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We will assume that the smiling times of an eight-week-old baby, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. What is the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds

User Gabie
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Answer:

P ( 2<x<18) = 16/23

Explanation:

Since it is a uniform distribution the f(x) is given by

f(x) = 1/b-a a ≤x ≤b [0,23]

f(x) = 1/23-0 = 1/23

Now we have to find the probability that a randomly chosen eight-week-old baby smiles between two and 18 seconds

P ( 2<x<18) = base * height

= ( b-a) * 1/ b-a

= 18-2 *1/23

=16/23

The uniform distribution is called rectangular distribution because its total probability is confined to a rectangular region with base equal to (b-a) and height (1/b-a).

We will assume that the smiling times of an eight-week-old baby, in seconds, follow-example-1
User Kyuubi
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