6. The sick-leave time of employees in a firm in a month is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. Find the probability that the sick-leave time of an employee in a month exceeds 130 hours.
6. The sick-leave time of employees in a firm in a month is normally distributed with a mean of 100 hours and a standard deviation of 20 hours. Find the probability that the sick-leave time of an employee in a month exceeds 130 hours.
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Transcribed Image Text:6. The sick-leave time of employees in a firm in a month is normally distributed with a
mean of 100 hours and a standard deviation of 20 hours. Find the probability that
the sick-leave time of an employee in a month exceeds 130 hours.

Transcribed Image Text:•Types of functions to express
probability distribution
(a) Probability Mass Function (PMF)
f(x) = P(x = x)
(b) Cumulative Distribution Functions (CDF)
f(x) = P(X ≤ x)
Mean
M = { xf(x)
Variance
0² = [x²f (x) - μ²
Standard Deviation.
0x Tot
Discrete Uniform Distribution
f(x) = //12
M =
0² =
b+ a
2
(b-a+1)²
12
•Binomial Distribution
(f(x) = (~) p² (1-P)^-x
с пр
np(1-P)
1• Hypergeometric Distribution
(*) (*)
(~)
f(x) =
и = пр
8² = np(1-P/(N=1)
where
K/N
P=
·Poisson Distribution
-^+ (^T)*
X!
f(x) = e
M = AT
0² = NT
Where
N = population size
sample size /no. of trials
P
probability
-
p = probability of success on a
single trial
K = no. of successes in the population
in the sample
of
X - no.
successes
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