Chapter 6.5, Problem 1MRPS

Solution

(a)

The mean of the given data.

The mean is $1.895$

Given:

The weight of $20$ marine landlocked salmon is given.

Calculation:

The mean of the data is the average of all the numbers in the given data. It can be calculated as:

  $\text{Mean}=\frac{{\displaystyle \sum _{i=1}^{i=n}{x}_i}}{n}$

The data given is;

The weight of salmon ( in lb)
$1.9$
$1.5$
$2.0$
$2.0$
$2.1$
$2.0$
$1.8$
$1.5$
$2.2$
$1.9$
$1.8$
$1.7$
$2.1$
$2.2$
$1.7$
$2.1$
$2.0$
$1.5$
$1.8$
$2.1$
Sum = $37.9$

The sum of the given data is: $37.9$

The number of data is : $20$

Hence, the mean of the data is:

  $\text{Mean}=\frac{37.9}{20}=1.895$

Conclusion:

The mean is $1.895$

(b)

The standard deviation of the given data.

The standard deviation of the given data is $0.219$

Given:

The weight of $20$ marine landlocked salmon is given.

Calculation:

1. Calculate the value of square of the difference between each data from the mean.

The weight of salmon ( in lb)	${\left|x_i-\overline{x}\right|}^2$
$1.9$	$0.005$
$1.5$	$0.156$
$2.0$	$0.011$
$2.0$	$0.011$
$2.1$	$0.042$
$2.0$	$0.011$
$1.8$	$0.009$
$1.5$	$0.156$
$2.2$	$0.093$
$1.9$	$0.00025$
$1.8$	$0.009$
$1.7$	$0.038$
$2.1$	$0.042$
$2.2$	$0.093$
$1.7$	$0.038$
$2.1$	$0.042$
$2.0$	$0.011$
$1.5$	$0.156$
$1.8$	$0.009$
$2.1$	$0.042$
Sum = $37.9$	Sum = $0.9742$

2. $\frac{\text{Sum}}{\text{n}}=\frac{0.9742}{20}=0.048$
3. Now, standard deviation can be find out as:

  $S.D=\sqrt{0.048}=0.219$

Conclusion:

The standard deviation is $0.219$

(c)

The number of fish expected having weight less than $1.443\text{ lb}$ when $3000$ fish caught in a year.

There is no possibility of fish having weight less than $1.443\text{ lb}$

Given:

The weight of $20$ marine landlocked salmon is given.

Calculation:

In the given data, among $20$ fish there is not even a single fish whose weight is less than $1.443\text{ lb}$ . Therefore, the possibility is zero.

Conclusion:

There is no possibility of fish having weight less than $1.443\text{ lb}$