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Sagot :
Using the normal approximation to the binomial, it is found that there is a 0.9319 = 93.19% probability that fewer than 260 are women.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
In this problem:
- Women comprise 83.3% of all elementary school teachers, hence [tex]p = 0.833[/tex]
- A sample of 300 teachers is taken, hence [tex]n = 300[/tex]
The mean and the standard deviation for the approximation are given by:
[tex]\mu = np = 300(0.833) = 249.9[/tex]
[tex]\sigma = \sqrt{np(1 - p)} = \sqrt{300(0.833)(0.167)} = 6.46[/tex]
Using continuity correction, as the binomial distribution is discrete and the normal distribution is continuous, the probability that fewer than 260 are women is P(X < 260 - 0.5) = P(X < 259.5), which is the p-value of Z when X = 259.5, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{259.5 - 249.9}{6.46}[/tex]
[tex]Z = 1.49[/tex]
[tex]Z = 1.49[/tex] has a p-value of 0.9319.
0.9319 = 93.19% probability that fewer than 260 are women.
A similar problem is given at https://brainly.com/question/25347055
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