Problem 17 - Algebra 2 and Trigonometry

Question

In \(14-17\) , use the normal approximation to estimate each probability. Round your answers to three decimal places. . \(P(\text { at most } 100 \text { successes }), p=\frac{6}{7}, n=125\)

Step-by-Step Solution

Answer

The probability of at most 100 successes is approximately 0.046.

Step 1: Define Variables

First, identify the values given in the problem. We have \(n = 125\) trials, each with a probability of success \(p = \frac{6}{7}\). We want to find the probability of at most 100 successes.

Step 2: Calculate Mean and Standard Deviation

For a binomial distribution, the mean \(\mu\) and standard deviation \(\sigma\) are given by \(\mu = np\) and \(\sigma = \sqrt{np(1-p)}\). Calculate these using \(n = 125\) and \(p = \frac{6}{7}\).- \(\mu = 125 \times \frac{6}{7} = 107.14\)- \(\sigma = \sqrt{125 \times \frac{6}{7} \times \frac{1}{7}} \approx 3.94\)

Step 3: Convert to Standard Normal Distribution

To find the approximate probability using the normal distribution, convert the binomial random variable to a standard normal variable. Use the continuity correction factor by adjusting for discrete outcomes: \(P(X \leq 100) \approx P\left(Z \leq \frac{100.5 - \mu}{\sigma}\right)\). Here, we choose 100.5 for continuity correction.- \(Z = \frac{100.5 - 107.14}{3.94} \approx \frac{-6.64}{3.94} \approx -1.685\)

Step 4: Find Probability Using Z-table

Using the standard normal Z-table, find \(P(Z \leq -1.685)\). Look up the Z-value \(-1.69\) (rounding to two decimal places for precision) in the table, which approximately gives 0.046. This is the probability that the number of successes is at most 100.

Key Concepts

Understanding the Binomial Distribution

The binomial distribution is a probability distribution that summarizes the likelihood of a given number of successes in a set number of trials. It is appropriate when each trial is independent and there are only two possible outcomes: success or failure. For example, in our exercise, we have 125 trials, each with a success probability of \( p = \frac{6}{7} \).
Broadly speaking, the binomial distribution is defined by two parameters:

• \( n \): the total number of trials or experiments.
• \( p \): the probability of achieving success in a single trial.

The probability of getting a specific number of successes can be calculated using the binomial probability formula, but as the number of trials increases, direct calculations become complex. This is where approximations, like the normal approximation, become useful.

Role of Continuity Correction in Approximations

When we use the normal approximation to estimate probabilities for a binomial distribution, we must adjust for the fact that the binomial distribution is discrete (only whole numbers can occur as outcomes) while the normal distribution is continuous. This adjustment is called the continuity correction.
The continuity correction suggests adding or subtracting 0.5 to the discrete variable when converting it to a continuous variable.
In our problem, we're interested in \( P(X \leq 100) \), so we use \( 100.5 \) instead of 100. This method smoothens out the transition between the two types of distributions and improves the accuracy of our approximation.

Using the Standard Normal Distribution for Probabilities

To calculate the probability and use the standard normal distribution, we transform our binomial variable into a standard normal variable. This involves converting the mean, \( \mu \), and standard deviation, \( \sigma \), calculated earlier to fit a standard normal curve with mean 0 and standard deviation 1.
The transformation formula is:

• \( Z = \frac{X - \mu}{\sigma} \)

In our exercise, the calculation was:

• \( Z = \frac{100.5 - 107.14}{3.94} \approx -1.685 \)

A negative \( Z \) value indicates that the observed value is below the mean. Once we have the \( Z \) value, we can refer to standard normal distribution tables (often known as Z-tables) to determine the probability of this outcome.

Estimating Probability with Z-tables and Interpretation

Z-tables are used to find the probability that a standard normal random variable is less than or equal to a particular value. In our example, once the \( Z \) value was calculated as approximately \( -1.685 \), we referred to a Z-table to find the corresponding probability.
The Z-table value of \( -1.685 \) is approximately 0.046, indicating the probability of achieving at most 100 successes in our trials.
This means there is a 4.6% chance that the outcome will be 100 successes or fewer, given that each trial has a 6/7 chance of success. Z-tables provide a crucial reference for interpreting these probabilities, helping to practically understand the likelihood of various outcomes without needing to perform extensive calculations.