Respuesta :
Answer:
1.22% of batteries last less than 5 years
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 5.9, \sigma = 0.4[/tex]
What proportion of batteries fail to meet the guarantee? In other words, what proportion of batteries last less than 5 years?
This is the pvalue of Z when X = 5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 5.9}{0.4}[/tex]
[tex]Z = -2.25[/tex]
[tex]Z = -2.25[/tex] has a pvalue of 0.0122
1.22% of batteries last less than 5 years
Answer:
[tex]P(X<5)=P(\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(Z<\frac{5-5.9}{0.4})=P(z<-2.25)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.25)=0.0122[/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 lifetimes of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(5.9,0.4)[/tex]
Where [tex]\mu=5.9[/tex] and [tex]\sigma=0.4[/tex]
We are interested on this probability
[tex]P(X<5)[/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(X<5)=P(\frac{X-\mu}{\sigma}<\frac{5-\mu}{\sigma})=P(Z<\frac{5-5.9}{0.4})=P(z<-2.25)[/tex]
And we can find this probability using the normal standard table or excel and we got:
[tex]P(z<-2.25)=0.0122[/tex]
