Suppose that the distribution of touchdown passes (in football) is normally distributed with a mean of 250 feet and a standard deviation of 50 feet. We randomly sample 49 touchdowns.What is the probability that the 49 touchdowns traveled an average of less than 245 feet? Please explain how you derived your answer.

Answer: 0.2420Step-by-step explanation:Given: Mean : [tex]\mu = 250 \text{ feet}[/tex]Standard deviation : [tex]\sigma =50\text{ inch}[/tex]Sample size : [tex]n=49[/tex]The formula to calculate z is given by :-[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]For x= 245[tex]z=\dfrac{245-250}{\dfrac{50}{\sqrt{49}}}=-0.7[/tex]The P Value =[tex]P(Z<245)=P(z<-0.7)=0.2419637\approx0.2420[/tex]Hence, the probability that the 49 touchdowns traveled an average of less than 245 feet= 0.2420