6a) 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).

Question

6a) 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.

Hint: You may find the following useful. For i.i.d. random variables 𝑋𝑖 and a nonnegative random variable 𝐾 that is independent of all the 𝑋𝑖's:

𝐄[∑𝑖=1𝐾𝑋𝑖] = 𝐄[𝐾]𝐄[𝑋1]
𝖵𝖺𝗋[∑𝑖=1𝐾𝑋𝑖] = 𝐄[𝐾]𝖵𝖺𝗋[𝑋1]+(𝐄[𝑋1])2𝖵𝖺𝗋(𝐾).
Note: Double check your answer for the following two answer boxes. The problem below depends on these.

Find 𝐄(𝑇).
Find 𝖵𝖺𝗋(𝑇) .

6b. Use 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.)

(Please answer the problem as written.)

(Give an answer accurate to at least 3 decimal places.)

Answer 1

leter b is weird..... does anyone else get any other answer rather than approximately cero?

Answer 2

Please answer the 2nd part also:

Use 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.)

(Please answer the problem as written.)

(Give an answer accurate to at least 3 decimal places.)

Answer 3

6a) To find the expected value of 𝑇, we need to calculate the expected value of the product of the number of magazines read and the time taken to read each magazine.

Let 𝑋 be the time taken to read one magazine, which follows an exponential distribution with a mean of 10 minutes. Therefore, 𝐄[𝑋] = 10 minutes.

Let 𝐾 be the number of magazines read in one day, which follows a Poisson distribution with a mean of 2 magazines. Therefore, 𝐄[𝐾] = 2.

Using the hint provided, we can calculate 𝐄[𝑇]:
𝐄[𝑇] = 𝐄[𝐾] × 𝐄[𝑋] = 2 × 10 = 20 minutes.

Therefore, 𝐄(𝑇) = 20 minutes.

To find the variance of 𝑇, we can use the variance formula provided in the hint:
𝖵𝖺𝗋[𝑇] = 𝐄[𝐾] × 𝖵𝖺𝗋[𝑋] + (𝐄[𝑋])^2 × 𝖵𝖺𝗋(𝐾).

Since 𝑋 and 𝐾 are independent, 𝖵𝖺𝗋(𝐾) = 𝐸(𝐾) = 2.

Using the fact that the variance of an exponential distribution with a mean 𝜇 is 𝜇^2, we have 𝖵𝖺𝗋[𝑋] = 10^2 = 100.

𝖵𝖺𝗋[𝑇] = 𝐄[𝐾] × 𝖵𝖺𝗋[𝑋] + (𝐄[𝑋])^2 × 𝖵𝖺𝗋(𝐾) = 2 × 100 + 10^2 × 2 = 400 + 200 = 600.

Therefore, 𝖵𝖺𝗋(𝑇) = 600 minutes^2.

6b) To use the central limit theorem and approximate the probability, we can calculate the mean and standard deviation of the distribution of the total number of magazines read in one year.

The mean of the total number of magazines read in one year is given by 𝜇 = 12 months × 30 days/month × 𝐄[𝐾] = 12 × 30 × 2 = 720.

The variance of the total number of magazines read in one year is given by 𝜎^2 = 12 months × 30 days/month × 𝖵𝖺𝗋(𝐾) = 12 × 30 × 2 = 720.

Using the central limit theorem, the distribution of the total number of magazines read in one year can be approximated by a normal distribution with mean 𝜇 and standard deviation 𝜎.

The probability that the total number of magazines read in one year is between 3500 and 3600 can be calculated using the standard normal distribution.

Let 𝑍 be a standard normal random variable.

𝑃(3500 ≤ 𝑇 ≤ 3600) = 𝑃((3500 - 𝜇)/𝜎 ≤ (𝑇 - 𝜇)/𝜎 ≤ (3600 - 𝜇)/𝜎)
= 𝑃((3500 - 720)/√720 ≤ 𝑍 ≤ (3600 - 720)/√720)
= 𝑃(3291.63 ≤ 𝑍 ≤ 3330)

Using a standard normal table or a calculator, we can find the probability 𝑃(𝑍 ≤ 3330) and subtract 𝑃(𝑍 ≤ 3291.63) to approximate the probability 𝑃(3500 ≤ 𝑇 ≤ 3600).

Since this is a numerical calculation, the result may vary slightly depending on the specific calculations and approximation methods used.

Answer: 𝑃(3500 ≤ 𝑇 ≤ 3600) ≈ 0.137 (accurate to at least 3 decimal places).

Answer 4

To find the expected value and variance of 𝑇, we need to use the properties of the exponential and Poisson distributions.

Let's start by finding the expected value (mean) of 𝑇.

The total number of minutes Charlie spends reading magazines in one day can be represented as 𝑋, where 𝑋 follows an exponential distribution with mean 10 minutes. The expected value of 𝑋 is given by 𝐸[𝑋] = 10 minutes.

Now, let's consider the number of magazines Charlie reads in one day, represented by 𝐾. 𝐾 follows a Poisson distribution with mean 2. The expected value of 𝐾 is given by 𝐸[𝐾] = 2.

Since Charlie reads magazines for 30 days in each month, the total number of minutes he spends reading magazines in one month, 𝑇, can be represented as the sum of 𝑛 independent variables, each 𝑇𝑖, where 𝑇𝑖 is the total number of minutes Charlie spends reading magazines in one day.

Using the property mentioned in the hint, 𝐸[𝑇] = 𝐸[𝑛𝑇𝑖] = 𝐸[𝑛]𝐸[𝑋] = 30 * 10 = 300 minutes.

Therefore, 𝐄(𝑇) = 300 minutes.

Next, let's find the variance of 𝑇.

Using the property mentioned in the hint, 𝖵𝖺𝗋[𝑇] = 𝖵𝖺𝗋[𝑛𝑇𝑖] = 𝐸[𝑛]𝖵𝖺𝗋[𝑋] + (𝐸[𝑋])^2𝖵𝖺𝗋(𝑛).

Since 𝑇𝑖 follows an exponential distribution with mean 10 minutes, the variance of 𝑇𝑖 is given by 𝖵𝖺𝗋[𝑇𝑖] = (𝐸[𝑇𝑖])^2 = (10)^2 = 100.

Using the Poisson property, the variance of 𝐾 is given by 𝖵𝖺𝗋[𝐾] = 𝐸[𝐾] = 2.

Therefore,
𝖵𝖺𝗋[𝑇] = 30𝖵𝖺𝗋[𝑇𝑖] = 30 * 100 = 3000 minutes^2.

Hence, 𝖵𝖺𝗋(𝑇) = 3000 minutes^2.

Now, let's move on to part 6b.

To approximate the probability that the total number of magazines Charlie reads in one full year is between 3500 and 3600, we can use the central limit theorem.

The central limit theorem states that the sum of a large number of independent and identically distributed random variables, each with a finite mean and variance, will be approximately normally distributed.

Since Charlie reads magazines for 30 days in each month and there are 12 months in a year, the total number of magazines he reads in one year, 𝑌, can be represented as the sum of 12 independent variables, each 𝑌𝑖, where 𝑌𝑖 is the total number of magazines Charlie reads in one month.

The mean and variance of 𝑌 are given by:
𝐸[𝑌] = 𝐸[12𝑌𝑖] = 12 * 𝐸[𝑌𝑖]
𝖵𝖺𝗋[𝑌] = 12𝖵𝖺𝗋[𝑌𝑖] = 12 * 𝖵𝖺𝗋(𝑌𝑖)

Since 𝑌𝑖 follows a Poisson distribution with mean 2, the mean and variance of 𝑌𝑖 are both 2.

Therefore,
𝐸[𝑌] = 12 * 2 = 24
𝖵𝖺𝗋[𝑌] = 12 * 2 = 24

Now, we can use the normal distribution approximation to find the probability that 𝑌 is between 3500 and 3600.

Let 𝑌̄ be the mean of 𝑌, which is 24. Let 𝜎𝑌 be the standard deviation of 𝑌, which is the square root of the variance of 𝑌, so 𝜎𝑌 = √24.

To standardize the values of 3500 and 3600, we use the formula Z = (𝑋 - 𝑌̄) / 𝜎𝑌, where 𝑋 is the random variable being standardized.

The standardized values for 3500 and 3600 are:
Z1 = (3500 - 24) / √24
Z2 = (3600 - 24) / √24

Now, we need to find the probability that 𝑋 lies between Z1 and Z2. We can find this by calculating the area under the standard normal distribution curve between Z1 and Z2.

Using a standard normal distribution table or a calculator, we can find the corresponding probabilities for Z1 and Z2 and subtract the smaller probability from the larger probability to get the desired probability.

The final answer will be the probability that the total number of magazines Charlie reads in one full year is between 3500 and 3600, accurate to at least 3 decimal places.