Question

The mean of the scores obtained by a class of students on a physics test is 42. The standard deviation is 8%. Students have to score at least 50 to pass the test.

Assuming that the data is normally distributed,  (______ %) of the students passed the test.



% of students passed ?
  choices:   5   16   34   58

Answer 1

B. 16%.

Explanation:

We have been given that the The mean of the scores obtained by a class of students on a physics test is 42. The standard deviation is 8%. Students have to score at least 50 to pass the test.

First of all, we will find z-score of sample score 50 by using z-score formula.

`z=(x-\mu)/(\sigma)`, where,

`z=\text{z-score}`,

`x=\text{Sample score}`,

`\mu=\text{Mean}`,

`\sigma=\text{Standard deviation}`.

Upon substituting our given values in z-score formula we will get,

`z=(50-42)/(8)`

`z=(8)/(8)`

`z=1`

Now, we will use normal distribution table to find the area above the z-score of 1.

Using normal distribution table we will get,

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

`P(z>1)=1-0.84134`

`P(z>1)=0.15866`

Now we will multiply our answer by 100 to convert it into percentage.

`0.15866* 100\%=15.866\%\approx 16\%`

Therefore, approximately 16% of the students passed the test and option B is the correct choice.

Answer 2

The correct answer is:
16%.

Explanation:
We use a z-score to answer this question.
The formula for a z-score is:
z=
`(x-mue)/(sigma)`,
where mue is the mean and sigma is the standard deviation.

Using our information, we have:
z=
`(50-42)/(8) = (8)/(8) = 1`.

Using a z-table, we see that the area to the left of, or less than, this score is 0.8413. We want to know how many students scored more than this, so we subtract from 1:
1-0.8413=0.1587, which rounds to 16%.