Answer:
0.0475
Step-by-step explanation:
We use the binomial approximation to the normal 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 distributions 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 a large company, the proportion of employees who were promoted during the last year was 0.10.
This means that [tex]p = 0.1[/tex]
100 employees
This means that [tex]n = 100[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 100*0.1 = 10[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.1*0.9} = 3[/tex]
What is the probability that at least 15 of them were promoted during the last year?
This is [tex]P(X \geq 15)[/tex], which is 1 subtracted by the pvalue of Z when X = 15. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{15 - 10}{3}[/tex]
[tex]Z = 1.67[/tex]
[tex]Z = 1.67[/tex] has a pvalue of 0.9525
1 - 0.9525 = 0.0475.
0.0475 is the answer.
