A stoplight is red for a random amount of time distributed N

A stoplight is red for a random amount of time distributed N(180, 361) seconds.

What is the 75^th percentile for the time the stoplight is red?

What two times mark the most likely (middle) 95% of the times the light is red?

Solution

a)

First, we get the z score from the given left tailed area. As          
          
Left tailed area =    0.75      
          
Then, using table or technology,          
          
z =    0.67448975      
          
As x = u + z * s / sqrt(n)          
          
where          
          
u = mean =    180      
z = the critical z score =    0.67448975      
s = standard deviation =    19      
          
Then          
          
x = critical value =    192.8153053   [ANSWER, 75TH PERCENTILE]

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b)

Note that              
              
Lower Bound = X - z(alpha/2) * s               
Upper Bound = X + z(alpha/2) * s           
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    180          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    19              
              
Thus,              
              
Lower bound =    142.7606843          
Upper bound =    217.2393157          
              
Thus, the confidence interval is              
              
(   142.7606843   ,   217.2393157   )

Thus the two times that mark the likely times are 142.76 and 217.24 [ANSWER]