Standardized Testing and Probabilities

The scores on a standardized test are normally distributed with a population mean of 720 and a population standard deviation of 60.

Test scores are discrete, so do not forget to make the continuity correction before using the standard normal distribution (which is continuous).

A. What is the probability that an individual, randomly-selected student will score less than 726?
z(726)= (726-720) / 60= 6/60= .1
P(x<726)= P(z<.1)=
z-score that corresponds to .1 is .0398
.5 - .0398 = .4602

B. If we take a sample of 4 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 4)= 6 / (60/2)= 6/30= 1/5= .2
z-score that corresponds to .2 is .0793
.5-.0793= .4207

C. Why are your answers to Parts A and B above in this problem different?
Please be specific! Do not simply describe the difference in the formulas. Explain WHY the formulas are different.
I have no idea… Because the values of P are almost complements???

D. If we take a sample of 100 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 100)= 6 / (60/10)= 6/6= 1
z-score that corresponds to 1 is .3413
.5-.3413= .1587

E. Why are your answers to Parts B and D in this problem different?
Please be specific and include the value of the mean and the number 726 in your explanation.
I have no idea… As P increased, the probability of the interval increased?

Romsek said:

You should be adding your looked up Z score to 0.5 not subtracting. 726 > 720.

Your confusion probably arises because due to this error the behavior appeared to be the opposite of what it actually is.
Redo this correcting this error and see if the results now agree with your intuition. Check back if still confused.

Click to expand...

I don't know why I have a hard time remembering when to add and when to subtract from .5! Thanks! So, my calculations would be:


A. What is the probability that an individual, randomly-selected student will score less than 726?
z(726)= (726-720) / 60= 6/60= .1
P(x<726)= P(z<.1)=
z-score that corresponds to .1 is .0398
.5 + .0398 = .5398

B. If we take a sample of 4 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 4)= 6 / (60/2)= 6/30= 1/5= .2
z-score that corresponds to .2 is .0793
.5+.0793= .5793

C. Why are your answers to Parts A and B above in this problem different?
Please be specific! Do not simply describe the difference in the formulas. Explain WHY the formulas are different.
Still not sure on this one... is it because the values of .1 and .2 are different? I know the formulas are different because you have the extra step of dividing by the square root of the sample of students, in the case of B, 4 students, but I don't know why we divide by the square root instead of just dividing by 4 for the 4 students?

D. If we take a sample of 100 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 100)= 6 / (60/10)= 6/6= 1
z-score that corresponds to 1 is .3413
.5+.3413= .8413

E. Why are your answers to Parts B and D in this problem different?
Please be specific and include the value of the mean and the number 726 in your explanation.
Same thing here... D. uses a sample of 100 rather than 4, so obviously the sample sizes are different. The mean and deviation are the same in all of these problems, just the sample size changes. I'm not sure what including "the value of the mean and number 726" has to do with anything if the value of the mean is 726. Maybe it's because the sample size is so much larger in D than B, so the probability is much higher?

Thanks! That really made sense seeing the visual.

I have one last problem I'm working on, but I can't figure out the last part.

Find the value of X such that 95% of the samples, size n=3, will have means greater than X. Start by drawing a sketch!

I found the mean of 6 and standard deviation of 1.632993162. How do I go about solving this? I drew a sketch with a normal distribution, .5 in the middle where the mean of 6 falls and shaded 95% of it and looked up the coordinating z-score of .95, which is .3289. Now I'm stuck as to where to go from here?

Romsek said:

This doesn't quite make sense. Samples don't have means, they are single values. Do you mean 95% of the samples chosen will fall below the underlying mean of X ?

Click to expand...

Well this is the last part of a question and I found the mean to be on one of the other parts. I guess it doesn't apply here?

Consider a box containing a POPULATION of 5 identical blocks numbered 2, 4, 6, 8, and 10. Samples of size n=3 are drawn WITH REPLACEMENT. Each draw is independent of other draws (the first block is drawn, observed, and returned to the box; then the second block is drawn, observed, and returned to the box; then the third block is drawn).

Any help with this?

I submitted my study guide with all of my work to my professor to prepare for my final next week and he said this part is wrong? He won't give the answer, but just says if it is right or wrong for me to fix. Any help on this?


The scores on a standardized test are normally distributed with a population mean of 720 and a population standard deviation of 60.

Test scores are discrete, so do not forget to make the continuity correction before using the standard normal distribution (which is continuous).

A. What is the probability that an individual, randomly-selected student will score less than 726?
z(726)= (726-720) / 60= 6/60= .1
P(x<726)= P(z<.1)=
z-score that corresponds to .1 is .0398
.5 + .0398 = .5398

B. If we take a sample of 4 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 4)= 6 / (60/2)= 6/30= 1/5= .2
z-score that corresponds to .2 is .0793
.5+.0793= .5793

D. If we take a sample of 100 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 100)= 6 / (60/10)= 6/6= 1
z-score that corresponds to 1 is .3413
.5+.3413= .8413

Can anybody help with the last post in this thread? Thanks!

prettylittlepixels said:

I submitted my study guide with all of my work to my professor to prepare for my final next week and he said this part is wrong? He won't give the answer, but just says if it is right or wrong for me to fix. Any help on this?


The scores on a standardized test are normally distributed with a population mean of 720 and a population standard deviation of 60.

Test scores are discrete, so do not forget to make the continuity correction before using the standard normal distribution (which is continuous).

A. What is the probability that an individual, randomly-selected student will score less than 726?
z(726)= (726-720) / 60= 6/60= .1
P(x<726)= P(z<.1)=
z-score that corresponds to .1 is .0398
.5 + .0398 = .5398

B. If we take a sample of 4 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 4)= 6 / (60/2)= 6/30= 1/5= .2
z-score that corresponds to .2 is .0793
.5+.0793= .5793

D. If we take a sample of 100 randomly-selected students, what is the probability that the sample mean will be less than 726?
z(726)= (726-720) / (60/square root of 100)= 6 / (60/10)= 6/6= 1
z-score that corresponds to 1 is .3413
.5+.3413= .8413

Click to expand...

There are two minor errors. The first is the "continuity correction." Since the normal distribution is continuous, any number less than 725.5 will round to be an integer less than 726. You need to use z(725.5), rather than 726.

The second error is a matter of semantics: the ratio (x-mu)/sigma IS the z-score. What you look up in the table is the probability corresponding to the z-score, not the z-score corresponding to z. Your table seems to give the area between 0 and z. Since any x<0 is also <z, you have to ADD the full area of the lower half of the distribution. That is, add 0.5.

ok - part A, I think the more precise value of the z-score is 0.917.
................The area between 0 and z is 0.0365.
................The area from -infinity to z = 0.5365

[I am sure I answered this question in a different thread.]

DrPhil said:

There are two minor errors. The first is the "continuity correction." Since the normal distribution is continuous, any number less than 725.5 will round to be an integer less than 726. You need to use z(725.5), rather than 726.

The second error is a matter of semantics: the ratio (x-mu)/sigma IS the z-score. What you look up in the table is the probability corresponding to the z-score, not the z-score corresponding to z. Your table seems to give the area between 0 and z. Since any x<0 is also <z, you have to ADD the full area of the lower half of the distribution. That is, add 0.5.

ok - part A, I think the more precise value of the z-score is 0.917.
................The area between 0 and z is 0.0365.
................The area from -infinity to z = 0.5365

[I am sure I answered this question in a different thread.]

Click to expand...

That's what I was forgetting! The continuity correction!

A. (725.5-720) / 60= 5.5/60= .0917, which corresponds to .0359 on my 0 to z table, which I will add .5 to to get .5359.

B. (725.5-720) / (60/(square root of 4)= 5.5/30= .1833, which corresponds to .0714+.5= .5714

C. (725.5-720) / (60/(square root of 100)= 5.5/6= .9167, which corresponds to .3212+.5= .8212

Now why is there a difference in the z-scores you found and the z-scores I found, such as I found .0359 in part A and you found .0365? Are we using different tables?

Nevermind... I got it now. Thanks again!

prettylittlepixels said:

That's what I was forgetting! The continuity correction!

A. (725.5-720) / 60= 5.5/60= .0917, which corresponds to .0359 on my 0 to z table, which I will add .5 to to get .5359.

B. (725.5-720) / (60/(square root of 4)= 5.5/30= .1833, which corresponds to .0714+.5= .5714

C. (725.5-720) / (60/(square root of 100)= 5.5/6= .9167, which corresponds to .3212+.5= .8212

Now why is there a difference in the z-scores you found and the z-scores I found, such as I found .0359 in part A and you found .0365? Are we using different tables?

Click to expand...

I interpolated in the table! Instead of rounding z to 0.092, I left it as z=0.0917 and interpolated between the table entries for x=0.91 and x=0.92. I am probably a little more fussy than necessary, but to quote the probability precise to 4 digits I think the interpolation would be needed. On the other hand, we seldom need to quote 4 decimal places! Could just round to 0.036.