Chapter 15.4, Problem 5ST

To determine

To calculate: The percent of lobsters that weigh more than $4.1\text{ kg}$ .

Answer to Problem 5ST

The percent of lobsters that weigh more than $4.1\text{ kg}$ is $1.39\%$ .

Explanation of Solution

Given information:

Themean weight is $3\text{ kg}$ of the lobsters in a certain bay and standard deviation of the same is $0.5\text{ kg}$ .

Formula used:

The number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Calculation:

Consider the provided information that mean weight is $3\text{ kg}$ of the lobsters in a certain bay and standard deviation of the same is $0.5\text{ kg}$ .

To compute percent of lobsters that weigh more than $4.1\text{ kg}$ .

Recall that the number x is known as standardized value to z where z is provided value of normal distribution, m is the mean and $\sigma$ is the standard deviation.

  $x=\frac{z-M}{\sigma }$

Here, zis $4.1$ , m is 3 and $\sigma$ is 0.5.

Apply it,

  $\begin{array}{c}x=\frac{4.1-3}{0.5}\\ x=2.2\end{array}$

The total area under the curve is 1 and standard normal curve is symmetric about y -axis.

The required percent is the area under the curve of standard normal to the right of $x=2.2$ that is,

  $\begin{array}{c}0.5-A\left(0\le x\le 2.2\right)=0.5-A\left(2.2\right)\\ =0.5-0.4861\\ =0.0139\end{array}$

Multiply the result by 100, to obtain the answer in percentage form,

  $\begin{array}{c}0.5-A\left(0\le x\le 2.2\right)=0.5-A\left(2.2\right)\\ =0.5-0.4861\\ =0.0139\\ =1.39\%\end{array}$

Thus, the percent of lobsters that weigh more than $4.1\text{ kg}$ is $1.39\%$ .