Answer to Question #225799 in Statistics and Probability for Ashvi

Question

(a) Packages of a certain candy vary slightly in weight. The weights of nine packages  are given as follows: 

1.6823 1.6844 1.6851 1.6866 1.6562 1.6848 1.6829 1.6835 1.6858 

(i) Find the mean and median of these weights. 

Answer 1

$solution \space \\ \space given \space data \space set\\ 1.6823, \space 1.6844, \space 1.6851, \space 1.6866, \space 1.6562, \space 1.6848, \space 1.6829, \space 1.6835, \space 1.6858\\ -----------------------\\ mean\\ mean=\frac{\sum{x_i}}{n}\\ where \space \\ x_i \space = \space values \space from \space data \space set \space \\ n= \space 9\\ put \space the \space values\\ mean=\frac{1.6823+ \space 1.6844+ \space 1.6851+ \space 1.6866+ \space 1.6562+ \space 1.6848+ \space 1.6829+ \space 1.6835+ \space 1.6858}{9}\\ mean=\frac{15.1316}{9}\\ mean=1.68128888889\\ -----------------------\\ median\\ we \space arrange \space data \space set \space as \space ascending \space order\\ 1.6562, \space 1.6823, \space 1.6829, \space 1.6835, \space 1.6844, \space 1.6848, \space 1.6851, \space 1.6858, \space 1.6866\\ we \space have \space total \space 9 \space terms \space in \space data \space set\\ so \space middle \space term \space of \space data \space set \space is \space 5th \space term\\ it \space 5th \space term \space is \space called \space median \space \\ \space hence \space median \space =1.6844\\$