Answer: 0.9525
Step-by-step explanation:
Given : The salaries of elementary school teachers in the united states are normally distributed with a mean of [tex]\mu=\$32,000[/tex] and a standard deviation of [tex]\sigma=\$3000[/tex].
Sample size : n= 100
Let x be a random variable that denotes the salaries of elementary school teachers in the united states.
Formula : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
Then, the probability that their mean salary is less than $32,500 will be :-
[tex]P(x<32500)=P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{32500-32000}{\dfrac{3000}{\sqrt{81}}})\\\\=P(z<\dfrac{500}{\dfrac{3000}{10}})\approx P(z<1.67)=0.9525\ \ [\text{Using z-value}][/tex]
Hence, the probability that their mean salary is less than $32,500 = 0.9525
