Question

Suppose that the weight distribution of monitor lizards is normal, with 10% of all
monitor lizards having a weight of more than 10.6340 kg and 5% having a weight
of less than 9.7565 kg. Find the mean and standard deviation of the weight dis-
tribution.

Answer 1

Answer on Question #67848 – Math – Statistics and Probability

Question

Suppose that the weight distribution of monitor lizards is normal, with $10\%$ of all monitor lizards having a weight of more than $10.6340\mathrm{kg}$ and $5\%$ having a weight of less than 9.7565 kg. Find the mean and standard deviation of the weight distribution.

Solution

Let $X$ be a random variable which corresponds to the weight distribution of monitor lizards. We have that

$$P \{X > 10.6340\} = \frac{10}{100} = 0.1$$

and

$$P \{X < 9.7565\} = \frac{5}{100} = 0.05.$$

If $X$ is normally distributed with the mean $\mu$ and the variance $\sigma^2$, then $\frac{X - \mu}{\sigma}$ has the standard normal distribution.

Therefore,

$$P \{X > 10.6340\} = P \left\{\frac{X - \mu}{\sigma} > \frac{10.6340 - \mu}{\sigma} \right\} = 0.1$$

and

$$P \{X < 9.7565\} = P \left\{\frac{X - \mu}{\sigma} < \frac{9.7565 - \mu}{\sigma} \right\} = 0.05.$$

By Excel we get that

$$\frac{10.6340 - \mu}{\sigma} = 1.2816$$

and

$$\frac{9.7565 - \mu}{\sigma} = -1.6449.$$

In order to find $\mu$ and $\sigma$ we have to solve the system of linear equations

$$\left\{ \begin{array}{l} 10.6340 - \mu = 1.2816 \sigma, \\ 9.7565 - \mu = -1.6449 \sigma. \end{array} \right.$$

Hence $\mu = 10.2498$ and $\sigma = 0.2998$.

**Answer**: the mean is $\mu = 10.2498$ and the standard deviation is $\sigma = 0.2998$.

Answer provided by https://www.AssignmentExpert.com