
Answer:
see explanation
Step-by-step explanation:
Given
- [ -
c + 6c + 1 ] - 3c
Distribute the square bracket by -
= c - 3c -
- 3c ← combine like terms
= c - 6c -
= c -
c -
= - c -