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Compute \(P(x)\) using the binomial probability formula. Then determine whether the normal distribution can be used as an approximation for the binomial distribution. If so, approximate \(P(x)\) and compare the result to the exact probability. $$ n=40, p=0.25, x=30 $$

Short Answer

Expert verified
Exact probability: \( P(30) \). Approximate probability: nearly 1.

Step by step solution

01

Define the Binomial Probability Formula

The binomial probability formula is given by \[ P(x) = \binom{n}{x} p^x (1-p)^{n-x} \] where \( n \) is the number of trials, \( p \) is the probability of success, and \( x \) is the number of successes.
02

Calculate the Binomial Coefficient

The binomial coefficient \( \binom{n}{x} \) can be calculated as \[ \binom{n}{x} = \frac{n!}{x! (n-x)!} \] For this problem, \( n = 40 \) and \( x = 30 \): \[ \binom{40}{30} = \frac{40!}{30! \, 10!} \] Computing this, we get \( \binom{40}{30} \)
03

Compute the Binomial Probability

Now substitute the values into the binomial probability formula: \[ P(30) = \binom{40}{30} (0.25)^{30} (0.75)^{10} \] Calculate the terms separately and then multiply them together to get the probability.
04

Check Normal Approximation Conditions

To use the normal approximation, the conditions \( np \geq 5 \) and \( n(1-p) \geq 5 \) must be satisfied. For this problem, \( n = 40 \) and \( p = 0.25 \): \[ np = 40 \times 0.25 = 10 \] and \[ n(1-p) = 40 \times 0.75 = 30 \] Both conditions are met, so we can use the normal distribution.
05

Find Mean and Standard Deviation for Normal Distribution

The mean \( \mu \) and standard deviation \( \sigma \) for the normal distribution are given by: \[ \mu = np \] and \[ \sigma = \sqrt{np(1-p)} \] For \( n = 40 \) and \( p = 0.25 \): \[ \mu = 40 \times 0.25 = 10 \] \[ \sigma = \sqrt{40 \times 0.25 \times 0.75} = \sqrt{7.5} \approx 2.74 \]
06

Convert Binomial to Standard Normal Variable

Convert \( x = 30 \) to a standard normal variable using: \[ Z = \frac{X - \mu}{\sigma} \] For this problem: \[ Z = \frac{30 - 10}{2.74} \approx 7.30 \]
07

Find the Normal Probability

Since \( Z \approx 7.30 \) is a very high value, the probability \( P(Z \leq 7.30) \) is nearly 1. Hence, the approximate probability using the normal distribution is 1.
08

Compare Exact and Approximate Probabilities

Compare the exact probability calculated using the binomial formula with the approximate probability obtained from the normal distribution. Note that they may not be exactly the same but should be close.

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Key Concepts

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

Binomial Probability Formula
The binomial probability formula is essential when calculating the probability of a given number of successes in a fixed number of trials. The formula is: ing \(P(x) = \binom{n}{x} p^x (1-p)^{n-x}\). \(n\) represents the total number of trials, \(p\) is the probability of success in each trial, and \(x\) is the desired number of successes. To solve an example where \(n = 40, p = 0.25, x = 30\), follow these steps:
  • Calculate the binomial coefficient \(\binom{n}{x}\): This can be found using \(\binom{n}{x} = \frac{n!}{x! (n-x)!}\). For our example, compute \(\binom{40}{30} = \frac{40!}{30!10!}\).
  • Substitute into the binomial probability formula: Plug the values into the formula: \(P(30) = \binom{40}{30} (0.25)^{30} (0.75)^{10}\).
Normal Approximation
When a binomial distribution meets certain conditions, it can be approximated using the normal distribution. This method simplifies calculations for large \(n\). The conditions required for this approximation are:
  • \(np \geq 5\): Here, \(n = 40\) and \(p = 0.25\), so \(np = 40 \times 0.25 = 10\).
  • \(n(1 - p) \geq 5\): Here, \(n = 40\) and \(1 - p = 0.75\), so \(n(1 - p) = 40 \times 0.75 = 30\).
Since both conditions are satisfied, the normal distribution can be used for approximating the binomial distribution.To approximate, compute the mean \(\mu\) and standard deviation \(\sigma\), with:
  • Mean: \(\mu = np\)
  • Standard Deviation: \(\sigma = \sqrt{np(1 - p)}\)
For the given values, \(\mu = 40 \times 0.25 = 10\) and \(\sigma = \sqrt{40 \times 0.25 \times 0.75} \approx 2.74\). To convert the binomial variable \(x\) to a standard normal variable \(Z\), use \(Z = \frac{X - \mu}{\sigma}\).Here, \(Z = \frac{30 - 10}{2.74} \approx 7.30\). Given the high Z-value, the probability \(P(Z \leq 7.30)\) is nearly 1, thus the approximate probability using the normal distribution is almost 1.
Statistical Calculation Steps
Performing statistical calculations step-by-step helps in grasping each component clearly. Here’s a concise blueprint to guide you through similar problems:
  • Step 1: Define the parameters (\(n\), \(p\), and \(x\)).
  • Step 2: Calculate the binomial coefficient \(\binom{n}{x}\).
  • Step 3: Compute the probability using the binomial formula
  • Step 4: Check if normal approximation conditions are met (\(np \geq 5\) and \(n(1 - p) \geq 5\)).
  • Step 5: Calculate the mean and standard deviation for the normal distribution.
  • Step 6: Convert the binomial variable \(x\) to a standard normal variable \(Z\).
  • Step 7: Determine the probability using the normal distribution.
  • Step 8: Compare the exact binomial probability with the approximated probability using the normal distribution.

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Most popular questions from this chapter

The following data represent the distribution of birth weights (in grams) for babies in which the pregnancy went full term (37-41 weeks). $$ \begin{array}{cc} \text { Birth Weight (g) } & \text { Number of Live Births } \\ \hline 0-499 & 22 \\ \hline 500-999 & 201 \\ \hline 1000-1499 & 1,645 \\ \hline 1500-1999 & 9,365 \\ \hline 2000-2499 & 92,191 \\ \hline 2500-2999 & 569,319 \\ \hline 3000-3499 & 1,387,335 \\ \hline 3500-3999 & 988,011 \\ \hline 4000-4499 & 255,700 \\ \hline 4500-4999 & 36,766 \\ \hline 5000-5499 & 3,994\\\ \hline \end{array} $$ (a) Construct a relative frequency distribution for birth weight. (b) Draw a relative frequency histogram for birth weight. Describe the shape of the distribution. (c) Determine the mean and standard deviation birth weight. (d) Use the normal model to determine the proportion of babies in each class. (e) Compare the proportions predicted by the normal model to the relative frequencies found in part (a). Do you believe that the normal model is effective in describing the birth weights of babies?

The birth weights of full-term babies are normally distributed with mean \(\mu=3400\) grams and \(\sigma=505\) grams. Source: Based on data obtained from the National Vital Statistics Report, Vol. \(48,\) No. 3 (a) Draw a normal curve with the parameters labeled. (b) Shade the region that represents the proportion of full-term babies who weigh more than 4410 grams. (c) Suppose the area under the normal curve to the right of \(x=4410\) is \(0.0228 .\) Provide two interpretations of this result.

Find the indicated areas. For each problem, be sure to draw a standard normal curve and shade the area that is to be found. Determine the area under the standard normal curve that lies to the right of (a) \(z=-3.49\) (b) \(z=-0.55\) (c) \(z=2.23\) (d) \(z=3.45\)

In games where a team is favored by more than 12 points, the margin of victory for the favored team relative to the spread is normally distributed with a mean of -1.0 point and a standard deviation of 10.9 points. Source: Justin Wolfers, "Point Shaving: Corruption in NCAA Basketball' (a) In games where a team is favored by more than 12 points, what is the probability that the favored team wins by 5 or more points relative to the spread? (b) In games where a team is favored by more than 12 points, what is the probability that the favored team loses by 2 or more points relative to the spread? (c) In games where a team is favored by more than 12 points, what is the probability that the favored team "beats the spread"? Does this imply that the possible point shaving spreads are accurate for games in which a team is favored by more than 12 points?

Explain why \(P(X<30)\) should be reported as \(<0.0001\) if \(X\) is a normal random variable with mean 100 and standard deviation \(15 .\)

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