1.14 =0.36(克) A.=1y 0.36=0.72 28:4-0.72≤x≤284+0.72即2768≤x≤2912(克) 以95.45%概率推算该批食品重量范围不符合规格要求 (2)p==10% p-p)_0.1×09 0.095=95 2×9.5%=19% 10%-19%≤p≤10%+19%即p≤29% 以95.45%的概率推算该批食品重量合格率在29%以下0.36 10 1.14 2 2 = = = = n S n x   (克)  = = 2 0.36 = 0.72 x x t 28.4 − 0.72  x  28.4 + 0.72 即 27.68  x  29.12 (克) 以 95.45%概率推算该批食品重量范围不符合规格要求 ⑵ 10% 10 1 p = = ( ) 0.095 9.5% 10 1 0.1 0.9 = =  = − = n p p  p  p = t p = 29.5% = 19% 10% −19%  p  10% +19% 即 p  29% 以 95.45%的概率推算该批食品重量合格率在 29%以下