Answer: 0.0062
Step-by-step explanation:
Given : The weight of trucks traveling on a particular section of I-475 have a population mean of [tex]\mu=15.8[/tex] tons and a population standard deviation of [tex]\sigma=4.2[/tex] tons.
Sample size : n=49
Let x be the random variable that represents the weight of trucks .
z-score : [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
For x= 14.3 tons
[tex]z=\dfrac{14.3-15.8}{\dfrac{4.2}{\sqrt{49}}}\approx-2.5[/tex]
Now, the probability a state highway inspector could select a sample of 49 trucks and find the sample mean to be 14.3 tons or less will be :-
[tex]P(X\leq14.3)=P(z<-2.5)= 0.0062097\approx 0.0062[/tex]
Hence, the probability a state highway inspector could select a sample of 49 trucks and find the sample mean to be 14.3 tons or less =0.0062
