Question

V. Money Magazine reported that the average price of gasoline in the United States during the first quarter of 2008 was $3.46. Assume that the price reported by Money is the population mean, and the standard deviation σ is $0.15. a. What is the probability that the mean price for a sample of 30 gas stations is within $0.03 of the population mean?

Answer 1

Answer:

$z=\frac{3.43 -3.46}{\frac{0.15}{\sqrt{30}}} = -1.095$

$z=\frac{3.49 -3.46}{\frac{0.15}{\sqrt{30}}} = 1.095$

And we can find this probability using the normal standard table and we got:

$P(-1.095<z<1.095) = P(z<1.095) -P(z<-1.095) =0.863 -0.137= 0.726$

Step-by-step explanation:

Let X the random variable that represent the price of a population, and for this case we know the distribution for X is given by:

$X \sim N(3.46,0.15)$  

Where $\mu=3.46$ and $\sigma=0.15$

And for this case we want to find the following probability:

$P(3.43 \leq \bar X \leq 3.49)$

And we can use the z score formula given by:

$z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

If we find the z score for the limits we got:

$z=\frac{3.43 -3.46}{\frac{0.15}{\sqrt{30}}} = -1.095$

$z=\frac{3.49 -3.46}{\frac{0.15}{\sqrt{30}}} = 1.095$

And we can find this probability using the normal standard table and we got:

$P(-1.095<z<1.095) = P(z<1.095) -P(z<-1.095) =0.863 -0.137= 0.726$