Question

Mary earned a score of 958 on a national achievement test. The mean test score was 850 with a standard deviation of 200. What proportion of students had a higher scores than Mary? Assume that the test scores are normally distributed. Group of answer choices

29.46%
39.19%
23.75%
17.36%​

Answer 1

29.46% of students scored higher than Mary.

Explanation:

Transform Mary's test score into z-score, using z-score equation.

Mean = 850

Mary's score = x = 958

Standard deviation = 200

`\boxed{\bf Z= (x-mean)/(Standard \ deviation)}`

`=(958-850)/(200)\\\\\\=(108)/(200)\\\\= 0.54`

Now, find the cumulative probability of the z-score using Standard normal distribution table. That is, we have to find P(z <0.54).

P(z<0.54) = 0.70540

Using this, we can find P(z > 0.54).

P(z > 0.54) = 1 - P(z < 0.54)

= 1 - 0.70540

= 0.2946

To find the percentage, multiply 0.2946 by 100.

0.2946 *100 = 29.46%

29.46% of students scored higher than Mary.