Answer:
The probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.
Step-by-step explanation:
Let the random variable X represent the miles-per-gallon rating of passenger cars.
It is provided that [tex]X\sim N(\mu=33.8,\ \sigma^{2}=3.5^{2})[/tex].
Compute the probability that a randomly selected passenger car gets more than 37.3 mpg as follows:
[tex]P(X>37.3)=P(\frac{X-\mu}{\sigma}>\frac{37.3-33.8}{3.5})[/tex]
[tex]=P(Z>1)\\\\=1-P(Z<1)\\\\=1-0.84134\\\\=0.15866\\\\\approx 0.1587[/tex]
Thus, the probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.
