Each magazine that Charlie reads taken 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 time.
Hint: You may find the following useful. For i.i.d random variables X_i and a non-negative random variable K that is independent of all the X_i's:
E[summation [I=1 to K] X_i = E[K]*E[X_1]
Variance [summation [I=1 to K] X_i]=E[K]*Variance[X_1]+(E[X_1])^2*Variance(K).
1. Find E[T]
2. Find Variance(T)
3. Using the central limit theorem to approximate the probability that the total number of magazines that Charlie reads in one full year (12 months of 30 days each) is between 3500 and 3600.
(Note that you are asked about the number of magazines, NOT the number of minutes of total reading time).
1. To find E[T], we can use the properties of the exponential distribution and the Poisson distribution.
Since the length of time to read each magazine follows an exponential distribution with mean 10 minutes, we have E[X_i] = 10 minutes.
The number of magazines read on any given day follows a Poisson distribution with mean 2. So E[K] = 2.
By the properties discussed in the hint, we can calculate E[T] as:
E[T] = E[K] * E[X_1] = 2 * 10 = 20 minutes.
Therefore, the expected total time Charlie spends reading magazines in one sitting is 20 minutes.
2. To find Variance(T), we can again use the properties of the exponential and Poisson distributions.
We already know that E[X_i] = 10 minutes and E[K] = 2.
The variance of the exponential distribution is equal to the square of its mean, so Variance[X_i] = (10)^2 = 100.
The variance of the Poisson distribution is equal to its mean, so Variance[K] = 2.
Using the properties discussed in the hint, we can calculate Variance(T) as:
Variance(T) = E[K] * Variance[X_1] + (E[X_1])^2 * Variance(K)
= 2 * 100 + (10)^2 * 2
= 200 + 100
= 300 minutes^2.
Therefore, the variance of the total time Charlie spends reading magazines in one sitting is 300 minutes^2.
3. To use the central limit theorem to approximate the probability, we need to make some assumptions.
Assuming that the number of magazines Charlie reads on different days are approximately normally distributed, we can use the formula for the mean and variance of the sum of normally distributed random variables.
The mean number of magazines read in one full year (12 months of 30 days each) is 2 * 30 * 12 = 720.
The variance of the number of magazines read in one full year is 2 * 30 * 12 = 720.
Using the central limit theorem, we can approximate the total number of magazines read in one full year as a normal distribution with mean 720 and variance 720.
To calculate the probability that the total number of magazines is between 3500 and 3600, we can standardize it by subtracting the mean and dividing by the square root of the variance:
Z = (3500 - 720) / sqrt(720)
= 2780 / 26.83
= 103.57
Z = (3600 - 720) / sqrt(720)
= 2880 / 26.83
= 107.30
We can then use a standard normal table or calculator to find the probability that Z is between 103.57 and 107.30.
P(103.57 < Z < 107.30) ≈ P(Z < 107.30) - P(Z < 103.57)
This will give us the approximate probability that the total number of magazines Charlie reads in one full year is between 3500 and 3600.
To find the expected value E[T], we need to determine the expected value of the total number of minutes Charlie spends reading one magazine and then multiply that by the expected number of magazines read in a day.
1. Expected value of the time spent reading one magazine:
Given that the time taken to read each magazine follows an exponential distribution with a mean of 10 minutes, we know that the expected value of this distribution is also 10 minutes.
2. Expected value of the number of magazines read in a day:
The number of magazines read in a day has a Poisson distribution with a mean of 2. The expected value of a Poisson distribution is equal to its mean. Therefore, the expected value of the number of magazines read in a day is 2.
Now we can calculate the expected value E[T] as follows:
E[T] = E[number of magazines read] * E[time spent reading one magazine]
= 2 * 10
= 20 minutes
Therefore, the expected value of the total number of minutes Charlie spends reading magazines in one time is 20 minutes.
To find the variance of T, we can use the formula mentioned in the hint:
Variance(T) = E[K] * Variance[X₁] + (E[X₁])² * Variance(K)
3. Variance of the time spent reading one magazine:
The time taken to read each magazine follows an exponential distribution with a mean of 10 minutes. The variance of an exponential distribution is equal to the square of its mean. Therefore, the variance of the time spent reading one magazine is (10)² = 100 minutes².
4. Variance of the number of magazines read in a day:
The number of magazines read in a day has a Poisson distribution with a mean of 2. The variance of a Poisson distribution is also equal to its mean. Therefore, the variance of the number of magazines read in a day is 2.
Now we can calculate the variance of T:
Variance(T) = E[K] * Variance[X₁] + (E[X₁])² * Variance(K)
= 2 * 100 + 10² * 2
= 200 + 200
= 400 minutes²
Therefore, the variance of the total number of minutes Charlie spends reading magazines in one time is 400 minutes².
To approximate the probability using the central limit theorem, we need to find the mean and standard deviation of the total number of magazines Charlie reads in a full year.
5. Mean of the total number of magazines read in a full year:
Since the number of magazines read on different days is independent, the mean of the total number of magazines read in a full year is simply the product of the mean number of magazines read in a day (2) and the number of days in a year (12 * 30 = 360). Therefore, the mean is 2 * 360 = 720 magazines.
6. Standard deviation of the total number of magazines read in a full year:
The standard deviation of the total number of magazines read in a full year can be found by taking the square root of the sum of variances of the number of magazines read in each individual day. Since the variance of the number of magazines read in a day is 2, the variance of the total number of magazines read in a full year is (2 * 360) = 720. Therefore, the standard deviation is √720 ≈ 26.87.
Now we can use the central limit theorem to approximate the probability that the total number of magazines Charlie reads in one full year is between 3500 and 3600.
Using z-scores for both upper and lower limits:
Lower limit: (3500 - 720) / 26.87 ≈ 108.68
Upper limit: (3600 - 720) / 26.87 ≈ 119.96
Using the standard normal distribution table, we can find the probability associated with these z-scores:
P(108.68 < Z < 119.96)
By finding the corresponding values from the table, we can determine the probability.