Question

The masses of a sample of male frogs taken from a pond can be modeled by a normal distribution with a mean mass of 70g and standard deviation 5g. Four male frogs are chosen at random. Find the probability that their mean mass is less than 65g.

Answer 1

Answer on Question #56458 – Math – Statistics and Probability

Question

The masses of a sample male frogs taken from a pond can be modeled by a normal distribution with a mean mass of $70\mathrm{g}$ and standard deviation $5\mathrm{g}$. Four male frogs are chosen at random. Find the probability that their mean mass is less than $65\mathrm{g}$.

Solution

Let $\xi_{k}$, where $k = 1,2,3,4$, be mass of a male frog. It is given that $\xi_{k} \sim N(70; 5)$. Assume that they are independent identically distributed random variables.

Then we obtain

$$\xi_{1} + \xi_{2} + \xi_{3} + \xi_{4} \sim N(70 + 70 + 70 + 70; 5 + 5 + 5 + 5) = N(280; 20),$$

hence $\frac{\xi_1 + \xi_2 + \xi_3 + \xi_4 - 280}{20} \sim N(0; 1)$.

So the required probability is equal to

$$\begin{array}{l} P\left\{\frac{\xi_{1} + \xi_{2} + \xi_{3} + \xi_{4}}{4} < 65\right\} = P\{\xi_{1} + \xi_{2} + \xi_{3} + \xi_{4} < 260\} = P\{\eta < 260\} = P\left\{\frac{\eta - 280}{20} < \frac{260 - 280}{20}\right\} = \\ = P\left\{\frac{\eta - 280}{20} < -1\right\} = 0.5 - \Phi(1) = 0.5 - 0.34134 = 0.15866. \end{array}$$

Here $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{0}^{x} e^{-\frac{u^2}{2}} du$ is a tabulated function of Laplace and the value $\Phi(1)$ was found from the table of Laplace.

Answer: 0.15866.

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