The Information Systems Audit and Control Association surveyed office workers to learn about the anticipated usage of office computers for personal holiday shopping (USA Today, November 11, 2009). Assume that the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution.
a. The study reported that there is a .53 probability that a worker uses an office computer for holiday shopping 5 hours or less. Is the mean time spent using an office computer for holiday shopping closest to 5.8, 6.2, 6.6, or 7 hours?
b. Using the mean time from part (a), what is the probability that a worker uses an office computer for holiday shopping more than 10 hours (to 4 decimals)?
c. What is the probability that a worker uses an office computer for holiday shopping between 4 and 8 hours (to 4 decimals)?
To answer these questions, we need to use the properties of the exponential distribution. The exponential distribution is commonly used to model the time between events that occur randomly and independently at a constant average rate.
In this case, the number of hours a worker spends doing holiday shopping on an office computer follows an exponential distribution.
a. We are given that the probability of a worker using an office computer for holiday shopping for 5 hours or less is 0.53. The exponential distribution is defined by its mean, denoted by μ, which represents the average time spent doing holiday shopping on an office computer. To find the mean, we can use the formula:
P(X ≤ x) = 1 - e^(-λx)
Where X is the random variable representing the time spent, λ is the rate parameter, and e is Euler's number (approximately 2.71828).
Given P(X ≤ 5) = 0.53, we can solve for λ:
0.53 = 1 - e^(-5λ)
Rearranging this equation, we have:
e^(-5λ) = 1 - 0.53
e^(-5λ) = 0.47
Taking natural logarithm on both sides:
-5λ = ln(0.47)
λ = -ln(0.47)/5
Now, to find the mean μ, we can use the relationship between the rate parameter λ and the mean:
μ = 1/λ
Therefore, the mean time spent using an office computer for holiday shopping is:
μ = 1/(-ln(0.47)/5) ≈ 6.6 hours
So, the closest option is 6.6 hours.
b. Now, using the mean time obtained in part (a) which is approximately 6.6 hours, we can find the probability that a worker uses an office computer for holiday shopping more than 10 hours. This can be calculated using the exponential distribution's survival function:
P(X > x) = e^(-λx)
Substituting the values:
P(X > 10) = e^(-6.6 * 10)
Calculating this, we find:
P(X > 10) ≈ 0.0088
Therefore, the probability that a worker uses an office computer for holiday shopping more than 10 hours is approximately 0.0088.
c. To find the probability that a worker uses an office computer for holiday shopping between 4 and 8 hours, we can subtract the probability of spending less than 4 hours from the probability of spending less than 8 hours:
P(4 < X < 8) = P(X < 8) - P(X < 4)
Using the exponential distribution's cumulative distribution function, we get:
P(X < 8) = 1 - e^(-6.6 * 8)
P(X < 4) = 1 - e^(-6.6 * 4)
Substituting the values and calculating, we find:
P(X < 8) - P(X < 4) ≈ 0.5143 - 0.0965 ≈ 0.4178
Therefore, the probability that a worker uses an office computer for holiday shopping between 4 and 8 hours is approximately 0.4178.