Answer to Question #329061 in Statistics and Probability for Ashley

Question

Directions: Read, analyze, and solve the problems below. Show your complete solutions.

a. What is the probability that the 36 tires will have an average of less than 16,000 miles until the tires begin to wear out?

b. What is the probability that the 36 tires will have an average of more than 18,000 miles until the tires begin to wear out?

3 The number of driving miles before a certain kind of tire begins to show wear is on the average, 16, 800 miles with a standard deviation of 3, 300 miles.

2 The average number of pages in a novel is 326 with a standard deviation of 24 pages. If a sample of 50 novels is randomly chosen, what is the probability that the average number of pages in these books is between 319 and 331?

1 There are 250 dogs at a dog show that weigh an average of 12 pounds, with a standard deviation of 8 pounds. If 4 dogs are chosen at random, what is the probability that the average weight is greater than 8 pounds?

Answer 1

3)We have that:

$\mu=16800$

$\sigma=3300$

$n=36$

a) $P(\bar X<16000)=P(Z<\frac{x-\mu}{\frac{\sigma}{\sqrt n}})=P(Z<\frac{16000-16800}{\frac{3300}{\sqrt {36}}})=P(Z<-1.45)=0.0735$

b) $P(\bar X>18000)=1-P(\bar X < 18000)=1-P(Z<\frac{18000-16800}{\frac{3300}{\sqrt {36}}})=1-P(Z<2.18)=1-0.9854=0.0146$

Answer:

a) 0.0735

b) 0.0146

2)Given, $\mu=326,\sigma=24,n=50$

$X\sim N(\mu,\sigma)$

$P(319\le X\le331)=P(\dfrac{319-326}{24/\sqrt{50}}\le z\le\dfrac{331-326}{24/\sqrt{50}})$

$=P(-0.04\le z\le0.03)=P(z\le0.03)-P(z \le-0.04)$

$=P(z\le0.03)-[1-P(z \le0.04)]=0.51197-[1-0.51595] \\=0.02792$

1) Given, $\mu=12,\sigma=8,n=4$

$X\sim N(\mu,\sigma)$

$P(X>8)=P(z>\dfrac{8-12}{8/\sqrt{4}})=P(z>-1) \\=1-P(z\le-1)=1-[1-P(z\le1)] \\=P(z\le1)=0.84134$