Question 914487
normally distributed with a mean value 31 psi and standard deviation 1 psi. 
a) What is the probability that the pressure for a randomly selected tire is between 30 psi and 32 psi? 
Empirical Rule(1SD from mean) 68%
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}

one  standard deviation from the mean accounts for about 68% of the set 
two standard deviations from the mean account for about 95%
and three standard deviations from the mean account for about 99.7%.
  
0r P = normalcdf(30,32,31,1) 
0r  {{{z =blue (x - mu)/blue(sigma)}}} P = normalcdf(-1,1)
........
 b) What is the maximum tire pressure X for the bottom 25% of the cars? 
1(invNorm(.25) + 31 = X
.................
 c) What is the median tire pressure of all cars of this model? 
Normal Distribution(Mean Median & Mode the same):   31 psi
.............
 d) If a sample of 25 tires is selected, what is the probability that the average air pressure of these tires is more than 31.5 psi?
 {{{z =blue (x - mu)/blue(sigma/sqrt(n)) = .5/(1/5)}}} = 2.5, 
P(x > 31.5) = normalcdf(2.5,100)