A volunteer ambulance service handles 0 to 5 service calls on any given day. The probability distribution for the number of service calls is as follows: one column has number of service calls (X) those numbers are 0 1 2 3 4 5. The next column Probability P (X) those numbers are 0.08, 0.16, 0.27, 0.2, 0.15, 0.14. The next column is XP(X), the next column (X-meuh)next column (X-meuh)to the second power, then P(X) (X-meuh) to the second power. The questions are: What is the mean number of service calls. Next question What is the variance of service calls.What is the standard deviation. I'm stuck on the process between each column to get the answers, this is set up as a chart. Help please this is my exam😱

Question

A volunteer ambulance service handles 0 to 5 service calls on any given day. The probability distribution for the number of service calls is as follows: one column has number of service calls (X) those numbers are 0 1 2 3 4 5. The next column Probability P (X) those numbers are 0.08, 0.16, 0.27, 0.2, 0.15, 0.14. The next column is XP(X), the next column (X-meuh)next column (X-meuh)to the second power, then P(X) (X-meuh) to the second power. The questions are: What is the mean number of service calls. Next question What is the variance of service calls.What is the standard deviation. I'm stuck on the process between each column to get the answers, this is set up as a chart. Help please this is my exam😱

Answer 1

I understand that you have a probability distribution for the number of service calls, and you need help calculating the mean, variance, and standard deviation. Let's go through the process step by step:

1. Mean (μ):
To calculate the mean, you need to multiply each number of service calls (X) by its corresponding probability (P(X)), and then sum up the results. The formula is:
μ = Σ (X * P(X))

Using your given probabilities, you can set up the calculation like this:
Mean (μ) = (0 * 0.08) + (1 * 0.16) + (2 * 0.27) + (3 * 0.2) + (4 * 0.15) + (5 * 0.14)
Mean (μ) = 0 + 0.16 + 0.54 + 0.6 + 0.6 + 0.7
Mean (μ) = 2.2

Therefore, the mean number of service calls is 2.2.

2. Variance (σ²):
To calculate the variance, you need to find the squared difference between each number of service calls (X) and the mean (μ). Multiply this squared difference by its corresponding probability (P(X)), and then sum up the results. The formula is:
σ² = Σ [(X - μ)² * P(X)]

Using your given probabilities, the calculation would be:
Variance (σ²) = ((0 - 2.2)² * 0.08) + ((1 - 2.2)² * 0.16) + ((2 - 2.2)² * 0.27) + ((3 - 2.2)² * 0.2) + ((4 - 2.2)² * 0.15) + ((5 - 2.2)² * 0.14)
Variance (σ²) = (4.84 * 0.08) + (1.96 * 0.16) + (0.04 * 0.27) + (0.64 * 0.2) + (2.44 * 0.15) + (7.84 * 0.14)
Variance (σ²) = 0.3872 + 0.3136 + 0.0108 + 0.128 + 0.366 + 1.0976
Variance (σ²) = 2.3022

Therefore, the variance of service calls is approximately 2.3022.

3. Standard Deviation (σ):
To calculate the standard deviation, you simply need to take the square root of the variance. The formula is:
σ = √(Variance)

Using the variance calculated above:
Standard Deviation (σ) ≈ √2.3022
Standard Deviation (σ) ≈ 1.5167

Therefore, the standard deviation of service calls is approximately 1.5167.

I hope this explanation helps you understand the process of calculating the mean, variance, and standard deviation. Good luck with your exam!