One study on managers’ satisfaction with management tools reveals that 58% of all managers use self-directed work teams as a management tool. Suppose 70 managers selected randomly in the United States are interviewed. What is the probability that fewer than 35 use self-directed work teams as a management tool?

6.94% probability that fewer than 35 use self-directed work teams as a management tool

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

The standard deviation of the binomial distribution is:

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that , .

In this problem, we have that:

So

What is the probability that fewer than 35 use self-directed work teams as a management tool?

Using continuity correction, this is P(X < 35 - 0.5) = P(X < 34.5), which is the pvalue of Z when X = 34.5. So

has a pvalue of 0.0694.

6.94% probability that fewer than 35 use self-directed work teams as a management tool

2.83941151 would be the answer you're looking for! : )

i would say b) is your answer

sorry if im wrong

step-by-step explanation: