Problem 16 - A First Course in Probability

Question

A.J. has 20 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. M.J. has 20 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. (a) Find the probability that A.J. finishes in less than 900 minutes. (b) Find the probability that M.J. finishes in less than 900 minutes. (c) Find the probability that A.J. finishes before M.J.

Step-by-Step Solution

Answer

(a) The probability that A.J. finishes in less than 900 minutes is approximately 0.0569. (b) The probability that M.J. finishes in less than 900 minutes is approximately 0.0068. (c) The probability that A.J. finishes before M.J. is approximately 0.0367.

Step 1: Calculate combined mean and standard deviation for A.J. and M.J.

For independent random variables, the sum's mean is the sum of the means, and the sum's variance is the sum of the variances. For A.J.: Mean of a job: \(\mu_A = 50\) Standard deviation of a job: \(\sigma_A = 10\) For M.J.: Mean of a job: \(\mu_M = 52\) Standard deviation of a job: \(\sigma_M = 15\) Combined mean and standard deviation for A.J.'s 20 jobs: \[\mu_{sumA} = 20 \times \mu_A = 20 \times 50 = 1000\] \[\sigma_{sumA} = \sqrt{20 \times \sigma_A^2} = \sqrt{20 \times 10^2} = 10\sqrt{20}\] Combined mean and standard deviation for M.J.'s 20 jobs: \[\mu_{sumM} = 20 \times \mu_M = 20 \times 52 = 1040\] \[\sigma_{sumM} = \sqrt{20 \times \sigma_M^2} = \sqrt{20 \times 15^2} = 15\sqrt{20}\] (a) Find the probability that A.J. finishes in less than 900 minutes. First, we need to standardize the value 900 for A.J.'s job-completion-time distribution. Then we will find the probability using the standard normal distribution table.

Step 2: Standardize 900 for A.J.

To standardize 900 using the mean and standard deviation of A.J.'s 20 jobs sum: \[Z_A = \frac{900 - \mu_{sumA}}{\sigma_{sumA}} = \frac{900 - 1000}{10\sqrt{20}} \approx -1.58\]

Step 3: Look up probability in z-table for A.J.

From the z-table, corresponding to a Z-score of -1.58, we find a probability of 0.0569. This is the probability that A.J. finishes in less than 900 minutes. (b) Find the probability that M.J. finishes in less than 900 minutes. We will follow the same steps as we did for A.J.

Step 4: Standardize 900 for M.J.

To standardize 900 using the mean and standard deviation of M.J.'s 20 jobs sum: \[Z_M = \frac{900 - \mu_{sumM}}{\sigma_{sumM}} = \frac{900 - 1040}{15\sqrt{20}} \approx -2.47\]

Step 5: Look up probability in z-table for M.J.

From the z-table, corresponding to a Z-score of -2.47, we find a probability of 0.0068. This is the probability that M.J. finishes in less than 900 minutes. (c) Find the probability that A.J. finishes before M.J. We need to find the probability that the difference in their completion times is positive, i.e., A.J. finishes faster. Since A.J.'s and M.J.'s jobs are independent, we can calculate the mean and standard deviation of the difference as follows:

Step 6: Calculate the mean and standard deviation of the difference

The mean and standard deviation of the difference in completion times: \[\mu_{diff} = \mu_{sumA} - \mu_{sumM} = 1000 - 1040 = -40\] \[\sigma_{diff} = \sqrt{\sigma_{sumA}^2 + \sigma_{sumM}^2} = \sqrt{(10\sqrt{20})^2 + (15\sqrt{20})^2} = 5\sqrt{20}\] Now, we want to find the probability of the difference being greater than zero:

Step 7: Standardize the difference

To standardize the value 0 using the mean and standard deviation of the difference: \[Z_{diff} = \frac{0 - (-40)}{5\sqrt{20}} \approx 1.79\]

Step 8: Look up the probability in the z-table for the difference

From the z-table, corresponding to a Z-score of 1.79, we find a probability of 0.9633. However, since we are interested in the probability that A.J. finishes before M.J., we need to consider the right tail of the distribution. Therefore, we need to subtract this probability from 1: \[P(A.J. finishes before M.J.) = 1 - 0.9633 = 0.0367\]

Key Concepts

Standard Deviation

Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. In the context of probability problems, understanding standard deviation is crucial in determining how spread out the values in a data set are relative to the mean. It's calculated by taking the square root of the variance, which is the average of the squared differences from the Mean.

For example, in a given problem, if the standard deviation is small, it suggests that the data points tend to be close to the mean, giving us a tighter data distribution. Conversely, a large standard deviation indicates that the data points are spread out over a wider range of values. This principle applies to A.J. and M.J.'s job completion times in the given exercise - where the standard deviation helps determine the probability of them finishing the jobs within a certain timeframe.

It's also important to note when dealing with a series of independent random variables, like the jobs A.J. and M.J. must complete, the total or combined standard deviation becomes critical. To find this, you sum the variances of each individual task (since the tasks are independent) and then take the square root of this sum to obtain the composite standard deviation for the entire sequence of tasks.

Normal Distribution

The normal distribution, also known as the Gaussian distribution, is a symmetric, bell-shaped distribution characterized by its mean and standard deviation. In the realm of statistics and probability, many datasets approximate this type of distribution, particularly when the number of data points is large.

When working with large sets of data, such as the time it takes for A.J. and M.J. to complete multiple tasks, the Central Limit Theorem tells us that the distribution of the sample means will tend to be normal, regardless of the shape of the source population. This is critical in solving probability problems because it allows us to standardize data and use the standard normal distribution (a normal distribution with a mean of 0 and standard deviation of 1) to calculate probabilities.

Understanding that the times to complete the jobs are normally distributed, given a large number of jobs, enables us to use Z-scores and normal distribution tables to find the probability of certain outcomes occurring, such as finishing all jobs within a particular time frame.

Z-score

The Z-score, also referred to as a standard score, measures the number of standard deviations a given data point is from the mean. If a Z-score is 0, it signifies that the data point's score is identical to the mean score. A Z-score can be positive or negative, indicating whether the data point lies above or below the mean, respectively.

In probability problems like those involving A.J. and M.J., we use the Z-score to standardize an observation, transforming it to a score that can be compared across different normal distributions. This standardization process involves subtracting the mean from the individual observation and then dividing the result by the standard deviation.

For instance, in the exercise, by finding the Z-scores for A.J. and M.J., we are able to determine the likelihood of them completing their tasks in less than 900 minutes. Moreover, the Z-score allows us to calculate the probability of one event happening before another by referencing the standard normal distribution table. This is key to solving part (c) of the problem, where we find how likely it is for A.J. to finish before M.J. This insight hinges on understanding the concept of Z-scores and being able to apply them effectively in statistical analysis.