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

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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:

$$z=\frac{x-\mu}{\sigma}$$

We obtain:

$$0.525=\frac{16-\mu}{\sigma}$$

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:

$$.5-0.15=0.35$$

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:

$$1.037=\frac{\mu-10}{\sigma}$$

Recall we found:

$$0.525=\frac{16-\mu}{\sigma}$$

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

$$1.562\sigma=6\implies \sigma\approx3.84\text{ cm}$$