he U.S. Department of Transportation reported the accompanying data on the number of speeding-related crash fatalities during holiday periods for the years from 1994 to 2003. Holiday Period 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 New Year's Day 141 142 178 72 219 138 171 134 210 70 Memorial Day 191 174 185 197 138 183 156 190 188 181 (a) Compute the standard deviation for the New Year's Day data. (Round the answer to three decimal places.) (b) Without computing the standard deviation of the Memorial Day data, determine whether the standard deviation for the Memorial Day data would be larger or smaller than the standard deviation of the New Year's Day data. larger smaller
Accepted Solution
A:
Answer:New year Standard deviation = 2255.25 Step-by-step explanation: Years: 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003New Year: 141 142 178 72 219 138 171 134 210 70Memorial: 191 174 185 197 138 183 156 190 188 181 Formula:[tex]Mean = \displaystyle\frac{\text{Sum of all observation}}{\text{Total number of observations}}[/tex][tex]S.D = \sqrt{\displaystyle\frac{\sum(x_i - \bar{x})^2}{n}}[/tex]a) Mean = [tex]\frac{1475}{10} = 147.5[/tex][tex]S.D = \sqrt{\frac{42.25+ 30.25+ 930.25+5700.25+ 5112.25+ 90.25+ 552.25+ 182.25+ 3906.25+ 6006.25}{10}}\\= \sqrt{\frac{22552.5}{10} }\\= 2255.25[/tex]b) The standard deviation of the memorial day will be less than the new years day because for memorial day, the data points are not varying as in case of new year. Basically, we can say the spread in case of new year is more as compared to memorial day. Hence, standard deviation of memorial year is less than the new years day.