The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let $X=$the number of colors on a national flag.

$\begin{array}{|cc|}\hline x& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

Calculate the following:

(a) $\overline{x}=\_\_\_\_\_$

(b) $s_x=\_\_\_\_\_$

(c) $n=\_\_\_\_\_$

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

The sample mean is computed using the formula given below,

$\overline{x}=\frac{\sum XF}{\text{total number of flags}}$

The calculation is given below,

$\overline{x}=\frac{1\times 1+2\times 7+3\times 18+4\times 7+5\times 6}{39}$

$=\frac{127}{39}$

$=3.2564$

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

$\begin{array}{|cc|}\hline x& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

The sample standard deviation is computed using the formula given below,

$s_x=\sqrt{\sum _{i=1}^k\frac{f_i{\left(x_i-\mu \right)}^2}{n-1}}$

The calculation is given below,

$s_x=\sqrt{\frac{1\times (1-3.26{)}^2+7\times (2-3.26{)}^2+18\times (3-3.26{)}^2+7\times (4-3.26{)}^2+6\times (5-3.26{)}^2}{39-1}}$

$=\sqrt{\frac{1\times (-2.26{)}^2+7\times (-1.26{)}^2+18\times (-0.26{)}^2+7\times (0.74{)}^2+6\times (1.74{)}^2}{38}}$

$=\sqrt{\frac{39.4364}{38}}$

$=1.019$

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

$\begin{array}{|cc|}\hline x& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

The number of survey of national flags from various countries is $39$ , that is,$n=39$