To answer this problem we have to have a <span>Standard Normal Distribution table because we need to look up z scores. Then we use this equation </span><span>z = (value - mean)/(standard deviation) = (250-200)/20 = 2.5. </span><span> The z score is then 2.5 </span>If so, look up z=2.5 on the left edge then you then determine from the table what the area to the left of that is. <span> Here are the conditions: </span><span>p(z>2.5)=1
p(z<2.5)=.621
1-.621= .379
100-37.9= 62.1</span><span>
Given that, we should go with </span><span>.621%.</span>