Generating a list of numbers such that $\mu = 75$ and $\text{SD} > 15$

Question

Problem

The list $\left( 60, 90, 60, 80, 75, 60, 100 \right)$ has the average $\mu = 75$ and standard deviation $\sigma \approx 15$.

Generate a list of 7 numbers such that the average is $75$ and the standard deviation is greater than 15.

My thoughts

I figure I can generate a list of numbers where one of them is $75$ and the rest of them are $0$ and $150$ equally spread out.

My problem is that this is simply an intuitive method, and without calculating it specifically, I can't give a reason why SD>15 is guaranteed.

Question

Is there a way for me to, without calculating the SD, guarantee that it is greater than 15?

Answer 1

I think they want you to use the given list to generate yours. Remember that the SD is a measure of how spread out your values are from the center (that is $\mu$). Hint: If you take the numbers less than 75 and decrease them, and increase the ones greater than 75 you'll get a greater spread. But you'll have to make sure that you do this in a way that keeps the $\mu$ at 75. Does that get you started?

Answer 2

Hint: You probably know that $sd=\sqrt{\frac1n \sum_{i=1}^n x_i^2-\overline x^2}$

with $n=7$ , $\overline x=75$ and $ sd>15$. Now you can solve for $\sum_{i=1}^n x_i^2$ to get the necessary condition.

I used $\overline x$ instead of $\mu$ because the average is calculated by given numbers.

Answer 3

Standard deviation is a "measure of the dispersion of a set of data from its mean". If the data points are further from the mean, there is higher deviation within the data set.

Try this set:

$\left( 00, 25, 50, 75, 100, 125, 150 \right)$

and then this one:

$\left( 40, 50, 60, 75, 90, 100, 110 \right)$

and compare results.