A factory worker productivity is normally distributed. worker A produces an average of 84 units per day with a standard deviation of 24. worker B produces at an average rate of 74 units per day with stanard deviation of 25. (a) what is the probability that in any single day worker A will produce more units than worker B? (b) what is the probablity that during one week. (5 weeking days). worker A will produce more units than worker B on average?
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To find the probability for each scenario, we can use the concept of z-scores and the normal distribution. The z-score measures the number of standard deviations away from the mean a particular value is.
(a) Probability that worker A will produce more units than worker B in a single day:
We need to compare the mean and standard deviation of both workers to calculate the probability. Let's denote:
- X_A as the random variable for worker A's daily production,
- X_B as the random variable for worker B's daily production,
- μ_A as the mean for worker A's production (84 units),
- μ_B as the mean for worker B's production (74 units),
- σ_A as the standard deviation for worker A's production (24),
- σ_B as the standard deviation for worker B's production (25).
To find the probability that worker A produces more units than worker B in a single day (P(X_A > X_B)), we can calculate the probability of the standardized variable (z-score):
P(X_A > X_B) = P((X_A - μ_A) > (X_B - μ_B))
= P((X_A - μ_A) - (X_B - μ_B) > 0)
= P((X_A - X_B) > (μ_A - μ_B))
Now, let's calculate the z-score for (μ_A - μ_B):
z = (μ_A - μ_B) / square root(σ_A^2 + σ_B^2)
z = (84 - 74) / square root(24^2 + 25^2) ≈ 0.377
To find the probability using the z-score, we can refer to the z-table or use statistical software. Looking up the z-table, we find that the probability associated with a z-score of approximately 0.377 is approximately 0.647. Therefore, the probability that worker A produces more units than worker B in a single day is approximately 0.647.
(b) Probability that worker A will produce more units than worker B on average during one week (5 working days):
Since we are interested in the average production over a week, we need to consider the distribution of the average production. The mean of the distribution of averages (μ_A-B) will be the difference of the means of both workers (μ_A - μ_B), and the standard deviation (σ_A-B) of the distribution of averages will be calculated using the formula:
σ_A-B = square root((σ_A^2)/n_A + (σ_B^2)/n_B),
where n_A and n_B are the sample sizes (number of days). In this case, both workers work for 5 days a week.
Using the given values:
μ_A-B = μ_A - μ_B = 84 - 74 = 10
σ_A-B = square root((24^2)/5 + (25^2)/5) ≈ 16.31
Now, we can calculate the z-score for (μ_A-B = 10):
z = (μ_A-B - 0) / σ_A-B
z = 10 / 16.31 ≈ 0.613
Using the z-score, we can calculate the probability associated with a z-score of approximately 0.613 by referring to the z-table or using statistical software. The probability is approximately 0.731. Therefore, the probability that worker A will produce more units than worker B on average during one week (5 working days) is approximately 0.731.