Question 853762
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Hi,
mean of 103 minutes with a standard deviation of 42 minutes.
Using TI: The syntax is normalcdf(smaller, larger, µ, &#963;).
Note: The -9999 is used as the smaller value to be at least 5 standard deviations from the mean.
P(x > 115) =  1 - normalcdf(-9999,115, 103,42)

Using TI: The syntax is normalcdf(smaller, larger, µ, &#963;).
P(100 &#8804; x &#8804; 120)= normalcdf(100,120, 103,42)