"Given \\; that, \u03bc=210, \u03c3=56, then,\\\\\n 1)P(x>300)=P(Z>\\frac{300-210}{56})\\\\\n=P(Z>1.61)=0.5-P(0<Z<1.61)\\\\\n=0.5-0.4463=0.0537\\\\\n2) P(x>100)=P(Z>\\frac{100-210}{56})\\\\\n=P(Z>-1.96)\\\\=0.5+P(0<Z<1.96)\\\\\n=0.5+0.4750=0.9750\\\\\n3)P(150<x<250)\\\\\n=P(\\frac{150-210}{56}<Z<\\frac{250-210}{56})\\\\\n=P(-1.07<Z<0.71)\\\\\n=P(0<Z<1.07)+P(0<Z<0.71)\\\\\n=0.3577+0.2611=0.6188\\\\"
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