Mrs Joness reading class can read an average of 180 words pe

Mrs. Jones’s reading class can read an average of 180 words per minute with a standard deviation of 15 words per minute. The top 3% of the class is to receive a special award. What is the minimum number of words per minute a student would need to read in order to get the award? Assume the data is normally distributed. Round your answer to the nearest whole number.

Solution

Normal Distribution
Mean ( u ) =180
Standard Deviation ( sd )=15
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
P ( Z > x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is 1.88
P( x-u/ (s.d) > x - 180/15) = 0.03
That is, ( x - 180/15) = 1.88
--> x = 1.88 * 15+180 = 208.215 ~ 209 should be the minimum to receive the award