Question 853778
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Hi, previously Posted
mean of 60 and a standard deviation of 8
Using TI: The syntax is normalcdf(smaller, larger, µ, &#963;).
P(44 &#8804; x &#8804; 68) = normalcdf(44,68,60,8)
P(52&#8804; x &#8804; 76) = normalcdf(52,76,60,8)
0r
P(44 &#8804; x &#8804; 68) = NORMSDIST(1) - NORMSDIST(-2) =  .8413 - .0228
P(52&#8804; x &#8804; 76) =  NORMSDIST(2) - NORMSDIST(-1) =  .9772 - .1587