Question

A tire manufacturer designed a new tread pattern for its all-weather tires. Repeated tests were conducted on cars of approximately the same weight traveling at 60 miles per hour. The tests showed that the new tread pattern enables the cars to stop completely in an average distance of 125 feet with a standard deviation of 6.5 feet and that the stopping distances are approximately normally distributed. a. (2pts) What is the 70th percentile of the distribution of stopping distances? (Show work, give units) b. (2pts) What is the probability that a randomly selected car will have a stopping distance less than 115 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places) c. (4pts) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places) d. (4pts) What is the probability that a randomly selected sample of 15 cars in the study will have a mean stopping distance between 120 and 130 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places)

Answer 1

Answer:

a) $X= 125+ 0.524(6.5)=128.406$ feet

b) $P(X$

c) $P(\bar X\geq 130)=1-0.9573=0.0427$

d) $P(120 \leq \bar X \leq 130)=0.9971$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

a. (2pts) What is the 70th percentile of the distribution of stopping distances? (Show work, give units)

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

$X \sim N(125,6.5)$  

Where $\mu=125$ and $\sigma=6.5$

So we are interested on a value c that satisfy the following condition:

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

So we can find a z score in the normal standard distribution that accumulates 0.7 of the area on the left and 0.3 on the right. And this value on this case is z=0.524. And now we can solve X from the z score formula:

$0.524=\frac{x-125}{6.5}$

$X= 125+ 0.524(6.5)=128.406$ feet

b. (2pts) What is the probability that a randomly selected car will have a stopping distance less than 115 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places)

On this case we want this probability:

$P(X$

And we can solve this using again the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply the z score formula we got:

$P(X$

And we can find this probability on this way:

$P(z$

c. (4pts) What is the probability that a randomly selected sample of 5 cars in the study will have a mean stopping distance of at least 130 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places)

Let $\bar X$ represent the sample mean, the distribution for the sample mean by the ceentral limit theorem is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case  $\bar X \sim N(125,\frac{6.5}{\sqrt{5}}=2.907)$

And we want this probability:

$P(\bar X \geq 130)$

And using the z score formula we got this:

$P(\bar X \geq 130)=P(Z>\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}})=P(Z>\frac{130-125}{2.907})=P(Z>1.720)$

And we can use the complement rule like this:

$P(Z>1.720)=1-P(Z$

d. (4pts) What is the probability that a randomly selected sample of 15 cars in the study will have a mean stopping distance between 120 and 130 feet? (Give the proper probability statements/notation, show work, and give value to 4 decimal places)

On this case  $\bar X \sim N(125,\frac{6.5}{\sqrt{15}}=1.678)$

And we want this probability:

$P(120 \leq \bar X \leq 130)$

And using the z score formula we got this:

$P(120 \leq \bar X \leq 130)=P(\frac{120-125}{1.678}$

And we can find this probability on this way:

$P(-2.9792.979)=P(Z$