Suppose the number of customer complaints per day on the morning shift 6-12 noon at Doys is distributed as
# of complaints f(x)
0 .3
1 ?
2 .2
what is the value of the missing probability?
What is the average number of complaints per day on the morning shift at Doys?
What is the standard deviation of the number of complaints per day on the morning shift at Doys?
If y is the number of inexperienced staff people at Doys on the morning shift, would the values of x and y be correlated?
To find the missing probability, we need to determine the total probability sum of all possible outcomes. In this case, we have three possible outcomes: 0 complaints, 1 complaint, and 2 complaints.
Given that the sum of the probabilities should equal 1, we can set up the equation:
0.3 + ? + 0.2 = 1
Simplifying the equation, we have:
0.5 + ? = 1
Subtracting 0.5 from both sides, we find:
? = 1 - 0.5
? = 0.5
Therefore, the missing probability is 0.5.
To calculate the average number of complaints per day on the morning shift at Doys, we need to find the expected value. The expected value is computed by multiplying each outcome by its respective probability and summing them up.
The calculations for the average are as follows:
(0 complaints * 0.3) + (1 complaint * 0.5) + (2 complaints * 0.2) = average
(0 * 0.3) + (1 * 0.5) + (2 * 0.2) = average
Average = 0 + 0.5 + 0.4 = 0.9
Therefore, the average number of complaints per day on the morning shift at Doys is 0.9.
To calculate the standard deviation of the number of complaints per day on the morning shift at Doys, we need to follow these steps:
1. Calculate the variance: Multiply each outcome by its squared difference from the mean, and multiply the result by its respective probability. Sum up these calculations.
(0-0.9)^2 * 0.3 + (1-0.9)^2 * 0.5 + (2-0.9)^2 * 0.2 = variance
(0.9)^2 * 0.3 + (0.1)^2 * 0.5 + (1.1)^2 * 0.2 = variance
0.729 * 0.3 + 0.01 * 0.5 + 1.21 * 0.2 = variance
Variance = 0.2187 + 0.005 + 0.242 = 0.4657
2. Take the square root of the variance to get the standard deviation:
Standard deviation = square root of variance
Standard deviation = square root of 0.4657
Standard deviation ≈ 0.682
Therefore, the standard deviation of the number of complaints per day on the morning shift at Doys is approximately 0.682.
Lastly, to determine if the values of x (number of complaints per day) and y (number of inexperienced staff people) are correlated, we would need more information or a correlation analysis. The existence of a correlation would depend on various factors, such as the staff's expertise, training, customer behavior, etc. Without further information, it is not possible to determine the correlation between x and y.