What is the probability that a randomly selected teacher’s salary is greater than $43,700?
For a sample of 90 teachers, what is the probability that the sample mean is greater than 36,917? Assume that the sample is taken from a large population and the correction factor can be ignored.
Answer :
Given :
The population mean $(μ) = 35,441$
The population standard deviation $(σ) = 5,100$
Solution :
a) The Probability of a Teacher’s Salary Being Greater Than $43,700 :
$$\therefore P(x > 43,700) = P\left(\frac{x – \mu}{\sigma} > \frac{43,700 – 35,441}{5,100}\right)$$ $$= P(z > 1.62)$$ $$= 0.0526$$
b) The Probability of Sample Mean Salary Being Greater Than $36,917 :
→ here we have The sample size (n)=90
$$\therefore P(\bar{x} > 36,917) = P\left(\frac{\bar{x} – \mu}{\sigma/\sqrt{n}} > \frac{36,917 – 35,441}{\frac{5,100}{\sqrt{90}}}\right)$$ $$= P(z > 2.75)$$ $$= 0.0030$$
