SOLUTION: 5. The air pressure in a randomly selected tire put on a certain model new car is normally distributed with a mean value 31 psi and standard deviation 1 psi. a) What is the

Algebra ->  Probability-and-statistics -> SOLUTION: 5. The air pressure in a randomly selected tire put on a certain model new car is normally distributed with a mean value 31 psi and standard deviation 1 psi. a) What is the      Log On


   



Question 914487: 5. The air pressure in a randomly selected tire put on a certain model new car is 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?
b) What is the maximum tire pressure for the bottom 25% of the cars?
c) What is the median tire pressure of all cars of this model?
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?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
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%

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+%28x+-+mu%29%2Fblue%28sigma%29 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+%28x+-+mu%29%2Fblue%28sigma%2Fsqrt%28n%29%29+=+.5%2F%281%2F5%29 = 2.5,
P(x > 31.5) = normalcdf(2.5,100)