A distribution of values is normal with a mean of 1986.1 and a standard deviation of 27.2. Find the probability that a randomly selected value is greater than 1914.8.
P(X> 1914.8) =
Enter your answer as a number accurate to 4 decimal places. *Note: all z-scores must be rounded to
the nearest hundredth.

I used the function normCdf(lower bound, upper bound, mean, standard deviation) on the graphing calculator to solve this.

Input in these values and it will result in:

normCdf(1914.8,9999999,1986.1,27.2) = 0.995621

So the probability that the value is greater than 1914.8 is about 99.5621%

I'm not sure if this is correct 0_o

step-by-step explanation:

hans did not square the 40-unit side; he merely multiplied it by 2.   incorrect.