/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 75 \- Find an algebraic formula for... [FREE SOLUTION] | 91Ó°ÊÓ

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\- Find an algebraic formula for the population standard deviation of a sample \(\\{x, y\\}\) of two scores \((x \leq y)\).

Short Answer

Expert verified
The algebraic formula for the population standard deviation of a sample with two scores, \(x\) and \(y \ (x \leq y)\), is: \[\sigma = \sqrt{\frac{(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}}{2}}\]

Step by step solution

01

Calculate the population mean

First, we need to calculate the population mean (\(\mu\)), which is the average of the two scores x and y. This can be computed with the formula: \[ \mu = \frac{x + y}{2} \]
02

Compute the difference of each score from the mean

Next, we will compute the difference between each score and the population mean: 1. For x: \(x - \mu = x - \frac{x + y}{2} \) 2. For y: \(y - \mu = y - \frac{x + y}{2} \)
03

Square the differences

Now, we need to square the differences computed in Step 2: 1. For x: \((x - \frac{x + y}{2})^{2}\) 2. For y: \((y - \frac{x + y}{2})^{2}\)
04

Calculate the sum of squared differences

We will now find the sum of squared differences obtained in Step 3: \[(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}\]
05

Divide the sum of squared differences by the number of scores

Since we have 2 scores in the sample, we will divide the sum of squared differences obtained in Step 4 by 2: \[\frac{(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}}{2}\]
06

Take the square root of the value obtained in Step 5

The final step is to take the square root of the value obtained in Step 5, which will give us the population standard deviation (\(\sigma\)) for the sample of two scores: \[\sigma = \sqrt{\frac{(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}}{2}}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Population Mean
To find the population standard deviation, we start by calculating the **Population Mean**, which is a fancy term for finding the average. Imagine you have two test scores: one for Math and one for Science. If you want to know your average, you'd add both scores and then divide by the number of scores you have. The formula for the population mean \( \mu \) is:
  • \( \mu = \frac{x + y}{2} \)
Here, \( x \) and \( y \) are your scores. By adding them and dividing by 2 (because there are 2 scores), you get a number that tells you where the center or balance point of your scores is on the number line.

Understanding the population mean helps us see how far each score is from this "center." It's like knowing the position of the median of a see-saw to understand where each end will tilt.
Sum of Squared Differences
Once we have the population mean, the next step is to determine how far each score lies from this mean. This is called finding the "**Sum of Squared Differences**." It's like checking how far every friend in a group is standing from a central point. For each score, \( x \) and \( y \), we calculate their differences from the mean, square those differences, and then add them.
  • For score \( x \), the difference is \( x - \mu \).
  • For score \( y \), the difference is \( y - \mu \).
Now square these differences:
  • \((x - \mu)^2 = (x - \frac{x+y}{2})^2\)
  • \((y - \mu)^2 = (y - \frac{x+y}{2})^2\)
Adding these squared differences together gives:
  • \((x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}\)
This squared step ensures that all differences are positive and emphasizes larger differences, just like squares in geometry make sizes look larger.

It's mathematical "magnifying." Sum of Squared Differences encapsulates the total deviation from the mean, showing the overall spread or variability in the dataset.
Square Root Calculation
Now, the tricky part, but don't worry, it's easy! We've found the average spread of your scores from the mean with Sum of Squared Differences, but it’s not yet the standard deviation. Why? Because to get there, we have to dive into the "**Square Root Calculation**." Imagine if measuring how children can jump, but then you make that figure normal-sized by taking the square root.From the Sum of Squared Differences, we divide by the number of scores (2, in our case) to "normalize" or average their impact:
  • \(\frac{(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}}{2}\)
Taking the square root of the result adjusts the squared scores back to their original scale:
  • \(\sigma = \sqrt{\frac{(x - \frac{x + y}{2})^{2} + (y - \frac{x + y}{2})^{2}}{2}}\)
This calculation gives us the population standard deviation, a handy statistic to summarize how your scores vary from the average.

Like evening out your raised pulse after a sprint, it tells you the typical distance of each score from the mean.

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Most popular questions from this chapter

Your friend Charlesworth claims that the median of a collection of data is always close to the mean. Is he correct? If so, say why; if not, give an example to prove him wrong.

Kent's Tents has four red tents and three green tents in stock. Karin selects four of them at random. Let \(X\) be the number of red tents she selects. Give the probability distribution and find \(P(X \geq 2)\).

Following is a sample of the day-byday change, rounded to the nearest 100 points, in the Dow Jones Industrial Average during 10 successive business days around the start of the financial crisis in October \(2008:^{20}\) $$ -100,400,-200,-500,200,-300,-200,900,-100,200 $$ Compute the mean and median of the given sample. Fill in the blank: There were as many days with a change in the Dow above \(\quad\) points as there were with changes below that.

The Blue Sky Flight Insurance Company insures passengers against air disasters, charging a prospective passenger \(\$ 20\) for coverage on a single plane ride. In the event of a fatal air disaster, it pays out \(\$ 100,000\) to the named beneficiary. In the event of a nonfatal disaster, it pays out an average of \(\$ 25,000\) for hospital expenses. Given that the probability of a plane's crashing on a single trip is \(.00000087,{ }^{32}\) and that a passenger involved in a plane crash has a \(.9\) chance of being killed, determine the profit (or loss) per passenger that the insurance company expects to make on each trip. HINT [Use a tree to compute the probabilities of the various outcomes.]

The probability of a plane crashing on a single trip in 1990 was .00000087. Find the approximate probability that in \(100,000,000\) flights, there will be more than 110 crashes.

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