Answer:
The probability that the sample mean is less than $56,000 is 0.0375.
Explanation:
Let X = teacher's salary in Connecticut.
The random variable X follows a Normal distribution with parameters μ = $57,337 and σ = $7,500.
A random sample of n = 100 teacher's salaries are selected.
The sample mean of every sample of 100 salaries will follow a normal distribution with mean μ and standard deviation σ/√n.
Compute the probability that the sample mean of the 100 salaries is less than $56,000 as follows:
[tex]P(\bar X<56000)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}<\frac{56000-57337}{7500/\sqrt{100}})\\=P(Z<-1.78)\\=1-P(Z<1.78)\\=1-0.9625\\=0.0375[/tex]
*Use a z-table for the probability.
Thus, the probability that the sample mean is less than $56,000 is 0.0375.
The probability that the sample mean is less than $56,000 is 0.0375.
Since average teacher's salary is $57,337. The standard deviation of $7,500. So,
Here,
= 7500/sqrt(100) = 750
Now
= Pr[x-bar < 56000]
= Pr[z < ((56000-57337)/750)]
= Pr[z < -1.78]
= 0.0375
hence, The probability that the sample mean is less than $56,000 is 0.0375.
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