# Thread: Help with a few Standard Normal Distribution problems

1. ## Help with a few Standard Normal Distribution problems

Hello,
I have a few questions that I'm wondering if I could get any help on:

1) A student recieved a standardized Z-score on a test that was -0.57. What does this score tell about how this student scored in relation to the rest of the class?

When I looked up -.57 on the Z-score table it gives me .28, or I'm assuming, 28%. So does this mean that she did better than 28% of her class, or worse than 28% of her class?

2) A large college class has 900 students, broken down into section meetings with 30 students each. On the final exam, scores followed a normal distribution with an average of 63 and a standard deviation of 20.

a) if you randomly select one of these students, what is the probability that the selected student scored between 56 and 70 on the exam?

b)if we consider a section of 30 students as a random sample from this population, will the probability that the average for the section is between 56 and 70 be higher or lower or the same as what you calculated in the previous question?
On this part, it would be the same right? Since it doesnt say anything about certain sections, I would assume the average and standard deviation apply the same way.

3)Find the mean of a normally distributed random variable with a standard deviation of 8 if 95.99% of all the values are less than 17.

I would appreciate any help I can get, I'm feeling lost on some of this

2. Originally Posted by strikemuffin
Hello,
I have a few questions that I'm wondering if I could get any help on:

1) A student recieved a standardized Z-score on a test that was -0.57. What does this score tell about how this student scored in relation to the rest of the class?

When I looked up -.57 on the Z-score table it gives me .28, or I'm assuming, 28%. So does this mean that she did better than 28% of her class, or worse than 28% of her class? Mr F says: Better. Draw the curve and see where she is.

2) A large college class has 900 students, broken down into section meetings with 30 students each. On the final exam, scores followed a normal distribution with an average of 63 and a standard deviation of 20.

a) if you randomly select one of these students, what is the probability that the selected student scored between 56 and 70 on the exam?

would the answer be 27%? Mr F says: Yes.

b)if we consider a section of 30 students as a random sample from this population, will the probability that the average for the section is between 56 and 70 be higher or lower or the same as what you calculated in the previous question?
On this part, it would be the same right? Since it doesnt say anything about certain sections, I would assume the average and standard deviation apply the same way.

Mr F says: The sample average ${\color{red}\bar{X}}$ follows a normal distribution with mean 63 and standard deviation ${\color{red}\frac{20}{\sqrt{30}}}$. Calculate ${\color{red} \Pr(56 < \bar{X} < 70 )}$.

3)Find the mean of a normally distributed random variable with a standard deviation of 8 if 95.99% of all the values are less than 17.

I would appreciate any help I can get, I'm feeling lost on some of this
3) From $Z = \frac{X - \mu}{\sigma}$ it follows that $z_{17} = \frac{17 - \mu}{8}$ where $\Pr(Z < z_{17}) = 0.9599$.

So get $z_{17}$, substitute and solve for $\mu$.

3. Originally Posted by strikemuffin
Hello,
I have a few questions that I'm wondering if I could get any help on:

1) A student recieved a standardized Z-score on a test that was -0.57. What does this score tell about how this student scored in relation to the rest of the class?

When I looked up -.57 on the Z-score table it gives me .28, or I'm assuming, 28%. So does this mean that she did better than 28% of her class, or worse than 28% of her class?

2) A large college class has 900 students, broken down into section meetings with 30 students each. On the final exam, scores followed a normal distribution with an average of 63 and a standard deviation of 20.

a) if you randomly select one of these students, what is the probability that the selected student scored between 56 and 70 on the exam?

b)if we consider a section of 30 students as a random sample from this population, will the probability that the average for the section is between 56 and 70 be higher or lower or the same as what you calculated in the previous question?
On this part, it would be the same right? Since it doesnt say anything about certain sections, I would assume the average and standard deviation apply the same way.

3)Find the mean of a normally distributed random variable with a standard deviation of 8 if 95.99% of all the values are less than 17.

I would appreciate any help I can get, I'm feeling lost on some of this

1) When you look up the Z-score and it said 0.28. That's the cumulative area. That's the 28th percentile, which means that she did better than 28% of her class, but 72% of the class did better than her

2) I believe the Central Limit Theorem applies here. You may have to use the formula mu/sqrt(n) to find the mean SD.

3) Now you are simply working in reverse. If you look at the Z-score table, the 95.99% percentile has a z-score of 1.75.

The formula is Z= (X-mean)/StdDev, which you must knew in order to have done #2.

Since the standard deviation is already given (8) and the X value (7), the finally equation you are trying to solve becomes...

1.75 = (17-mean)/8
doing some algebra...
14 = 17-mean
mean = 3 = mu

You can check your answer by now working forward using a = -infinity b= 17 mu = 3 sigma=8 and it will give you 95.99%, which means the mean of 3 is the correct answer.

4. Originally Posted by TitaniumX
[snip]
3) Now you are simply working in reverse. If you look at the Z-score table, the 95.99% percentile has a z-score of 1.75.

The formula is Z= (X-mean)/StdDev, which you must knew in order to have done #2.

Since the standard deviation is already given (8) and the X value (7), the finally equation you are trying to solve becomes...

1.75 = (17-mean)/8
doing some algebra...
14 = 17-mean
mean = 3 = mu

You can check your answer by now working forward using a = -infinity b= 17 mu = 3 sigma=8 and it will give you 95.99%, which means the mean of 3 is the correct answer.
*Ahem* ..... I was hoping that the OP might have made an attempt based on what I posted before being given the complete solution.

As for the part of the quote I snipped, how did it add to what I already said?

5. Originally Posted by mr fantastic
*Ahem* ..... I was hoping that the OP might have made an attempt based on what I posted before being given the complete solution.

As for the part of the quote I snipped, how did it add to what I already said?
When I replied to the post, there were no replies yet. I went out to grab a bite, came back, type, and when I hit submit... you beat me to it. LOL