Thread: Normal distribution problem

Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!

The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.

Originally Posted by Fnus

Hey, I've always been bad at these, and I don't even know what to plot into the calculator, so any help is great appreciated, thanks!

The weights of adult males of a type og dog may be assumed to be normally distributed with mean 25 kg and a standard deviation of 3 kg.
Given that 30% of the weights lie between 25 kg and x kg, where x > 25, find the value of x.

which is now just a routine inverse normal problem.

First thing you do it simply picture a normal distribution curve. Right in the middle is the average, which is 25 kg in this case. You are looking for an area under the curve in between x and the mean, and since x > 25 it will be the right side of the mean.

Since both on the left an the right side of the mean you have exactly 50% you know that left of the mean you have 50%, then you have the area mean to x which is 30% and then you have a remaining 20%.

I assume you have a TI calculator, if not, what I'll say is useless. You press [2ND] [Vars] [2] [ENTER] and you'll have the program 'normalcdf'. If you enter a left boundy, a right boundry, the mean, the standard deviation and it will give you the area contained withing the boundaries. (for instance: normalcdf(25,10^90,25,3) will give exactly .5 because that's half of the whole area.)

You are looking for not halve, but 30% = .3 of the area. Go to [tex] solver and enter eqn: 0= normalcdf(25,x,25,3) - 0,3 and then press down [APLHA] [ENTER] and it will give you the answer.

I hope you understand. And also, you mentioned plotting, which is what most schools teach, but using the solver is a LOT easier and faster.