Question

Can you break it down how I should do it?

Scores on a university exam are Normally distributed with a mean of 70 and a standard deviation of 10. The professor teaching the class declares that a student will receive a “F” if his or her score is below 50.

1. Using the 68-95-99.7 rule, what percent of students will receive “F”?

2. Using the 68-95-99.7 rule, what percent of students will score between 60 and 90?

3. Using the standard Normal distribution tables, the area under the standard Normal curve corresponding to –1.0 < Z < 1.0 is:

a. 0.3085.

* b. 0.6828.

c. 0.5328.

d. 0.2815.

Answer 1

Given that, mean = 70 and standard deviation = 10

The professor teaching the class declares that a student will receive a “F” if his or her score is below 50.

According to 68-95-99.7 rule,

i) Approximately 68% of the data fall within 1 standard deviations of the mean.

(mean - sd = 70 - 10 = 60 and mean + sd = 70 + 10 = 80)

ii) Approximately 95% of the data fall within 2 standard deviations of the mean.

(mean-2sd = 70 -(2 * 10) =50 and mean+2sd =70 + (2 * 10) =90)

iii) Approximately 99.7% of the data fall within 3 standard deviations of the mean.

(mean-3sd = 70 -(3 * 10) =40 and mean+3sd =70 + (3 * 10) =100)

1) By rule ii) 95% of the students will score between 50 and 90.

Therefore, 100 - 95 = 5% of students will score outside 50 and 90. And 5/2 = 2.5% of students will score below 50 and 2.5% of students will score above 90.

Hence, 2.5% of students will receive "F".

2) By rule i) 68/2 = 34% of students will score between 60 and 70 (mean). And 95/2 = 47.5% of students will score between 70 (mean) and 90.

Therefore, 34 + 47.5 = 81.5% of students will score between 60 and 90.

3) Using standard normal distribution tables, the area under the standard normal curve corresponding to -1 < Z < 1 is 0.6828