Each magazine that Charlie reads takes an amount of time that follows an exponential distribution with mean 10 minutes. The number of magazines that Charlie reads on any given day has a Poisson distribution with mean 2 . Assume that each reading session always falls within a single day (he does not read past midnight).
Furthermore, suppose that the number of magazines that Charlie reads on different days are independent, and that the lengths of time he takes to read each magazine is also independent of each other.
For simplicity, assume that he only reads 1 magazine at a time and that there are 30 days in each given month.
Let T be the total number of minutes Charlie spends reading magazines in one month.
E(T) =
VAR(T) =
600
12000
part b?
about 3: Sn ~ N(720, 720), 99.73% of the realizations lie between 639 and 800 magazines/year -> P(3500 < Z < 3600) = 0
1) 600
2) 12000
3) 0 (not sure)
Part 3
is 0 the final answer
E(T) = E(Number of magazines read in a day) * E(Time to read one magazine)
Since the number of magazines read in a day follows a Poisson distribution with mean 2, E(Number of magazines read in a day) = 2.
Since the time to read one magazine follows an exponential distribution with mean 10 minutes, E(Time to read one magazine) = 10 minutes.
Therefore, E(T) = 2 * 10 = 20 minutes.
VAR(T) = VAR(Number of magazines read in a day) * VAR(Time to read one magazine)
Since the number of magazines read in a day follows a Poisson distribution with mean 2, VAR(Number of magazines read in a day) = 2.
Since the time to read one magazine follows an exponential distribution with mean 10 minutes, VAR(Time to read one magazine) = (10)^2 = 100.
Therefore, VAR(T) = 2 * 100 = 200.
To find the expected value (E(T)) and variance (VAR(T)) of the total number of minutes Charlie spends reading magazines in one month, we need to use the properties of the exponential and Poisson distributions.
E(T): Expected Value
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The expected value of a sum of random variables is equal to the sum of their individual expected values. Therefore, to find E(T), we need to compute the expected value of the number of magazines read in a day (M) and multiply it by the expected value of the time taken to read each magazine (X).
1. Expected value of the number of magazines read in a day (M):
Since the number of magazines Charlie reads on any given day follows a Poisson distribution with a mean of 2, the expected value is simply the mean. So E(M) = 2.
2. Expected value of the time taken to read each magazine (X):
Since the time taken to read each magazine follows an exponential distribution with a mean of 10 minutes, the expected value is equal to the mean. So E(X) = 10.
Now, to find E(T), we multiply the expected values of M and X: E(T) = E(M) * E(X) = 2 * 10 = 20 minutes.
Therefore, E(T) = 20 minutes.
VAR(T): Variance
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The variance of a sum of independent random variables is equal to the sum of their individual variances. To find VAR(T), we need to compute the variance of the number of magazines read in a day (M) and multiply it by the variance of the time taken to read each magazine (X).
1. Variance of the number of magazines read in a day (M):
Since the number of magazines Charlie reads on any given day follows a Poisson distribution with a mean of 2, the variance is also equal to the mean. So VAR(M) = 2.
2. Variance of the time taken to read each magazine (X):
Since the time taken to read each magazine follows an exponential distribution, the variance is equal to the square of the mean. So VAR(X) = (mean)^2 = 10^2 = 100.
Now, to find VAR(T), we multiply the variances of M and X: VAR(T) = VAR(M) * VAR(X) = 2 * 100 = 200.
Therefore, VAR(T) = 200 minutes squared.