Question

A company is collecting data on number of call-outs employees are making yearly. These call outs also known as occurrences are outside the normal six days each employee gets a year. Below is a representation of that data showcasing the number of this occurrences tied to the age of the employee. Create a probability distribution table. Use the probability distribution to find the mean, variance, and standard deviation of the probability distribution.

Age	19	20	21	22	23	24	25	26
Number of occurrences	33	46	31	35	25	52	9	12

Answer 1

dividing each number of occurence with total frequency of 243:

below is probability distribution of Age(x):

x	P(x)
19	0.1358
20	0.1893
21	0.1276
22	0.1440
23	0.1029
24	0.2140
25	0.0370
26	0.0494

from above

expected value =E(X)=ΣxP(x)=19*0.1358+20*0.1893+21*0.1276+22*0.1440+23*0.1029+24*0.2140+25*0.0370+26*0.0494 =21.9259

E(X²)=Σx²P(x)=19²*0.1358+20²*0.1893+21²*0.1276+22²*0.1440+23²*0.1029+24²*0.2140+25²*0.0370+26²*0.0494 =484.9300

Variance =E(x2)-(E(x))²=4.1838

standard deviation =√σ =2.0454