Normal distribution question: determine sigma

[muffymuff]

The lengths of a certain species of worm follow a normal distribution. Thirty percent of the worms are at least
16cm long, and 15% of the worms are less than 10cm long. Find, to 2 decimal places, the standard deviation of
the lengths of the worms.

 

[MarkFL]

Hello and welcome to MHB! :D

We normally ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

So, let's look at what we've been given. First, we are told:

Thirty percent of the worms are at least 16 cm long.

This tells us that the $z$-score corresponding to the $x$-value of 16 cm has an associated area of 0.2. This is because 70% of the data will be to the left of it, and so 20% is between the mean and this value. Consulting a table, we then find the $z$-score is about 0.525. And so using the equation:

\(\displaystyle z=\frac{x-\mu}{\sigma}\)

We obtain:

\(\displaystyle 0.525=\frac{16-\mu}{\sigma}\)

Now, let's look at the other information we are given:

15% of the worms are less than 10cm long.

Can you obtain a second equation, so that you have two equations and two unknowns and both can be determined?

 

[MarkFL]

If "15% of the worms are less than 10cm long" then we need the $z$-score associated with an area of:

\(\displaystyle .5-0.15=0.35\)

and consulting a table, we find this is about $-z=1.037$ (we negate $z$ because we are below the mean), and so we may write:

\(\displaystyle 1.037=\frac{\mu-10}{\sigma}\)

Recall we found:

\(\displaystyle 0.525=\frac{16-\mu}{\sigma}\)

Multiplying both equations by $\sigma$ and then adding, we obtain:

\(\displaystyle 1.562\sigma=6\implies \sigma\approx3.84\text{ cm}\)