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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 

Verified Answer :

The correct answer is:
16%.

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

Using our information, we have:
z=\(\frac{50-42}{8} = \frac{8}{8} = 1\).

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%.

Answer:

B. 16%.

Step-by-step 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.

\(z=\frac{x-\mu}{\sigma}\), 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,

\(z=\frac{50-42}{8}\)

\(z=\frac{8}{8}\)

\(z=1\)

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.

\(0.15866\times 100\%=15.866\%\approx 16\%\)

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

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