Question

Suppose that 21.84% of all vehicles on the road get a fuel economy of at least 35 miles per gallon (MPG). You conduct an observational study on a stretch of road outside of campus and record the make and model of 729 vehicles over a month long period. What is the probability that between 21.81% and 23.99% of the cars you observed get at least 35 MPG? options: 1) 1.4122 2) 0.4278 3) 35.7818 4) 0.5722 5) -16.3499

Answer 1

Here, μ = 0.2184, σ = 0.0153, x1 = 0.2181 and x2 = 0.2399. We need to compute P(0.2181<= X <= 0.2399). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (0.2181 - 0.2184)/0.0153 = -0.02
z2 = (0.2399 - 0.2184)/0.0153 = 1.41

Therefore, we get
P(0.2181 <= X <= 0.2399) = P((0.2399 - 0.2184)/0.0153) <= z <= (0.2399 - 0.2184)/0.0153)
= P(-0.02 <= z <= 1.41) = P(z <= 1.41) - P(z <= -0.02)
= 0.9208 - 0.493

= 0.4278