A large group of test scores is normally distributed with mean 78.2 and standard deviation 4.3

eddy2017

eddy2017

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Hi, i would need a hint for this one.
what to zero in on first?. just a hint.

A large group of test scores is normally distributed with mean 78.2 and standard deviation 4.3. What percent of the students scored 85 or better (nearest whole percent)?
Possible Answers:
6%
4%
5%
8%
7%
 
Hi, i would need a hint for this one.
eddy2017 said:
what to zero in on first?. just a hint.

A large group of test scores is normally distributed with mean 78.2 and standard deviation 4.3. What percent of the students scored 85 or better (nearest whole percent)?
Possible Answers:
6%
4%
5%
8%
7%
Click to expand...
Did you learn how to calculate the z-score or standardize the random variable?
 
Morning, I saw your replies now. I'm going to work on your hints and like always post back what I have found out. Thank for the hints.
 
BigBeachBanana said:
Hi, i would need a hint for this one.

Did you learn how to calculate the z-score or standardize the random variable?
Click to expand...
BBB, this is the formula for z score:
[math]z= x-mean/standard deviation[/math]If this applies to our question,
what is observation in our problem?. I know is represented by x but I don't undertand what observation means in the context. Can you explain or drop a hint or tell me if this formula is not okay for our question?. Tx
 
What I have learned
A z-score measures exactly how many standard deviations above or below the mean a data point is.
Here's the formula for calculating a z-score:

z = data point - mean / standard deviation

[math]z= x-μ/σ[/math]
data given
μ= 78.2
σ=4.3
data points=?
z=?

how do I find the amount of data point?.
I'm stuck here.
 
eddy2017 said:
BBB, this is the formula for z score:
[math]z= x-mean/standard deviation[/math]If this applies to our question,
what is observation in our problem?. I know is represented by x but I don't undertand what observation means in the context. Can you explain or drop a hint or tell me if this formula is not okay for our question?. Tx
Click to expand...
1) The formula is currently incorrect, without parenthesis. Please fix it.
2) We're interested in seeing scored 85 or better, the data point x= 85.
3) Did you learn how to use the standard normal distribution table or calculator? It's hard to suggest when I don't know what you know.
 
okay, got it
[math]z= (x - mean)/ standard deviation[/math][math]z= (85- 78.2)/4.3[/math][math]z score ≈ 1.58[/math]
Did you learn how to use the standard normal distribution table?.
I will wok on that. I'll get back with work. thanks
 
eddy2017 said:
okay, got it
[math]z= (x - mean)/ standard deviation[/math][math]z= (85- 78.2)/4.3[/math][math]z score ≈ 1.58[/math]
Did you learn how to use the standard normal distribution table?.
I will wok on that. I'll get back with work. thanks
Click to expand...
Before you blindly look up value from the table, follow Harry’s advice from post #2. Determine the relevant area.
 
eddy2017 said:
earn how to use the standard normal distribution table?.
Click to expand...
okay, I do not know what you mean by relevant area but this is what I have found out
the z-score was approximately 1.58
From a z-score table, in a normal distribution table, I could see that the p value for 1.58=.9429
so, the value is less than the z-score and we want the percent of students whose test score is 85 or better
so we need P(z ≥ 1.58).
I am stuck right here. Don't know how to proceed annd how to get to a percentage. the FIND is in percentage.
(besides, the question does not provide any z-score or normal distribution table. i had to search it online. I wonder if when taking a test that has one of these questions they must provide the tables or is there any other way to get to the solution without the table)
1645982834211.png
 

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eddy2017 said:
okay, I do not know what you mean by relevant area but this is what I have found out
the z-score was approximately 1.58
From a z-score table, in a normal distribution table, I could see that the p value for 1.58=.9429
so, the value is less than the z-score and we want the percent of students whose test score is 85 or better
so we need P(z ≥ 1.58).
I am stuck right here. Don't know how to proceed annd how to get to a percentage. the FIND is in percentage.
(besides, the question does not provide any z-score or normal distribution table. i had to search it online. I wonder if when taking a test that has one of these questions they must provide the tables or is there any other way to get to the solution without the table)
View attachment 31370
Click to expand...
By the rule of complement,
[math]\Pr(z<1.58)+\Pr(z \ge 1.58)=1[/math]
 
BigBeachBanana said:
By the rule of complement,
[math]\Pr(z<1.58)+\Pr(z \ge 1.58)=1[/math]
Click to expand...
Oh, The complement rule is stated as "the sum of the probability of an event and the probability of its complement is equal to 1
So,
[math]P(z≥1.58)= 1-P(z<1.58)[/math][math]=1-0.9429[/math][math]=0.0571[/math]that would be about or about 5.7 % then the correct choice is 6%..
 
Harry_the_cat said:
I always draw a quick diagram and shade the relevant area. Can you do that?
Click to expand...
Harry, would you mind explaining that?. Sounds interesting to me. How would ou do the diagram and how would you know what area to shade.
 
no, but Hary said this immediately after i posted the problem
quote 'I always draw a quick diagram and shade the relevant area. Can you do that?'.
What information should I use to shade the relevant area. I have seen the NDT that I posted.
0 to Z
Up to Z
Z onwards
what does that mean?
What info given in the problem would be useful to shade the relevant area?
 
Last edited:
eddy2017 said:
no, but Hary said this immediately after i posted the problem
quote 'I always draw a quick diagram and shade the relevant area. Can you do that?'.
What information should I use to see the relevant area. I have seen the NDT that i posted
0 to Z
Up to Z
Z onwards
what does that mean?
Click to expand...
It's a similar idea. Instead of using the standard normal with mean =0 and sd=1, your diagram would have the mean=78.2 and sd = 4.3. You would shade the area to the right of 85.

0 to z is the area from 0 (the mean) up to your specified z value.
Up to z is the area from the left, up to your specified z value.
Z onwards is the area from your specified z value to the right.
Try it out yourself. It's an interactive diagram.
 
0 to z is the area from 0 (the mean) up to your specified z value.
1645989130884.png


Up to z is the area from the left, up to your specified z value.

1645989239565.png

Z onwards is the area from your specified z value to the right.

1645989320323.png
wowww!!!! Eurekaaaaaa!
 
Yes that's what I mean by a diagram. 3 standard deviations both sides of the mean takes you out to the tails. Remember also that the total area under the curve is 1 or 100%. The table you provided in post #10 gives the area to the left of z (like the second graph in your last post). So you need to find 1 - 0.9429.
Are you meant to use these tables (they are a bit outdated now)? A lot of calculators have inbuilt functions that will do these calculations.
 
Are these calculators online rouse?
that interactive tool was fun to use!
 
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