Question

The class average on a math test was an 82 with a standard deviation of 5. If the 

data is normally distributed, which statements are true?

1. A student scoring a 74  would have a z -score of -1.6.
2. About 68% of the students  scored below an 87.
3. A z-score of 2.6 would be a test 
score of 95.
4. About 16% of the class  scored a 77 or below.
45. About 25% of the class  scored above a 92.

Answer 1

Option 3 and 4.

Explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

`z=(x-\mu)/(\sigma) \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation`

Given that mean (μ) = 82, standard deviation (σ) = 5

1) For x = 74:

`z=(x-\mu)/(\sigma) \\\\z=(74-82)/(5) =-1.6`

Option 1 is incorrect

2) For x < 87

`z=(x-\mu)/(\sigma) \\\\z=(87-82)/(5) =1`

From the normal distribution table, P(x < 87) = P(z < 1) = 0.8413 = 84.13%

Option 2 is incorrect

3) For x = 95:

`z=(x-\mu)/(\sigma) \\\\z=(95-82)/(5) =2.6`

Option 3 is correct

4) For x < 77

`z=(x-\mu)/(\sigma) \\\\z=(77-82)/(5) =-1`

From the normal distribution table, P(x < 77) = P(z < -1) = 0.16 = 16%

Option 4 is correct

5) For x > 92

`z=(x-\mu)/(\sigma) \\\\z=(92-82)/(5) =2`

From the normal distribution table, P(x > 92) = P(z > 2) = 1 - P(z < -2) = 1 - 0.9772 = 0.0228 = 2.28%

Option 5 is incorrect