A factory's worker productivity is normally distributed. one worker produces an average of 77 units per day with a standard deviation of 22. another worker produces at an average rate of 66 units per day with a standard deviation of 19. a. what is the probability that in a single day worker 1 will outproduce worker 2?

To determine the probability that Worker 1 will outproduce Worker 2 on any given day, we need to consider the distribution of the difference in their production levels. Since productivity is normally distributed for both workers, the difference in productivity will also be normally distributed.

We can define a new variable, D, as the difference between the production of Worker 1 and Worker 2:

D = X1 - X2

Where: - X1 is the number of units produced by Worker 1 on a given day - X2 is the number of units produced by Worker 2 on a given day

The mean (μD) of D would therefore be the difference between the means of X1 and X2:

μD = μX1 - μX2

Given: - μX1 = 77 units (mean for Worker 1) - μX2 = 66 units (mean for Worker 2)

μD = 77 - 66 = 11 units

The standard deviation (σD) of D is not simply the difference between the two standard deviations but is calculated using the following equation:

σD^2 = σX1^2 + σX2^2

(applying the fact that for independent variables, variances add up, but standard deviations do not)

Given: - σX1 = 22 units (standard deviation for Worker 1) - σX2 = 19 units (standard deviation for Worker 2)

σD^2 = 22^2 + 19^2 = 484 + 361 = 845

σD = √845 ≈ 29.07 units

Now, we want to find the probability that Worker 1 will outproduce Worker 2 which equates to P(D > 0). Since D is normally distributed, we can standardize the problem by converting D to a standard normal variable Z:

Z = (D - μD) / σD

We need the Z-score for D = 0:

Z = (0 - 11) / 29.07 ≈ -0.378

We can now find the probability that Z > -0.378 by looking at the standard normal distribution table or using a statistical function in a calculator or software. This will give us the area to the right of Z = -0.378, which represents the probability that Worker 1 outproduces Worker 2:

P(Z > -0.378) = 1 - P(Z < -0.378)

Looking up the value in standard normal distribution tables or using statistical software: P(Z < -0.378) ≈ 0.3526

Therefore, the probability that Worker 1 will outproduce Worker 2 is:

P(Z > -0.378) = 1 - 0.3526 ≈ 0.6474

So, there is approximately a 64.74% chance that Worker 1 will outproduce Worker 2 on any given day.

Extra: Understanding normal distributions and probabilities is a fundamental part of statistics. A normal distribution, often called a "bell curve," is a type of continuous probability distribution for a real-valued random variable. The normal distribution is determined by two parameters: the mean and the standard deviation. The mean determines the center of the distribution, while the standard deviation determines its width. The greater the standard deviation, the more spread out the distribution is.

When comparing two normally distributed variables, the combination of their differences also forms a normal distribution if the original variables are independent. The probability of one variable exceeding the other can be found by analyzing this difference distribution and calculating the area under the curve for the desired range of values. This area represents the likelihood, or probability, of the event occurring.

Calculating Z-scores is a common method to standardize a value, allowing comparison across different normal distributions. A Z-score represents the number of standard deviations away from the mean a particular value is. Standard normal distribution tables or software can be used to find probabilities associated with particular Z-scores. These probabilities are crucial for making inferences in many fields, including business, economics, social sciences, and natural sciences.