Question

Suppose that random variable X is uniformly distributed between 7 and 23. Draw a graph of the density function, and then use it to heip find the following probabilities:

A. P(X>23)= ________
B. P(X<11.7)= __________
C. P(9D. P(16

Answer 1

To find the probabilities, we need to use the formula for the probability density function (pdf) for a uniform distribution. The pdf is given by f(x) = 1/(b-a) for a ≤ x ≤ b, where a and b are the lower and upper limits of the distribution. The probabilities are: A. P(X > 23) = 0, B. P(X < 11.7) = 0.29375, C. P(9 ≤ X ≤ 16) = 0.4375, and D. P(X = 16) = 0.

Step-by-step explanation:

To find the probabilities, we need to use the formula for the probability density function (pdf) for a uniform distribution. The pdf is given by:

f(x) = 1/(b-a) for a ≤ x ≤ b

where a and b are the lower and upper limits of the distribution.

In this case, a = 7 and b = 23. So the pdf is:

f(x) = 1/16 for 7 ≤ x ≤ 23

A. P(X > 23)

Since the random variable X is uniformly distributed between 7 and 23, the probability of X being greater than 23 is 0, because X can only take values between 7 and 23. Therefore, P(X > 23) = 0.

B. P(X < 11.7)

To find the probability that X is less than 11.7, we need to calculate the area under the pdf curve from 7 to 11.7. This area is given by:

P(X < 11.7) = (11.7-7) * (1/16) = 0.29375

C. P(9 ≤ X ≤ 16)

To find the probability that X is between 9 and 16, we need to calculate the area under the pdf curve from 9 to 16. This area is given by:

P(9 ≤ X ≤ 16) = (16-9) * (1/16) = 0.4375

D. P(X = 16)

Since the random variable X is a continuous distribution, the probability of X being equal to 16 is 0, because the pdf is defined as a constant function between 7 and 23. Therefore, P(X = 16) = 0.