1. ## Symmetrical about the mean question

I had a question like this yesterday but this one is a little different it seems i don't really get these ones...

The masses of 17 year old males in a school are normally distributed with a mean of 70 kg and a standard deviation of 6 kg.
c) Find the range of measurements that is symmetrical about the mean and contains 50% of these students

I don't even know how to start this question.. if someone could just give me some hints to get started...

I thought that the mean is already symmetrical about the mean by 50% in a normal distribution...

the answer in the back of the book is 66.0 kg is less than m is less than 74.0kg

2. Originally Posted by tmas
I had a question like this yesterday but this one is a little different it seems i don't really get these ones...

The masses of 17 year old males in a school are normally distributed with a mean of 70 kg and a standard deviation of 6 kg.
c) Find the range of measurements that is symmetrical about the mean and contains 50% of these students

I don't even know how to start this question.. if someone could just give me some hints to get started...

I thought that the mean is already symmetrical about the mean by 50% in a normal distribution...

the answer in the back of the book is 66.0 kg is less than m is less than 74.0kg
By symmetry, you need to find the value of a such that $\Pr(X > a) = 0.25$. Draw a picture to see why. Then the required interval will be $70 - a \leq X \leq 70 + a$.

3. 50% symmetrical around the mean breaks the interval into 25-50-25. Is this clear?

Therefore you are looking to find the number(s) on the curve that separates these breaks.

Calling this number 'a' and the fact that symmetry gives the answer in standard normal as $\pm a$ then find

$P\left( Z< \frac{a-70}{6}\right) = 0.25$