A Bureau of National Affairs survey found that 79% of employers provide their wi

Question

A Bureau of National Affairs survey found that 79% of employers provide their with a two-day paid Thanksgiving holiday with workers off both Thursday and (USA Today, November 12, 2009). Nineteen percent of employers provide a one-day holiday with off Thanksgiving Day. Two percent of employers do not paid Thanksgiving holiday. Consider a sample of 120 employers. a. What is the probability that at least 85 of the employers provide a two-day Thanksgiving holiday? b. What is the probability that between 90 and 100 employers provide a two-day Thanksgiving holiday? That is, what is P(90 lessthanorequalto x lessthanorequalto 100)? c. What is the probability that less than 20 employers provide a one-day paid Thank giving holiday?

Explanation / Answer

a)
n=120 (number of employers)

p = probability that an employer provides a two-day paid Thanksgiving holiday = 0.79

x = 85 or more (number of employers providing a two-day paid Thanksgiving holiday out of 120

This is a binomial probability but should be solved by normal approximation.

Mean = np = (120)(0.79) = 94.8

Standard deviation = sqrt [ np(1-p) ] = sqrt [ (120)(0.79)(0.21) ] = 4.4618

P( x >= 85) is required.

Using continuity correction P( x >= 85) becomes P( x > 84.5)

= 94.8 , = 4.4618

standardize x to z = (x - ) /

P(x > 84.5) = P( z > (84.5 - 94.8) / 4.4618)

= P(z > -2.3085) = 0.9896
(From Normal probability table)

b)

P( 90 <= x <= 100) = P( 89.5 < x < 100.5)

= 94.8 , = 4.4618

standardize x to z = (x - ) /

P( 89.5 < x < 100.5) = P[( 89.5 - 94.8) / 4.4618 < Z < ( 100.5 - 94.8) / 4.4618]

P( -1.1879 < Z < 1.2775) = .8997 - 0.1170 = 0.7827

find the area below 1.2775; find the area below -1.1879; subtract.

(From Normal probability table)

c)

= 94.8 , = 4.4618

standardize x to z = (x - ) /

P(x < 19.5) = P( z < (19.5-94.8) / 4.4618)

= P(z < -16.8766) = 0.0000

(From Normal probability table)

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