Problem 51 - Algebra 2

Question

Evaluate 2\(\sqrt{\frac{p(1-p)}{n}}\) for the given values of \(p\) and \(n\) Round to the nearest thousandth if necessary. $$ p=0.25, n=500 $$

Step-by-Step Solution

Answer

The evaluated expression rounds to 0.039.

Step 1: Identify the given values

We are given the values of \( p \) and \( n \). In this exercise, \( p = 0.25 \) and \( n = 500 \). We will use these values to evaluate the expression.

Step 2: Substitute the values into the expression

Substitute \( p = 0.25 \) and \( n = 500 \) into the expression \( 2\sqrt{\frac{p(1-p)}{n}} \). This becomes: \[ 2\sqrt{\frac{0.25(1-0.25)}{500}} \].

Step 3: Simplify inside the square root

First, calculate \( 1 - 0.25 = 0.75 \). Then multiply \( 0.25 \times 0.75 = 0.1875 \). Now, divide by \( 500 \): \[ \frac{0.1875}{500} = 0.000375 \].

Step 4: Calculate the square root

Find the square root of \( 0.000375 \): \( \sqrt{0.000375} \approx 0.019364916731 \).

Step 5: Multiply by 2

Multiply the result from the previous step by 2 to complete the evaluation of the expression: \[ 2 \times 0.019364916731 \approx 0.038729833462 \].

Step 6: Round the result

Round the value \( 0.038729833462 \) to the nearest thousandth, which gives \( 0.039 \).

Key Concepts

Square Root Calculation

Calculating square roots can initially seem daunting, but it is essentially finding a number that, when multiplied by itself, gives the original number. It's like asking, "what number times itself equals this number?"
In our exercise, after substituting the values of \( p \) and \( n \) and simplifying inside the square root, we arrive at \( \sqrt{0.000375} \). To calculate this:

• Use a calculator, as manual computation can be complex.
• The result for \( \sqrt{0.000375} \) is approximately \( 0.019364916731 \).

Calculators make this process effortless by quickly deriving precise results. Understanding how this calculation fits into larger problems helps refine your mathematical skills. Remember that square root calculations often appear in statistics and probability, signifying relationships such as standard deviations.

Substitution in Expressions

Substitution involves replacing variables in an expression with their given numerical values. It seems simple, but it's crucial for solving algebraic expressions.
To evaluate \( 2\sqrt{\frac{p(1-p)}{n}} \), we substitute \( p = 0.25 \) and \( n = 500 \):

• The expression becomes \( 2\sqrt{\frac{0.25(1-0.25)}{500}} \).
• This transforms a general equation into a specific one you can solve numerically.

Always double-check substitutions to ensure accuracy. Minor mistakes in substitution can lead to incorrect outcomes, so be meticulous in this step.

Rounding Numbers

Rounding is simplifying a number while retaining its approximate value. It's commonly used to make large or complex numbers more manageable and readable.
In mathematics, rules for rounding include looking at the digit just after the one you wish to round to. For our exercise, rounding is to the nearest thousandth:

• The result \( 0.038729833462 \) should be examined at the fourth decimal place.
• When the next digit (fourth place) is 5 or more, round up.

Thus, you round \( 0.0387... \) to \( 0.039 \). Precision matters in scientific calculations, and rounding ensures the results are clear and concise.

Simplifying Fractions

This involves reducing the numbers inside a fraction to their smallest form without changing its value. It's a vital step in preparing expressions for further calculation or evaluation.
For the expression \( \frac{0.1875}{500} \), simplifying means:

• Divide the numerator and the denominator by their greatest common divisor.
• Here, directly calculating gives \( \frac{0.1875}{500} = 0.000375 \), which is already a simple form.

Simplifying ensures correctness in calculations and eases subsequent operations. It allows you to present the problem in a more digestible form, making further steps like taking square roots or substituting straightforward.