AAAAAAAAAAAAAA ORRRRRRRR DDDDDDDDDD
2 answers:
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Answer:
Option D
Step-by-step explanation:
Given that a producer ships boxes of produce to individual customers.(say x)
X is a normal random variable with mean = 36 and std dev =4 lbs
By definition of std normal variate we know that
is N(0,1)
75th percentile of std normal distribution is
0.675
Hence corresponding x would be
Option D matches with this value
Hence answer is option D
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