The time a student spends in the shower on any school day is an exponential random variable with a mean of 298 seconds.
a)What is the probability that the student spends more than 310 seconds in the shower on one day?
b) What is the probability that the student spends an average of over 310 seconds in the shower per school day, taken over one calendar year (assume that one calendar year has 190 school days).
Don't know if my answers are correct, but I got for (a)0.484 and for (b) 0.555 (extremely skeptical of this answer)
What did you get so we may reference each others answers and help each other out? :)
correction, my hopefully correct yet still sketchy answer for (b) is now 0.342.
To solve these questions, we need to use the properties of exponential random variables.
An exponential random variable is characterized by a rate parameter λ, which corresponds to the mean time between events. In this case, the mean time spent in the shower is 298 seconds, so the rate parameter λ is the inverse of this value: λ = 1/298.
a) To find the probability that the student spends more than 310 seconds in the shower on one day, we can use the cumulative distribution function (CDF) of the exponential distribution.
The CDF of an exponential distribution with rate parameter λ is given by the equation:
CDF(x) = 1 - e^(-λx)
Substituting λ = 1/298 and x = 310 into the equation, we get:
CDF(310) = 1 - e^(-1/298 * 310)
Calculating this value, we find that CDF(310) ≈ 0.096.
However, we are interested in the probability of spending more than 310 seconds, which is equal to 1 - CDF(310). Therefore, the probability that the student spends more than 310 seconds in the shower on one day is approximately 1 - 0.096 = 0.904.
b) To find the probability that the student spends an average of over 310 seconds in the shower per school day, taken over one calendar year (190 school days), we need to calculate the probability of the sample mean being greater than 310 seconds.
The mean of the exponential distribution is given by 1/λ. Therefore, the mean time spent in the shower for 190 school days is equal to 298 * 190 seconds.
The standard deviation of the sample mean is given by σ/√n, where σ is the standard deviation of the individual observations (which is equal to the mean) and n is the sample size. In this case, σ = 298 and n = 190.
To find the probability that the sample mean is greater than 310 seconds, we can use the Central Limit Theorem to approximate the distribution of the sample mean as a normal distribution.
The z-score formula can be used to calculate the probability:
P(X > 310) = P[(X - μ) / (σ/√n) > (310 - μ) / (σ/√n)]
Substituting the values into the formula, we get:
P(X > 310) = P(Z > (310 - (298 * 190)) / (298 / √190))
Calculating this value gives us the probability that the sample mean is greater than 310 seconds.
Note: This calculation involves finding the area under the standard normal distribution curve to the right of the calculated z-score. A standard normal table or calculator can be used to find this probability.
Keep in mind that this is an approximate calculation based on the Central Limit Theorem, which assumes that the sample size is large enough (in this case, 190) for the sample mean to be normally distributed.