Question

Carbon monoxide​ (CO) emissions for a certain kind of car vary with mean 2.842 ​g/mi and standard deviation 0.7 g/mi. A company has 80 of these cars in its fleet. Let y (overbar) represent the mean CO level for the​ company's fleet.

​a) What's the approximate model for the distribution of y (overbar)​? Explain.

​b) Estimate the probability that y (overbar) is between 2.9 and 3 g/mi.

​c) There is only a 5% chance that the​ fleet's mean CO level is greater than what​ value?

Answer 1

a)

~ N ( , / sqrt(n) ) = N ( 2.842 , 0.7/sqrt(80 ) )

b)

P ( 2.9 < < 3 ) = ?
Standardizing the value
Z = ( y - µ ) / ( σ / √(n))
Z = ( 2.9 - 2.842 ) / ( 0.7 / √(80))
Z = 0.74
Z = ( 3 - 2.842 ) / ( 0.7 / √(80))
Z = 2.02
P ( 0.74 < Z < 2.02 )
P ( 2.9 < < 3 ) = P ( Z < 2.02 ) - P ( Z < 0.74 )
P ( 2.9 < < 3 ) = 0.9783 - 0.7704
P ( 2.9 < < 3 ) = 0.2079

c)

P ( > y ) = 1 - P ( < y ) = 1 - 0.05 = 0.95
To find the value of y
Looking for the probability 0.95 in standard normal table to calculate Z score = 1.6449
Z = ( y - µ ) / ( σ / √(n) )
1.6449 = ( y - 2.842 ) / (0.7/√(80))
= 2.971