Chapter 1, Problem 1.91AP

(a)

Interpretation Introduction

Interpretation: The data for the distribution if trail mix is given. The mean and standard deviation in the percentage of peanut and that of raisins in the four samples is to be calculated. Also the $90\%$ confidence interval of the percentages is to be checked to include the target composition of peanuts and raisins.

Concept introduction: The standard deviation is calculated as,

$\text{Standard}\text{deviation}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\text{Number}\text{of}\text{samples}}}$

The $90\%$ confidence interval is calculated by the formula,

$\text{Mean}\pm 1.645\times \frac{\text{Standard}\text{deviation}}{\sqrt{\text{Number}\text{of}\text{samples}}}$

To determine: The mean and standard deviation in the percentage of peanut and that of raisins in the four samples.

Answer to Problem 1.91AP

Solution

The value of mean and standard deviation of percentage of peanut is $\underset{\_}{62.81}$ and $\underset{\_}{3.61}$, respectively.

The value of mean and standard deviation of percentage of peanut is $\underset{\_}{37.19}$ and $\underset{\_}{3.61}$, respectively.

Explanation of Solution

Explanation

Given

The data for the amount of peanuts and raisins are given as,

$\begin{array}{ccc}\text{Day}& \text{Peanuts}& \text{Raisins}\\ 1& 50& 32\\ 11& 56& 26\\ 21& 48& 34\\ 31& 52& 30\end{array}$

The total content of the peanuts and raisins on day $1$ is $50+32=82$.

Therefore, the percentage of the peanuts on day one is $\frac{50}{82}\times 100=61\%$ and that of raisins is $100-61=39\%$.

The total content of the peanuts and raisins on day $2$ is $56+26=82$.

Therefore, the percentage of the peanuts on day one is $\frac{56}{82}\times 100=68.3\%$ and that of raisins is $100-68.3=31.7\%$.

The total content of the peanuts and raisins on day $3$ is $48+34=82$.

Therefore, the percentage of the peanuts on day one is $\frac{48}{82}\times 100=58.54\%$ and that of raisins is $100-58.54=41.46\%$.

The total content of the peanuts and raisins on day $1$ is $52+30=82$.

Therefore, the percentage of the peanuts on day one is $\frac{52}{82}\times 100=63.4\%$ and that of raisins is $100-63.4=36.6\%$.

Therefore, the table for the calculation of the standard deviation for peanut is,

$\begin{array}{cccc}\text{Day}& x\left(\%\text{Peanut}\right)& x-\overline{x}& {\left(x-\overline{x}\right)}^2\\ 1& 61& -1.81& 3.2761\\ 2& 68.3& 5.49& 30.1401\\ 3& 58.54& -4.27& 18.2329\\ 4& 63.4& 0.59& 0.3481\\ \text{Mean}\left(\overline{x}\right)& 62.81& & \end{array}$

The mean of the percentage of peanut is given by the formula,

$\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{percentages}}{\text{total}\text{number}\text{of}\text{samples}}$

Substitute the values of all the percentages and the number of samples of peanuts in the above equation.

$\begin{array}{c}\text{Mean}=\frac{61+68.3+58.54+63.4}{\text{4}}\\ =62.81\end{array}$

The standard deviation is calculated as,

$\text{Standard}\text{deviation}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\text{Number}\text{of}\text{samples}}}$

Substitute the value of ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ and the number of samples in the above equation.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{\frac{3.2761+30.1401+18.2329+0.3481}{\text{4}}}\\ =\sqrt{\frac{51.9972}{4}}\\ =\sqrt{12.9993}\\ =3.61\end{array}$

Therefore, the value of mean and standard deviation of percentage of peanut is $\underset{\_}{62.81}$ and $\underset{\_}{3.61}$, respectively.

The table for the calculation of the standard deviation for raisins is,

$\begin{array}{cccc}\text{Day}& x\left(\%\text{Peanut}\right)& x-\overline{x}& {\left(x-\overline{x}\right)}^2\\ 1& 39& 1.81& 3.2761\\ 2& 31.7& -5.49& 30.1401\\ 3& 41.46& 4.27& 18.2329\\ 4& 36.6& -0.59& 0.3481\\ \text{Mean}\left(\overline{x}\right)& 37.19& & \end{array}$

The mean of the percentage of raisins is given by the formula,

$\text{Mean}=\frac{\text{Sum}\text{of}\text{all}\text{the}\text{percentages}}{\text{total}\text{number}\text{of}\text{samples}}$

Substitute the values of all the percentages and the number of samples of raisins in the above equation.

$\begin{array}{c}\text{Mean}=\frac{\text{39+31}\text{.7+41}\text{.46+36}\text{.6}}{\text{4}}\\ =37.19\end{array}$

The standard deviation is calculated as,

$\text{Standard}\text{deviation}=\sqrt{\frac{{\displaystyle \sum {\left(x-\overline{x}\right)}^2}}{\text{Number}\text{of}\text{samples}}}$

Substitute the value of ${\displaystyle \sum {\left(x-\overline{x}\right)}^2}$ and the number of samples in the above equation.

$\begin{array}{c}\text{Standard}\text{deviation}=\sqrt{\frac{3.2761+30.1401+18.2329+0.3481}{\text{4}}}\\ =\sqrt{\frac{51.9972}{4}}\\ =\sqrt{12.9993}\\ =3.61\end{array}$

Therefore, the value of mean and standard deviation of percentage of raisins is $\underset{\_}{37.19}$ and $\underset{\_}{3.61}$, respectively.

(b)

Interpretation Introduction

To determine: If the $90\%$ confidence interval of the percentages is include the target composition of peanuts and raisins.

Answer to Problem 1.91AP

Solution

The $90\%$ confidence interval of peanut is $\underset{\_}{65.78\%-59.84\%}$ and it does not include the target composition of $67\%$ of peanuts.

The $90\%$ confidence interval of raisins is $\underset{\_}{40.16\%-34.22\%}$ and it does not include the target composition of $33\%$ of raisins.

Explanation of Solution

Explanation

The $90\%$ confidence interval is calculated by the formula,

$\text{Mean}\pm 1.645\times \frac{\text{Standard}\text{deviation}}{\sqrt{\text{Number}\text{of}\text{samples}}}$ (1)

Therefore, $90\%$ confidence interval of peanut is calculated by substituting its value for mean and standard deviation in the above equation.

$\text{62}\text{.81\%}\pm 1.645\times \frac{\text{3}\text{.61}}{\sqrt{\text{4}}}\%=\text{62}\text{.81\%}\pm 2.97\%$

Therefore, the $90\%$ confidence interval of peanut is $\underset{\_}{65.78\%-59.84\%}$ and it does not include the target composition of $67\%$ of peanuts.

$90\%$ confidence interval of raisins is calculated by substituting its value for mean and standard deviation in the above equation.

$\text{37}\text{.19\%}\pm 1.645\times \frac{\text{3}\text{.61}}{\sqrt{\text{4}}}\%=\text{37}\text{.19\%}\pm 2.97\%$

Therefore, the $90\%$ confidence interval of raisins is $\underset{\_}{40.16\%-34.22\%}$ and it does not include the target composition of $33\%$ of raisins.

Conclusion

1. a. The value of mean and standard deviation of percentage of peanut is $\underset{\_}{62.81}$ and $\underset{\_}{3.61}$, respectively.

   The value of mean and standard deviation of percentage of peanut is $\underset{\_}{37.19}$ and $\underset{\_}{3.61}$, respectively.

2. b. The $90\%$ confidence interval of peanut is $\underset{\_}{65.78\%-59.84\%}$ and it does not include the target composition of $67\%$ of peanuts.

   The $90\%$ confidence interval of raisins is $\underset{\_}{40.16\%-34.22\%}$ and it does not include the target composition of $33\%$ of raisins