question 5

Provide an appropriate response. 5) A variable is normally distributed with a mean of 100 and a 5) standard deviation of 10. A student wanted to find the percentage of observations of the variable lying between 106 and 110. What is wrong with his solution? Student's solution: 106-100 110-100 Z-scores: = 0.6 = 1.0 10 10 Percentage of scores lying between 106 and 110 = difference between z-scores = 1.0- 0.6 = 0.4= 40%