Question

The chief chemist for a major oil and gasoline production company claims that the regular unleaded gasoline produced by the company contains on average 4 ounces of a certain ingredient. The chemist further states that the distribution of this ingredient per gallon of regular unleaded gasoline is normal and has a standard deviation of 1.2 ounces. What is the probability of finding an average less than 3.85 ounces of this ingredient from 64 randomly inspected 1-gallon samples of regular unleaded gasoline?

.1587

.1357

.8643

.8413

Answer 1

Solution :

Given that ,

mean = = 4

standard deviation = = 1.2

n = 64

= 4

=  / n = 1.2/ 64=0.15

P( <3.85 ) = P[( - ) / < (3.85 -4) / 0.15]

= P(z < -1)

Using z table  

=0.1587

probability=0.1587