Question

The students scored an average final exam score of 83 with a standard deviation of 2. Assume that the scores are approximated by a normal distribution.

• What percent of students scored higher than an 86 on the final exam?
• What percent of students scored less than a 79 on the final exam?
• What percent of students scored between 79 and 86?
• What happens when you try to find the percent of students that scored less than 60?

Answer 1

a) P(X > 86)

= P((X - )/ > (86 - )/)

= P(Z > (86 - 83)/2)

= P(Z > 1.5)

= 1 - P(Z < 1.5)

= 1 - 0.9332

= 0.0668 = 6.68%

b) P(X < 79)

= P((X - )/ < (79 - )/)

= P(Z < (79 - 83)/2)

= P(Z < -2)

= 0.0228 = 2.28%

c) P(79 < X < 86)

= P((79 - )/ < (X - )/ < (86 - )/)

= P((79 - 83)/2 < Z < (86 - 83)/2)

= P(-2 < Z < 1.5)

= P(Z < 1.5) - P(Z < -2)

= 0.9332 - 0.0228

= 0.9104 = 91.04%

d) P(X < 60)

= P((X - )/ < (60 - )/)

= P(Z < (60 - 83)/2)

= P(Z < -11.5)

= 0.00 = 0%