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:

dc=6 square root of 3 and ac is 12 square root 3

c and d

step-by-step explanation:

square root of 3/2 = 18/c

cross multiply to get 18*2=√3*c

36/√3 * √3/√3 = c

c= 36*√3/3

c=12√3