Answer:
The probability is [tex]P(\$20,000 < X < \$35,000) = 0.6[/tex]
Step-by-step explanation:
From the question we are told that
Workers' salaries in your company are uniformly distributed between $15,000 and $40,000 per year
Generally the probability that a randomly chosen worker earns an annual salary between $20,000 and $35,000 is mathematically represented as
[tex]P(\$20,000 < X \$35,000) = \frac{ 35000 - 20000}{ 40000 - 150000}[/tex]
=> [tex]P(\$20,000 < X < \$35,000) = 0.6[/tex]
Using the uniform distribution, it is found that there is a 0.6 = 60% probability that a randomly chosen worker earns an annual salary between $20,000 and $35,000.
An uniform distribution has two bounds, a and b.
The probability of finding a value between c and d is:
[tex]P(c \leq X \leq d) = \frac{d - c}{b - a}[/tex]
In this problem, uniformly distributed between $15,000 and $40,000, hence [tex]a = 15, b = 40[/tex].
The desired probability is:
[tex]P(20 \leq X \leq 35) = \frac{35 - 20}{40 - 15} = 0.6[/tex]
0.6 = 60% probability that a randomly chosen worker earns an annual salary between $20,000 and $35,000.
A similar problem is given at https://brainly.com/question/13547683
