Respuesta :
Answer:
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 70, p = 0.58[/tex]
So
[tex]\mu = E(X) = np = 70*0.58 = 40.6[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{70*0.58*0.42} = 4.13[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{34.5 - 40.6}{4.13}[/tex]
[tex]Z = -1.48[/tex]
[tex]Z = -1.48[/tex] has a pvalue of 0.0694.
6.94% probability that fewer than 35 use self-directed work teams as a management tool
