Question

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards (according to GolfWeek). Assume that the driving distance for these golfers is uniformly distributed over this interval. a. Give a mathematical expression for the probability density function of driving distance. b. What is the probability the driving distance for one of these golfers is less than 290 yards

Answer 1

Answer:

a) $f(x) = \frac{1}{25.9}$

b) 0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards

Step-by-step explanation:

Uniform probability distribution:

An uniform distribution has two bounds, a and b.

The probability of finding a value of at lower than x is:

$P(X < x) = \frac{x - a}{b - a}$

The probability of finding a value between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The probability of finding a value above x is:

$P(X > x) = \frac{b - x}{b - a}$

The probability density function of the uniform distribution is:

$f(x) = \frac{1}{b-a}$

The driving distance for the top 100 golfers on the PGA tour is between 284.7 and 310.6 yards.

This means that $a = 284.7, b = 310.6$.

a. Give a mathematical expression for the probability density function of driving distance.

$f(x) = \frac{1}{b-a} = \frac{1}{310.6-284.7} = \frac{1}{25.9}$

b. What is the probability the driving distance for one of these golfers is less than 290 yards?

$P(X < 290) = \frac{290 - 284.7}{310.6-284.7} = 0.2046$

0.2046 = 20.46% probability the driving distance for one of these golfers is less than 290 yards