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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.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 13PPS
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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: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.
•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
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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