Answer to Question #224150 in Statistics and Probability for Payal

Question

A work-standards expert observes the amount of time required to prepare a sample of 10 business letters in an office with the following results listed in ascending order to the nearest minute: 5, 5, 5, 7, 9, 14, 15, 15, 16, 18. Determine the
(ii) Middle 70 percent for the sample.
(iii) Sample variance of the preparation-time data
(iv) Compute the coefficient of skewness for the data analyzed
(v) Determine the coefficient of variation for the data analyzed

Answer 1

$solution\\ given \space data \space set \space 5, \space 5, \space 5, \space 7, \space 9, \space 14, \space 15, \space 15, \space 16, \space 18. \space \\ ------------------------\\ (i)range \space for \space the \space central \space 70\% \space of \space data \space \\ \\step \space (1) \space mean=\frac{5+ \space 5+ \space 5+ \space 7+ \space 9+ \space 14+ \space 15+ \space 15+ \space 16+ \space 18}{10}=10.9\\ step \space (2)sd=σ=\sqrt{ \space \frac{Σ(xi \space - \space μ)^2}{N}}\\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =\sqrt{\frac{ (5 \space - \space 10.9)^2 \space + \space ... \space + \space (18 \space - \space 10.9)^2}{10}} \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \space 4.93\\ step(3)area \space of \space range \space =central \space .70\\ it \space mean \space we \space find \space z \space score \\ \space for \space 15\% \space to \space 85\%\\ we \space know \space that \space z \space score \\ \space for \space 15\% \space to \space 85\% \space is \space \space -1.036 \space to \space \space 1.036 \space respectively\\ \\ step(4)now \space we \space find \space central \space 70\% \space range \space (raw \space score \space of \space central \space 70\%)\\ formula \space is \space \\ x=z.σ+μ\\ x_1=(-1.036 \space × \space 4.93)+10.9\\ x_1=5.79252\\ x_2=(1.036 \space × \space 4.93)+10.9\\ x_2=16.00748\\ hence \space central \space 70\% \space range \space of \space data \space =(5.79252 \space to \space 16.00748)\\ ------------------------\\ (ii) \space Sample \space variance \space of \space the \space preparation-time \space data\\ s^2=\frac{∑_{i=1}^n \space (x_i−x̄)^2}{n−1}\\ s^2=\frac{∑_{i=1}^n \space (x_i−10.9)^2}{10−1}\\ put \space the \space values \space and \space solve, \space we \space get\\ s^2=26.988889\\ ------------------------------\\ (iii)Compute \space the \space coefficient \space of \space skewness \space for \space the \space data \space analyzed\\ Skewness \space Coefficient \space = \space 3 \space x \space \frac{(mean \space - \space median)}{ \space σ} \space \\ we \space have \space \\ mean \space =10.9\\ median \space =11.5\\ σ=4.9285\\ Skewness \space Coefficient \space = \space 3 × \space \space \frac{(10.9-11.5)}{ \space 4.9285} \space \\ Skewness \space Coefficient \space = \space -0.3652\\ ------------------------------\\ (iv) \space Determine \space the \space coefficient \space of \space variation \space for \space the \space data \space analyzed\\ N= \space 10\\ mean \space =10.9\\ Squared \space Dev.=242.9\\ σ^2=\frac{Squared \space Dev.}{N}\\ σ^2=\frac{242.9}{10}\\ σ=4.93\\ CV \space ( \space coefficient \space of \space variation)= \space \frac{σ}{Mean}×100 \space \\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space = \space \frac{4.93}{10.9}×100 \space \\ \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space =45.22\\$