Answer:
[tex]P(X\geq90) =0.00878[/tex]
Step-by-step explanation:
This situation can be modeled by a binomial distribution of parameters:
[tex]p = 0.8\\\\q = 0.2\\\\n = 100[/tex]
We want to find the probability that at least 90 are in repair.
We can approximate this problem to a normal distribution, where:
[tex]\mu = np[/tex]
[tex]\mu = 80[/tex]
[tex]\sigma = \sqrt{npq}\\\\\sigma = 4[/tex]
Then we look for
[tex]P(X\geq90) = P(X>90 -0.5)\\\\P(X\geq90) = P(X>89.5)[/tex]
Then we must find the normal standard statistic Z-score
[tex]Z = \frac{X-\mu}{\sigma}[/tex]
Therefore:
[tex]P(X>89.5) = P(\frac{X-\mu}{\sigma}>\frac{89.5-80}{4})\\\\P(X>89.5) =P(Z>2.375)[/tex]
Looking in the standard normal table we obtain:
[tex]P(Z>2.375) = 0.00878[/tex]
