The duration of a random smile (in seconds) of an 8-week-old baby can be modeled by a uniform distribution with a = 0 and b = 23 seconds. What is the probability a randomly chosen smile of an 8-week-old baby lasts between 2 and 18 seconds?
It is uniform? Then between 2 and 18 is 16/23 = .69565
To find the probability that a randomly chosen smile of an 8-week-old baby lasts between 2 and 18 seconds, you can calculate the area under the probability density function (PDF) of the uniform distribution between 2 and 18.
The PDF of a uniform distribution is given by:
f(x) = 1 / (b - a)
where a and b are the lower and upper bounds of the distribution, respectively.
In this case, a = 0 seconds and b = 23 seconds.
So, the PDF of the smile duration for the 8-week-old baby is:
f(x) = 1 / (23 - 0) = 1 / 23
To find the probability, you need to calculate the area under this PDF curve between 2 and 18 seconds.
The formula to calculate the area under a PDF curve is:
P(a ≤ X ≤ b) = ∫(a to b) f(x) dx
where X is the random variable representing the smile duration.
In this case, you need to calculate:
P(2 ≤ X ≤ 18) = ∫(2 to 18) (1 / 23) dx
Integrating the expression, you get:
P(2 ≤ X ≤ 18) = (1 / 23) * (18 - 2) = (1 / 23) * 16 = 16 / 23
Therefore, the probability that a randomly chosen smile of an 8-week-old baby lasts between 2 and 18 seconds is 16/23, or approximately 0.6957.