Problem 79 - Applied Statistics and Probability for Engineers

Question

The speed of a file transfer from a server on campus to a personal computer at a student's home on a weekday evening is normally distributed with a mean of 60 kilobits per second and a standard deviation of four kilobits per second. (a) What is the probability that the file will transfer at a speed of 70 kilobits per second or more? (b) What is the probability that the file will transfer at a speed of less than 58 kilobits per second? (c) If the file is one megabyte, what is the average time it will take to transfer the file? (Assume eight bits per byte.)

Step-by-Step Solution

Answer

(a) 0.0062; (b) 0.3085; (c) 136.53 seconds.

Step 1: Define the Variables

Given a normal distribution with a mean \( \mu = 60 \) kbps and a standard deviation \( \sigma = 4 \) kbps. We will use these to find probabilities using the Z-score formula \( Z = \frac{X - \mu}{\sigma} \).

Step 2: Calculate Z-score for 70 kbps

The Z-score is computed using \( X = 70 \). Apply the formula: \[ Z = \frac{70 - 60}{4} = 2.5 \].

Step 3: Find Probability for 70 kbps or More

Using the Z-table, find the probability that \( Z \geq 2.5 \). The value for \( Z = 2.5 \) gives a cumulative probability of 0.9938. Thus, \( P(X \geq 70) = 1 - 0.9938 = 0.0062 \).

Step 4: Calculate Z-score for 58 kbps

Use \( X = 58 \) in the Z-score formula: \[ Z = \frac{58 - 60}{4} = -0.5 \].

Step 5: Find Probability for Less than 58 kbps

Consult the Z-table for \( Z = -0.5 \), giving cumulative probability of 0.3085. Hence, \( P(X < 58) = 0.3085 \).

Step 6: Calculate Average Time to Transfer 1 MB

First convert 1 megabyte to kilobits: \( 1 \text{ MB} = 1,024 \times 8 = 8,192 \) kb. Use the average speed: \[ \text{Time} = \frac{8,192}{60} \approx 136.53 \text{ seconds} \].

Key Concepts

Z-score

The Z-score is a statistical measure that describes a value's position relative to the mean of a group of values. It tells us how many standard deviations away a particular value is from the mean. In the context of normal distribution, using the Z-score can help determine the probability associated with a particular outcome.

To calculate the Z-score, the formula used is:

• \[ Z = \frac{X - \mu}{\sigma} \]

In this formula, \( X \) represents the value you want to assess (such as 70 kbps in the example), \( \mu \) is the mean of the distribution (60 kbps), and \( \sigma \) is the standard deviation (4 kbps).

Once you have the Z-score, you can use a Z-table to find the probability that a value is greater or less than the value corresponding to the Z-score. This is an essential step for calculating probabilities in a normal distribution.

Probability Calculation

Calculating probabilities with the normal distribution involves using the Z-score and a Z-table. This table provides the probability that a normally distributed random variable is less than or equal to a given Z-score. Here’s how the calculation works for two examples:

• To find the probability that the transfer speed is 70 kilobits per second or more, first calculate the Z-score for 70 kbps. The Z-score is 2.5. Using a Z-table, we find that the cumulative probability up to 70 kbps is 0.9938. The probability for speeds of 70 kbps or more is given by the complement: 1 - 0.9938 = 0.0062, or 0.62%.
• For speeds less than 58 kbps, the Z-score calculates to -0.5. The cumulative probability for this Z-score from the Z-table is 0.3085, indicating that there is a 30.85% chance the speed will be below 58 kbps.

Understanding these procedures helps in various contexts, including estimating probabilities for different values in any normally distributed data set.

File Transfer Speed

File transfer speed is the rate at which data is transmitted from one place to another. In this example, the speed is normally distributed with a mean of 60 kilobits per second (kbps). This speed plays a critical role in determining how long it takes to send or receive data.

To calculate how long it would take on average to transfer a file of 1 megabyte (MB), first convert megabytes to kilobits, because the transfer speed is measured in kbps. Thus, 1 MB is equivalent to 8,192 kilobits (since 1 MB equals 1,024 kilobytes and 1 kilobyte is 8 kilobits).

The average transfer time is computed using:

• \[ \text{Time} = \frac{8,192 \text{ kb}}{60 \text{ kbps}} \approx 136.53 \text{ seconds} \]

This means that, on average, it would take about 136.53 seconds to complete the file transfer given the mean speed of the connection. This calculation is highly useful for understanding network speeds and estimating transfer times in practical scenarios.