The probability that a worker chosen at random works at least 8 hours is Option C: 0.84 approx.
<h3>How to evaluate the probability of a random variable getting at least some fixed value?</h3>
Suppose the random variable in consideration be X, and it is discrete.
Then, the probability of X attaining at least 'a' is written as:

It is evaluated as:

The probability distribution of X is:
x f(x) = P(X = x)
6 0.02
7 0.11
8 0.61
9 0.15
10 0.09
Worker working at least 8 hours means X attaining at least 8 as its values.
Thus, probability of a worker chosen at random working 8 hours is
P(X ≥ 8) = P(X = 8) + P(X = 9) +P(X = 10) = 0.85 ≈ 0.84 approx.
By the way, this probability distribution seems incorrect because sum of probabilities doesn't equal to 1.
The probability that a worker chosen at random works at least 8 hours is Option C: 0.84 approx.
Learn more about probability distributions here:
brainly.com/question/14882721
Answer:
3
Step-by-step explanation:
15
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12
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3
Answer:
D) f(x) = 6x
Step-by-step explanation:
84/6 is 14 could you give me more details to help you with the distributive property? In case you forgot it's like this 2(a+3)=2a+6
So he bought 8.5 pounds, and you have to find out how many ounces he bought. That would mean that you have to convert pounds to ounces. There are 16 ounces in 1 pound. In 8.5 pounds, there are 136 ounces.