Respuesta :
Answer:
a) [tex]X= 125+ 0.524(6.5)=128.406[/tex] feet
b) [tex]P(X<115)=0.0620[/tex]
c) [tex]P(\bar X\geq 130)=1-0.9573=0.0427[/tex]
d) [tex]P(120 \leq \bar X \leq 130)=0.9971[/tex]
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:
[tex]X \sim N(125,6.5)[/tex]
Where [tex]\mu=125[/tex] and [tex]\sigma=6.5[/tex]
So we are interested on a value c that satisfy the following condition:
[tex]P(X<c)=0.7[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
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:
[tex]0.524=\frac{x-125}{6.5}[/tex]
[tex]X= 125+ 0.524(6.5)=128.406[/tex] 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:
[tex]P(X<115)[/tex]
And we can solve this using again the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply the z score formula we got:
[tex]P(X<115)=P(\frac{X-\mu}{\sigma}<\frac{115-\mu}{\sigma})=P(Z<\frac{115-125}{6.5})=P(z<-1.538)[/tex]
And we can find this probability on this way:
[tex]P(z<-1.538)=0.0620[/tex]
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 [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean by the ceentral limit theorem is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(125,\frac{6.5}{\sqrt{5}}=2.907)[/tex]
And we want this probability:
[tex]P(\bar X \geq 130)[/tex]
And using the z score formula we got this:
[tex]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)[/tex]
And we can use the complement rule like this:
[tex]P(Z>1.720)=1-P(Z<1.720)=1-0.9573=0.0427[/tex]
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 [tex]\bar X \sim N(125,\frac{6.5}{\sqrt{15}}=1.678)[/tex]
And we want this probability:
[tex]P(120 \leq \bar X \leq 130)[/tex]
And using the z score formula we got this:
[tex]P(120 \leq \bar X \leq 130)=P(\frac{120-125}{1.678}<Z<\frac{130-125}{1.678})=P(-2.979<Z<2.979)[/tex]
And we can find this probability on this way:
[tex]P(-2.979<Z>2.979)=P(Z<2.979)-P(Z<-2.979)=0.9986-0.00145=0.9971[/tex]
