A company is studying the number of monthly absences among its 125 employees. The following probability distribution shows the likelihood that people were absent 0, 1, 2, 3, 4, or 5 days last month. Number of days Absent | Probability 0 0.60 1 0.20 2 0.12 3 0.04 4 0.04 5 0.001. What is the mean number of days absent? What are the variance and standard deviation?2. For the following probability distribution: a) x|10|11|12|13|14b) P(x)|.1|.25|.3|.25|.13. Find the mean, variance and standard deviation for the following probability distribution.

(1)

(2)

Step-by-step explanation:

Solving (1):

(a): Mean

This is calculated as:

So, we have:

Solving (b): The variance

This is calculated as:

Where:

and

So, we have:

So, we have:

Solving (c): The standard deviation.

This is calculated as:

Solving (2):

(a): Mean

This is calculated as:

So, we have:

Solving (b): The variance

This is calculated as:

Where:

and

So, we have:

So, we have:

Solving (c): The standard deviation.

This is calculated as:

the cat is blue because the probability of extension towards the prefix of the diameter of the circle could only mathematically equal to the sum of the roots of the carrots from   your garden.

step-by-step explanation: