Statistics/Standard Deviation Word Problem
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Statistics/Standard Deviation Word Problem

[From: ] [author: ] [Date: 11-07-08] [Hit: ]
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.Lets look at the math test.......
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.

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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!

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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

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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
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