The weight, in grams, of beans in a tin is normally distributed with mean m and standard deviation 7.8 grams. Given that 10% of tins contain less than 200 grams, explain why

Assuming this question "The weight, in grams, of beans in a tin is normally distributed with mean m and standard deviation 7.8 grams. Given that 10% of tins contain less than 200 grams, Find the mean m. explain why. ?

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 =m, 7.8)[/tex]

Where [tex]\mu =m[/tex] and [tex] \sigma =7.8[/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>200) =0.9[/tex] (a)

[tex] P(X<200) = 0.1[/tex] (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value m.

As we can see on the figure attached the z value that satisfy the condition with 0.1 of the area on the left and 0.9 of the area on the right it's z=-1.28.. On this case P(Z<-1.28)=0.1 and P(z>-1.28)=0.9

If we use condition (b) from previous we have this:

[tex]P(X<200)=P(\frac{X-\mu}{\sigma}<\frac{200-\mu}{\sigma})=0.1[/tex]

[tex]P(z<\frac{200-\mu}{\sigma})=0.1[/tex]

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

[tex]z=-1.28<\frac{200-\mu}{7.8}[/tex]

And if we solve for [tex]\mu[/tex] we got

[tex]\mu=200 +1.28*7.8=209.984[/tex]