Respuesta :
So we are given that the mean is 42% and the sd (standard deviation) is 8%
Assuming our data is normal we can use the 68-95-99 rule
So one thing you should realize is that 42% + 8% is 50% which is passing. That is one standard deviation higher. So we use:
100 - 68 - 13.5 - 2.35 - 0.15 = 16. That means 16% of students passed the test. Which is terrible. They probably need to hit the books more.
Anyways if you have any question feel free to message me!
Hopes this helps!
Assuming our data is normal we can use the 68-95-99 rule
So one thing you should realize is that 42% + 8% is 50% which is passing. That is one standard deviation higher. So we use:
100 - 68 - 13.5 - 2.35 - 0.15 = 16. That means 16% of students passed the test. Which is terrible. They probably need to hit the books more.
Anyways if you have any question feel free to message me!
Hopes this helps!
Using the normal distribution, it is found that 16% of students pass the test.
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem:
- The mean is of 42%, hence [tex]\mu = 42[/tex].
- The standard deviation is of 8%, hence [tex]\sigma = 8[/tex].
The proportion is 1 subtracted by the p-value of Z when X = 50, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{50 - 42}{8}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.84.
1 - 0.84 = 0.16
0.16 x 100% = 16%
16% of students pass the test.
You can learn more about the normal distribution at https://brainly.com/question/24663213
