Respuesta :
Answer:
100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 38000, \sigma = 1500[/tex]
What is the probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles
This is the pvalue of X when X = 44000 subtracted by the pvalue of Z when X = 32000. So
X = 44000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{44000 - 38000}{1500}[/tex]
[tex]Z = 4[/tex]
[tex]Z = 4[/tex] has a pvalue of 1.
X = 32000
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{32000 - 38000}{1500}[/tex]
[tex]Z = -4[/tex]
[tex]Z = -4[/tex] has a pvalue of 0.
1 - 0 = 100%
100% probability that the lifespan of a set of tires will be between 32,000 miles and 44,000 miles
Answer:
[tex]P(32000<X<44000)=P(\frac{32000-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{44000-\mu}{\sigma})=P(\frac{32000-38000}{1500}<Z<\frac{44000-38000}{1500})=P(-4<z<4)[/tex]
And we can find this probability with this difference:
[tex]P(-4<z<4)=P(z<4)-P(z<-4)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-4<z<4)=P(z<4)-P(z<-4)=0.999968-0.0000317=0.999937[/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".
Solution to the problem
Let X the random variable that represent the lifespan of a population, and for this case we know the following parameters
Where [tex]\mu=65.5[/tex] and [tex]\sigma=2.6[/tex]
We are interested on this probability
[tex]P(32000<X<44000)[/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]
If we apply this formula to our probability we got this:
[tex]P(32000<X<44000)=P(\frac{32000-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{44000-\mu}{\sigma})=P(\frac{32000-38000}{1500}<Z<\frac{44000-38000}{1500})=P(-4<z<4)[/tex]
And we can find this probability with this difference:
[tex]P(-4<z<4)=P(z<4)-P(z<-4)[/tex]
And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.
[tex]P(-4<z<4)=P(z<4)-P(z<-4)=0.999968-0.0000317=0.999937[/tex]
