A Bureau of National Affairs survey found that 79 of employe

A Bureau of National Affairs survey found that 79% of employers provide their workers a two-day paid Thanksgiving holiday with workers off both Thursday and Friday (USA Today, November 12, 2009). Nineteen percent of employers provide a one-day paid holiday with workers off Thanksgiving Day. Two percent of employers do not provide a paid Thanksgiving holiday. Consider a sample of 120 employers.

Round your answers to four decimal places.

a. What is the probability that at least 85 of the employers provide a two-day paid Thanksgiving holiday?

c. What is the probability that less than 20 employers provide a one-day paid Thanksgiving holiday?

Solution

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
? = 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)