The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for personal holiday shopping. Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution.
Round your answers to four decimal places.
a.)The study reported that there is a .53 probability that a worker uses the office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to 5.5, 6, 6.6, or 7 hours?
b.)Using the mean time from part (a), what is the probability that a worker uses the office computer for holiday shopping more than 10 hours?
c.) What is the probability that a worker uses the office computer for holiday shopping between four and eight hours?
fmflsdf
To answer these questions, we need to use the exponential distribution, which is characterized by a parameter λ (lambda), representing the average number of events in a specific time period. In this case, it represents the average number of hours a worker spends doing holiday shopping on an office computer.
a.) To find the mean time spent using an office computer for holiday shopping, we can use the fact that the mean of an exponential distribution is equal to 1/λ. We need to find the value of λ that corresponds to a probability of 0.53 that a worker uses the office computer for 5 hours or less.
Let X be a random variable representing the number of hours spent on holiday shopping. We are given that P(X ≤ 5) = 0.53. The cumulative distribution function (CDF) of the exponential distribution is given by F(x) = 1 - e^(-λx).
At X = 5, we have F(5) = 1 - e^(-5λ) = 0.53. Solving this equation for λ will give us the average number of hours spent on holiday shopping.
0.53 = 1 - e^(-5λ)
0.47 = e^(-5λ)
ln(0.47) = -5λ
λ = -ln(0.47) / 5 ≈ 0.125
Therefore, the mean time spent using an office computer for holiday shopping is 1 / λ = 1 / 0.125 ≈ 8 hours.
To determine the closest option from the given choices, the mean time is closest to 6 hours.
b.) Using the mean time of 8 hours, we can find the probability that a worker uses the office computer for holiday shopping more than 10 hours by calculating P(X > 10).
The probability density function (PDF) of the exponential distribution is given by f(x) = λ * e^(-λx).
Therefore, P(X > 10) = 1 - P(X ≤ 10) = 1 - (1 - e^(-10λ)) = e^(-10 * λ) = e^(-10 * 0.125) ≈ 0.3499.
So, the probability that a worker uses the office computer for holiday shopping more than 10 hours is approximately 0.3499.
c.) The probability that a worker uses the office computer for holiday shopping between four and eight hours can be calculated by finding P(4 ≤ X ≤ 8).
Using the PDF of the exponential distribution, we can integrate f(x) over the range [4, 8] to obtain the desired probability:
P(4 ≤ X ≤ 8) = ∫[4, 8] λ * e^(-λx) dx
Performing the integration:
P(4 ≤ X ≤ 8) = [-e^(-λx)] [4, 8] = -e^(-8λ) + e^(-4λ)
Substituting the value of λ from part (a), we get:
P(4 ≤ X ≤ 8) = -e^(-8 * 0.125) + e^(-4 * 0.125) ≈ 0.4030
Therefore, the probability that a worker uses the office computer for holiday shopping between four and eight hours is approximately 0.4030.