The Wall Street Journal studied for National Football League games during a rece

Question

The Wall Street Journal studied for National Football League games during a recent reasons to determine the number of minutes of actual playing time that occurred during a 3-hour game. Surprisingly, the average playing time was only 10.7 minutes per game! Assume the actual playing time per game follows the normal distribution with a standard deviation of 0.8 minutes. Determine the probability that the actual playing time for the next game will be between:

a. 9.8 and 11.9 minutes

b. 11.0 and 12.1 minutes

c. 9.7 and 10.3 minutes

Explanation / Answer

It is given that the mean is 10.7 and standard deviation is 0.8

a) P(9.8<x<11.9)=P((9.8-10.7)/0.8<z<(11.9-10.7)/0.8)=P(-1.13<z<1.5). this is P(z<1.5)-(1-P(z<1.13)), from normal distribution table it gives 0.9332-(1-0.8708)=0.804

b) P(11<x<12.1)=P((11-10.7)/0.8<z<(12.1-10.7)/0.8)=P(0.38<z<1.75), from normal distribution table it is 0.9599-0.6480=0.3119

c) P(9.7<x<10.3)=P((9.7-10.7)/0.8<z<(10.3-10.7)/0.8)=P(-1.25<z<-0.5) =P(z<1.25)-P(z<0.5), from the normal table it gives us 0.8944-0.6915=0.2029

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