Answer:
(a) 0.13894 (b) 13,89% (c) 0.08688 (e) 3006 gal/hr or less
Step-by-step explanation:
(a) The fuel consumption is normally distributed N(μ=3245,σ=186).
The z-scores for 3105 and 3498 are
[tex]z=\frac{(X-\mu)}{\sigma }\\\\z1=(3105-3245)/186 = 0.7527\\z2=(3498-3245)/186 = 1.3602\\[/tex]
We can calculate the probability of fuel consumption bieng between 3105 and 3498 as:
P(0.7527<z<1.3602) = P(z<1.3602) - P(z<0.7527)
The probability of the fuel consumption being smaller than 3498 is
P(z<1.3602) = 0.91312
and the probability of the fuel consumption being smaller than 3105 is
P(z<0.7527) = 0.77418
Then
P(0.7527<z<1.3602) = P(z<1.3602) - P(z<0.7527)
P(0.7527<z<1.3602) = 0.91312-0.77418 = 0.13894 or 13.89%
(c) For a fuel consumption of 3498, we calculate z=1.3602
P(fuel consumption>3498) = P(z>1.3602)=1-P(z<1.3602)
P(fuel consumption>3498) = 1-0.91312 = 0.08688
(e) The 10% most efficient planes satisfy the next equation:
P(fuel consumption<X) = P(z<z1) = 0,1
Looking at the probabilities table, z1=-1.28155
Knowing z1, we can calculate X as
[tex]z=\frac{(X-\mu)}{\sigma }\\\\X=\mu+\sigma*z\\X=3245+186*(-1.28155)= 3006[/tex]
The fuel consumption of the 10% most efficient Boeing 747 jets is 3006 gallons per hour or less