Respuesta :
Answer:
[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]
[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]
And if we solve for the mean we got
[tex]\mu =4 +0.842*1.1=4.926[/tex]
Step-by-step explanation:
Assuming this question "Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of 1.1 years. Suppose also that exactly 20% of the tubes die before 4 years. Find the mean lifetime of TV tubes. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place. ?
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 TV tubes of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu,1.1)[/tex]
Where [tex]\mu[/tex] and [tex]\sigma=1.1[/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]
For this part we know the following condition:
[tex]P(X>4)=0.8[/tex] (a)
[tex]P(X<4)=0.2[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(z>-0.842)=0.8
If we use condition (b) from previous we have this:
[tex]P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2[/tex]
[tex]P(z<\frac{4-\mu}{\sigma})=0.2[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=-0.842<\frac{4-\mu}{1.1}[/tex]
And if we solve for the mean we got
[tex]\mu =4 +0.842*1.1=4.926[/tex]
