Hello! I'm having problem with an equation & appreciate all the help I can get. Thanks in advance!
Question: A student scores a 60 on a mathematics test and an 80 on a science test. The data items for both tests are normally distributed. The mathematics test has a mean of 50 and a standard deviation of 5. The science test has a mean of 72 and a standard deviation of 6. On which test did the student have a better score? Explain why this is so.
Question: A student scores a 60 on a mathematics test and an 80 on a science test. The data items for both tests are normally distributed. The mathematics test has a mean of 50 and a standard deviation of 5. The science test has a mean of 72 and a standard deviation of 6. On which test did the student have a better score? Explain why this is so.
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Hello again!
We are given information relating to the data distribution, but in order to see how good/bad we did, we need to standardize it, that is, put it in terms of the Z distribution.
Let's look at the math test.
X ~ Normal(mu=50, sigma=5) and x1 = 60
Z ~ Normal(0, 1) and z1 = ? = (x1 - mu) / sigma = (60-50) / 5 = 10/5 = 2.
This means that the students math test was 2 standard deviations above the mean (that's pretty good!)
Do the same for the Science test, and whichever one is more standard deviations above the mean is the better score!
We are given information relating to the data distribution, but in order to see how good/bad we did, we need to standardize it, that is, put it in terms of the Z distribution.
Let's look at the math test.
X ~ Normal(mu=50, sigma=5) and x1 = 60
Z ~ Normal(0, 1) and z1 = ? = (x1 - mu) / sigma = (60-50) / 5 = 10/5 = 2.
This means that the students math test was 2 standard deviations above the mean (that's pretty good!)
Do the same for the Science test, and whichever one is more standard deviations above the mean is the better score!
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I guess if you are asking relative to the rest of the class, which test did the student rank higher in, it would be the math test.
The student is 2 standard deviations from the mean. This means 5% of the class falls outside of this range. Because it's normally distributed, we know, then that 2.5% scored higher and 2.5% scored lower. Thus the student is in the 97.5%-ile. This means only 2.5% of students did better.
In the science test, the student is not even 2 standard deviation above the mean. This means they are only maybe in the 90%-ile.
To show this mathematically:
Math: (60-50)/5 = 2. In your calculator: normalcdf(2, infinity, 0,1) = .023%
Science: (80-72)/6 = 1.333333. In your calculator: normalcdf(1.333333, infinity, 0,1) = .0912
This means only 2.3% of students did better than the student in the math test but over 9% did better than the student on the science test.
Thus, I would say the student did "better" on the math test
The student is 2 standard deviations from the mean. This means 5% of the class falls outside of this range. Because it's normally distributed, we know, then that 2.5% scored higher and 2.5% scored lower. Thus the student is in the 97.5%-ile. This means only 2.5% of students did better.
In the science test, the student is not even 2 standard deviation above the mean. This means they are only maybe in the 90%-ile.
To show this mathematically:
Math: (60-50)/5 = 2. In your calculator: normalcdf(2, infinity, 0,1) = .023%
Science: (80-72)/6 = 1.333333. In your calculator: normalcdf(1.333333, infinity, 0,1) = .0912
This means only 2.3% of students did better than the student in the math test but over 9% did better than the student on the science test.
Thus, I would say the student did "better" on the math test
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Math's score
M ∼ n(50; 5)
z = (60 - 50)/5 = 2
S ∼ n(72; 6)
z = (80 - 72)/6 = 4/3
The values z are pure, without units of measure; the student has a better score on Math because 2 > 4/3
M ∼ n(50; 5)
z = (60 - 50)/5 = 2
S ∼ n(72; 6)
z = (80 - 72)/6 = 4/3
The values z are pure, without units of measure; the student has a better score on Math because 2 > 4/3