Question

The quartiles of a normal distribution are 10 and 18 respectively. Find the approximate mean and standard deviation of the distribution?

Answer 1

Answer on Question #34559 – Math – Statistics and Probability

The quartiles of a normal distribution are 10 and 18 respectively. Find the approximate mean and standard deviation of the distribution?

Solution

We know that $Q_{1} = 10$ and $Q_{3} = 18$ are quartiles of normal distribution.

Since normal distribution is symmetric, the mean equals

$$\mu = \frac {Q _ {1} + Q _ {3}}{2} = \frac {1 0 + 1 8}{2} = 1 4$$

Now we know that

$$P (\xi < Q _ {1}) = 0. 2 5$$

Here $\xi$ has normal distribution with parameters which we want to find.

$$\begin{array}{l} P (\xi < Q _ {1}) = \Phi \left(\frac {Q _ {1} - \mu}{\sigma}\right) = \Phi \left(\frac {1 0 - 1 4}{\sigma}\right) = 0. 2 5 \\ - \frac {4}{\sigma} = \Phi^ {- 1} (0. 2 5) = - 0. 6 7 4 4 9 \\ \end{array}$$

Value of $\Phi^{-1}(0.25)$ can be found in any table of standard normal distribution.

So

$$\sigma = \frac {4}{0 . 6 7 4 4 9} = 5. 9 3 0 4 1$$

So parameters of normal distribution are: mean = 14 and

standard deviation = 5.93041.

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