A widget factory's worker productivity is normally distributed. one worker produces an average of 75 widgets per day with a standard deviation of 20. another worker produces at an average rate of 65 per day with standard deviation of 21. 1)what is the probability that in any single day worker 1 will outproduce worker 2? 2) what is the probability that during one week (5 working days) worker 1 will outproduce worker 2?
To find the probabilities in both scenarios, we can use the normal distribution and the concept of z-scores. A z-score represents how many standard deviations a specific value is away from the mean.
1) Probability of worker 1 outproducing worker 2 in a single day:
First, we need to calculate the z-score for worker 2's production rate using the formula: z = (x - μ) / σ
where x is the value we want to convert to a z-score, μ is the mean, and σ is the standard deviation.
For worker 2: x = 75 (worker 1's average production), μ = 65 (worker 2's average production), σ = 21 (worker 2's standard deviation).
z = (75 - 65) / 21 ≈ 0.4762
Next, we can find the probability corresponding to this z-score using a standard normal distribution table or a calculator. The table gives the probability to the left of the z-score.
Using a standard normal distribution table, the probability associated with a z-score of 0.4762 is approximately 0.6826 (rounded to four decimal places).
Therefore, the probability that worker 1 will outproduce worker 2 in any single day is 0.6826 or approximately 68.26%.
2) Probability of worker 1 outproducing worker 2 during one week (5 working days):
To find the probability over multiple days, we need to consider the production over the entire week. Since the mean and standard deviation for both workers are given for a single day, we need to find the mean and standard deviation for a week.
The mean for a week is simply the sum of the daily means: μ_week = 5 * μ_day = 5 * 75 = 375.
To find the standard deviation for a week, we use the fact that the standard deviation scales linearly with the square root of the sample size when sample sizes are independent. So, the standard deviation for a week (σ_week) is the square root of the sum of the variances for the individual days.
The variance for a single day for worker 1: σ_day1^2 = 20^2 = 400
The variance for a single day for worker 2: σ_day2^2 = 21^2 = 441
The variance for a week: σ_week^2 = 5 * (σ_day1^2) + 5 * (σ_day2^2)
σ_week = √(5 * 400 + 5 * 441) ≈ √3925 ≈ 62.65
Now we calculate the z-score for worker 2's production rate during the week:
z = (375 - 325) / 62.65 ≈ 0.7965
Using the standard normal distribution table or calculator, we find that the probability associated with a z-score of 0.7965 is approximately 0.7857 (rounded to four decimal places).
So, the probability that worker 1 will outproduce worker 2 during one week is 0.7857 or approximately 78.57%.
To calculate the probabilities, we need to standardize the individual worker productivity values using the z-score formula. Given the average productivity (μ) and the standard deviation (σ) for each worker:
1) Probability that worker 1 will outproduce worker 2 in a single day:
Step 1: Calculate the z-score for worker 1:
z1 = (x1 - μ1) / σ1
= (75 - 75) / 20
= 0
Step 2: Calculate the z-score for worker 2:
z2 = (x2 - μ2) / σ2
= (65 - 65) / 21
= 0
Step 3: Calculate the probability that worker 1 will outproduce worker 2:
P(worker 1 > worker 2) = P(z1 > z2)
Since both z-scores are 0, the probabilities will be equal.
The probability that worker 1 will outproduce worker 2 in any single day is 0.5.
2) Probability that worker 1 will outproduce worker 2 during one week (5 working days):
Since the daily productivity values are independent, we can sum the probabilities for a single day to find the probability for a week.
Let's denote the probability that worker 1 outproduces worker 2 on a single day as P(daily). Then, the probability that worker 1 outproduces worker 2 during a week is:
P(weekly) = P(daily) + P(daily) + P(daily) + P(daily) + P(daily)
= 5 * P(daily)
Therefore, the probability that worker 1 will outproduce worker 2 during one week is 5 times the probability for a single day.
P(weekly) = 5 * P(daily)
P(weekly) = 5 * 0.5
P(weekly) = 2.5
The probability that worker 1 will outproduce worker 2 during one week is 2.5.