Answer :
The correct answer is:
16%.
Explanation:
We use a z-score to answer this question.
The formula for a z-score is:
z=[tex] \frac{x-mue}{sigma} [/tex],
where mue is the mean and sigma is the standard deviation.
Using our information, we have:
z=[tex] \frac{50-42}{8} = \frac{8}{8} = 1[/tex].
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%.
16%.
Explanation:
We use a z-score to answer this question.
The formula for a z-score is:
z=[tex] \frac{x-mue}{sigma} [/tex],
where mue is the mean and sigma is the standard deviation.
Using our information, we have:
z=[tex] \frac{50-42}{8} = \frac{8}{8} = 1[/tex].
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.
[tex]z=\frac{x-\mu}{\sigma}[/tex], where,
[tex]z=\text{z-score}[/tex],
[tex]x=\text{Sample score}[/tex],
[tex]\mu=\text{Mean}[/tex],
[tex]\sigma=\text{Standard deviation}[/tex].
Upon substituting our given values in z-score formula we will get,
[tex]z=\frac{50-42}{8}[/tex]
[tex]z=\frac{8}{8}[/tex]
[tex]z=1[/tex]
Now, we will use normal distribution table to find the area above the z-score of 1.
Using normal distribution table we will get,
[tex]P(z>1)=1-P(z<1)[/tex]
[tex]P(z>1)=1-0.84134[/tex]
[tex]P(z>1)=0.15866[/tex]
Now we will multiply our answer by 100 to convert it into percentage.
[tex]0.15866\times 100\%=15.866\%\approx 16\%[/tex]
Therefore, approximately 16% of the students passed the test and option B is the correct choice.