Chapter 7, Problem 76CE

The length of a Colorado brook trout is normally distributed (a) What is the probability that a brook trout’s length exceeds the mean? (b) Exceeds the mean by at least 1 standard deviation? (c) Exceeds the mean by at least 2 standard deviations? (d) Is within 2 standard deviations?

To determine

Find the probability that the length of a brook trout exceeds the mean.

Answer to Problem 76CE

Theprobability that the length of a brook trout exceeds the meanis 0.5.

Explanation of Solution

Calculation:

It is given that the lengths of a population of brook trout are normally distributed.

Normal distribution:

A continuous random variable X is said to follow normal distribution if the probability density function of X is,

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{\left(x-\mu \right)}^2}{2{\sigma }^2}},-\infty <x<\infty$, with mean $\mu$ and standard deviation $\sigma$.

Denote X as the length of a randomly selected brook trout. It is taken from a population of brook trout that are normally distributed.

Denote $\mu$ as the mean of the population of the normally distributed lengths and $\sigma$ as the standard deviation.

The normal distribution is symmetrically distributed about its mean. From the concept of symmetry of a random variable X about $\mu$, it is known that:

$\begin{array}{c}P\left(X<\mu \right)=P\left(X>\mu \right)\\ =\mathrm{0.5.}\end{array}$

Thus, the probability that the length of a brook trout exceeds the mean is:

$P\left(X>\mu \right)=0.5$.

Hence, the probability that the length of a brook trout exceeds the mean is 0.5.

To determine

Find the probability that the length of a brook trout exceeds the mean by at least 1 standard deviation.

Answer to Problem 76CE

Theprobability that the length of a brook trout exceeds the mean by at least 1 standard deviationis 0.1587.

Explanation of Solution

Calculation:

Empirical Rule:

The Empirical Rule for a Normal model states that:

• • Within 1 standard deviation of mean, 68.26% of all observations will lie.
• • Within 2 standard deviations of mean, 95.44% of all observations will lie.
• • Within 3 standard deviations of mean, 99.73% of all observations will lie.

Consider the property regarding 1 standard deviation difference from mean of theEmpirical Rule. It indicates that:

$\begin{array}{l}P\left(\left|X-\mu \right|<\sigma \right)=0.6826\\ P\left(-\sigma <X-\mu <\sigma \right)=0.6826\\ P\left(X-\mu <\sigma \right)-P\left(X-\mu <-\sigma \right)=0.6826\\ P\left(X-\mu <\sigma \right)-\left[1-P\left(X-\mu <\sigma \right)\right]=0.6826\\ \left[\text{as for normal distribution, }P\left(X<-x\right)=1-P\left(X<x\right)\right]\end{array}$

$\begin{array}{c}2P\left(X-\mu <\sigma \right)-1=0.6826\\ 2P\left(X-\mu <\sigma \right)=1.6826\\ P\left(X-\mu <\sigma \right)=\frac{1.6826}{2}\\ =0.8413\\ 1-P\left(X-\mu >\sigma \right)=0.8413\left[\text{as }P\left(X<x\right)=1-P\left(X>x\right)\right]\\ P\left(X-\mu >\sigma \right)=1-0.8413\\ =\mathrm{0.1587.}\end{array}$

Hence, the probability that the length of a brook trout exceeds the mean by at least 1 standard deviationis 0.1587.

To determine

Find the probability that the length of a brook trout exceeds the mean by at least 2 standard deviations.

Answer to Problem 76CE

Theprobability that the length of a brook trout exceeds the mean by at least 2 standard deviations is 0.0228.

Explanation of Solution

Calculation:

Consider the property regarding 2 standard deviations difference from mean of theEmpirical Rule. It indicates that:

$\begin{array}{l}P\left(\left|X-\mu \right|<2\sigma \right)=0.9544\\ P\left(-\sigma <X-\mu <2\sigma \right)=0.9544\\ P\left(X-\mu <2\sigma \right)-P\left(X-\mu <-2\sigma \right)=0.9544\\ P\left(X-\mu <2\sigma \right)-\left[1-P\left(X-\mu <2\sigma \right)\right]=0.9544\end{array}$

$\begin{array}{c}2P\left(X-\mu <2\sigma \right)-1=0.9544\\ 2P\left(X-\mu <2\sigma \right)=1.9544\\ P\left(X-\mu <2\sigma \right)=\frac{1.9544}{2}\\ =0.9772\\ 1-P\left(X-\mu >2\sigma \right)=0.9772\\ P\left(X-\mu >2\sigma \right)=1-0.9772\\ =\mathrm{0.0228.}\end{array}$

Hence, the probability that the length of a brook trout exceeds the mean by at least 2 standard deviations is 0.0228.

To determine

Find the probability that the length of a brook trout is within 2 standard deviations of the mean.

Answer to Problem 76CE

Theprobability that the length of a brook trout is within2 standard deviations of the mean is 0.9544.

Explanation of Solution

Calculation:

Consider the property regarding 2 standard deviations difference from mean of theEmpirical Rule. It indicates that:

$P\left(\left|X-\mu \right|<2\sigma \right)=0.9544$.

Hence, the probability that the length of a brook trout is within 2 standard deviations of the mean is 0.9544.