Question

Suppose that the lifetimes of TV tubes are normally distributed with a standard deviation of years. Suppose also that exactly of the tubes die before 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.

Answer 1

Answer:

$P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2$  

$P(z<\frac{4-\mu}{\sigma})=0.2$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842<\frac{4-\mu}{1.1}$

And if we solve for the mean we got

$\mu =4 +0.842*1.1=4.926$

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:

$X \sim N(\mu,1.1)$  

Where $\mu$ and $\sigma=1.1$

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

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

For this part we know the following condition:

$P(X>4)=0.8$   (a)

$P(X<4)=0.2$   (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:

$P(X<4)=P(\frac{X-\mu}{\sigma}<\frac{4-\mu}{\sigma})=0.2$  

$P(z<\frac{4-\mu}{\sigma})=0.2$

But we know which value of z satisfy the previous equation so then we can do this:

$z=-0.842<\frac{4-\mu}{1.1}$

And if we solve for the mean we got

$\mu =4 +0.842*1.1=4.926$