Question

lextime is a term used to describe formalized flexible working hours on the job. Many managers believe that flextime reduces worker tardiness, absenteeism, and turnover, and increases job satisfaction and performance. (Michigan State University Business Topics, Summer 1979). Assume that the distribution of job satisfaction ratings of all flextime workers has a population mean of =35 and a standard deviation of =10. Suppose a random sample of 35 flextime workers was taken and the job satisfaction rating was determined for each person in the sample. Calculate the following probabilities.

a. P( > 39) =

b. P( < 37) =

c. P(33 < < 39 ) =

Answer 1

A)

Here, μ = 35, σ = 1.6903 and x = 39. We need to compute P(X >= 39). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (39 - 35)/1.6903 = 2.37

Therefore,
P(X >= 39) = P(z <= (39 - 35)/1.6903)
= P(z >= 2.37)
= 1 - 0.9911 = 0.0089

B)

Here, μ = 35, σ = 1.6903 and x = 37. We need to compute P(X <= 37). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (37 - 35)/1.6903 = 1.18

Therefore,
P(X <= 37) = P(z <= (37 - 35)/1.6903)
= P(z <= 1.18)
= 0.8810

C)

Here, μ = 35, σ = 1.6903, x1 = 33 and x2 = 39. We need to compute P(33<= X <= 39). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (33 - 35)/1.6903 = -1.18
z2 = (39 - 35)/1.6903 = 2.37

Therefore, we get
P(33 <= X <= 39) = P((39 - 35)/1.6903) <= z <= (39 - 35)/1.6903)
= P(-1.18 <= z <= 2.37) = P(z <= 2.37) - P(z <= -1.18)
= 0.9911 - 0.119
= 0.8721