Respuesta :
Answer:
(A) The percent of the problems takes more than 5 minutes to resolve is 60.87%.
(B) The problem solving time will be 50% as long as the difference between the two end points is 5.75.
(C) a = 0.50 minutes, b = 12 minutes.
(D) Mean = 6.25 minutes, Standard deviation = 3.32 minutes
Step-by-step explanation:
Let the random variable X represent the time it takes the technician to resolve the problem.
The random variable X follows an Uniform distribution with parameters a = 0.50 minutes and b = 12 minutes.
The probability density function of X is:
[tex]f_{X}(x)=\frac{1}{b-a};\ a<X<b[/tex]
(A)
Compute the probability that the problems takes more than 5 minutes to resolve as follows:
[tex]P(X>5)=\int\limits^{12}_{5}{\frac{1}{12-0.50}}\, dx[/tex]
[tex]=\frac{1}{11.50}\cdot \int\limits^{12}_{5}{1}\, dx \\\\=\frac{1}{11.50}\cdot [x]\limits^{12}_{5}\\\\=\frac{1}{11.50}\cdot [12-5]\\\\=\frac{7}{11.50}\\\\=0.608696\\\\\approx 0.6087[/tex]
Thus, the percent of the problems takes more than 5 minutes to resolve is 60.87%.
(B)
Let the middle 50% of the problem-solving times be between u and v.
Then,
P (u < X < v) = 0.50
[tex]\int\limits^{v}_{u}{\frac{1}{12-0.50}}\, dx=0.50\\\\\frac{1}{11.50}\cdot \int\limits^{v}_{u}{1}\, dx=0.50\\\\\frac{v-u}{11.50}=0.50\\\\(v-u)=5.75[/tex]
Thus, the problem solving time will be 50% as long as the difference between the two end points is 5.75.
(C)
The interval in which the technician can solve the problem is 30 seconds to 12 minutes.
So, the values of a and b in minutes is:
a = 30 seconds = 0.50 minutes
b = 12 minutes.
(D)
The mean time is:
[tex]\mu=\frac{a+b}{2}=\frac{0.50+12}{2}=6.25\ \text{minutes}[/tex]
The standard deviation of the time is:
[tex]\sigma=\sqrt{\frac{(b-a)^{2}}{12}}=\sqrt{\frac{(12-0.50)^{2}}{12}}=3.3197\approx 3.32\ \text{minutes}[/tex]
