Question

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.

Answer 1

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

• Lower bound = 1914.8
• Upper bound = 999999
• Mean = 1986.1
• Standard deviation = 27.2

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