"1) Given \\;that,\\\\\nx_{1}=66,x_{2}=75,Z_{1}=-0.6,Z_{2}=1.1,then\\\\\nZ=\\frac{x-\\mu}{\\sigma}\\\\\n\\therefore -0.6=\\frac{66-\\mu}{\\sigma}\\implies -0.6\\sigma=66-\\mu,\\\\\n\\therefore 1.1=\\frac{75-\\mu}{\\sigma}\\implies 1.1\\sigma=75-\\mu,\\\\\n\\text{by multiplying the first equation by (-1) }\\\\\n\\text{and then adding the result to the second equation}\\\\\n1.7 \\sigma=-66+75=9\\implies \\sigma=\\frac{9}{1.7}\\approx 5.294,\\\\\n\\mu=66+0.6(5.294)=69.1764\\\\\n2)Given \\; that, \u03bc=97.6, \u03c3=10.4, then,\\\\\na) P(x>100) = P( Z >\\frac{100-97..6}{10.4})\\\\\n=P(Z>0.23)\\\\=0.5- P(0<Z<0.23)=0.5-0.0910=0.409\\\\\n\\text{number of bulbs }=1600\\times 0.409\\\\\n=654.4\\approx655\\;vehicles \\\\"
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