One study on managers satisfaction with management tools rev

One study on manager’s satisfaction with management tools reveals that 70% of managers use self-directed work teams as a management tool. Suppose 150 managers selected at random in the United States are interviewed.

What is the mean of the distribution?

What is the standard deviation of the distribution?

What is the probability that 100 managers use self-directed work teams?

What is the probability that between 4 and 9 use self-directed work teams as a management tool?

What is the probability that fewer than 6 use self-directed work teams as a management tool?

Solution

Prob for a Manager to be successful p = 0.70

Each manager is independent, and there are two outcomes.

Hence X no of managers successful is binomial with n = 150 and p = 0.70

Mean of the distribution = np = 150(0.7) = 105

Variance = npq = 150(0.7)(0.3) = 31.5

Std dev = 5.612

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probability that 100 managers use self-directed work teams

= P(X=100)

= 0.04674

probability that between 4 and 9 use self-directed work teams as a management tool

= P(4<=x<=9) = 6.470x10^-66-2.614x10^-72

probability that fewer than 6 use self-directed work teams as a management tool

=P(X<6) = 1.5364(10^-68)