Explanations 
Textbooks 
Exams 
GCSE 
IGCSE 
AS 
A Level 
International A Level 
University Admissions Tests 
Flashcards 
Vaia AI 
Notes 
Study Plans 
Study Sets 
Find a job 
Find a degree 
Student Deals 
Magazine 
Mobile App Chapter 15: Problem 10
Using both the binomial distribution and the normal approximation. A true coin is tossed \(10^{4}\) times. (a) Find the probability of getting exactly 5000 heads. (b) Find the probability of between 4900 and 5075 heads.
Short Answer
Expert verified
Part (a): Approximately 0.008. Part (b): Approximately 0.9104.
Step by step solution
01
- Define the Binomial Distribution
The binomial distribution is defined as follows: If a random experiment consists of n independent trials, each with two possible outcomes (success or failure) and the probability of success in each trial is p, then the probability of getting exactly k successes is given by the formula \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]For a fair coin, p = 0.5 and n = 10^4.
02
- Calculate for Part (a)
We need to find the probability of getting exactly 5000 heads (k = 5000).Using the formula:\[ P(X = 5000) = \binom{10^4}{5000} (0.5)^{5000} (0.5)^{5000} \]This simplifies to:\[ P(X = 5000) = \binom{10^4}{5000} (0.5)^{10000} \]Given the large n, directly calculating this is complex. We'll use normal approximation in the next step.
03
- Normal Approximation for Part (a)
With the binomial distribution, for large n, we can approximate it to a normal distribution with mean \( \text{μ} = np \) and variance \( \text{σ}^2 = np(1-p) \).Here, mean \( \text{μ} = 10^4 \times 0.5 = 5000 \) and variance \( \text{σ}^2 = 10^4 \times 0.5 \times 0.5 = 2500 \).So, standard deviation \( \text{σ} = \text{√2500} = 50 \).Therefore, we use the normal distribution N(5000, 50).Now calculate the probability that X = 5000 in normal terms.\[ P(4999.5 < X < 5000.5) \]Find corresponding z-scores:\[ Z = \frac{5000.5 - 5000}{50} = 0.01 \]\[ Z = \frac{4999.5 - 5000}{50} = -0.01 \]Using z-tables, the probability P(4999.5 < X < 5000.5) is approximately 0.008.
04
- Calculate for Part (b) With Normal Approximation
We need to find the probability of getting between 4900 and 5075 heads.Convert 4900 and 5075 to z-scores:\[ Z = \frac{4900 - 5000}{50} = -2 \]\[ Z = \frac{5075 - 5000}{50} = 1.5 \]From z-tables, P(Z < -2) is approximately 0.0228 and P(Z < 1.5) is approximately 0.9332.Therefore, the probability of getting between 4900 and 5075 heads is:\[ P(-2 < Z < 1.5) = 0.9332 - 0.0228 = 0.9104 \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Approximation
The binomial distribution can be difficult to work with, especially when dealing with a large number of trials like in our exercise. That is where normal approximation comes into play. Normal approximation is a method used to simplify the calculation of probabilities in a binomial distribution, using the normal distribution instead.
For a binomial distribution with parameters n (number of trials) and p (probability of success), the mean or expected value is given by: \( \text{μ} = np \). And the variance, which tells us how much the data is spread out, is given by: \( \text{σ}^2 = np(1-p) \).
To transform the binomial distribution into a normal distribution, use the following: \( \text{mean (μ)} \) and \( \text{standard deviation (σ)} = \text{√(np(1-p))} \). This way, you can use the normal distribution N(μ, σ) to find probabilities.
For example, in our exercise, tossing 10,000 coins, we have a mean of 5000 and a standard deviation of 50. Instead of calculating the binomial probability directly, we use the mean and standard deviation to approximate it using normal distribution.
For a binomial distribution with parameters n (number of trials) and p (probability of success), the mean or expected value is given by: \( \text{μ} = np \). And the variance, which tells us how much the data is spread out, is given by: \( \text{σ}^2 = np(1-p) \).
To transform the binomial distribution into a normal distribution, use the following: \( \text{mean (μ)} \) and \( \text{standard deviation (σ)} = \text{√(np(1-p))} \). This way, you can use the normal distribution N(μ, σ) to find probabilities.
For example, in our exercise, tossing 10,000 coins, we have a mean of 5000 and a standard deviation of 50. Instead of calculating the binomial probability directly, we use the mean and standard deviation to approximate it using normal distribution.
Probability Calculation
When using the normal approximation, calculating probabilities is straightforward. You start by converting the binomial problem into a normal one.
For part (a) of our example, attempting to find exactly 5000 heads can be complex with a binomial calculation. However, with normal approximation, it becomes a lot easier.
With normal distribution N(5000, 50), you look for the probability around 5000. Because it's a continuous distribution, we look for the range 4999.5 to 5000.5.
This makes it easier to handle using the standard normal table (z-table) as once converted, z-table provides the probability values directly.
In our exercise, converting 5000.5 and 4999.5 to z-scores gives us z-values of 0.01 and -0.01 respectively. Using the z-table allows us to find the probability between these z-values, which results in approximately 0.008.
For part (b), looking for the probability of between 4900 and 5075 heads involves converting these to z-scores and using the z-table.
We find z-scores as follows: -2 for 4900 and 1.5 for 5075. According to the z-table, the cumulative probabilities give us a final probability of 0.9104.
For part (a) of our example, attempting to find exactly 5000 heads can be complex with a binomial calculation. However, with normal approximation, it becomes a lot easier.
With normal distribution N(5000, 50), you look for the probability around 5000. Because it's a continuous distribution, we look for the range 4999.5 to 5000.5.
This makes it easier to handle using the standard normal table (z-table) as once converted, z-table provides the probability values directly.
In our exercise, converting 5000.5 and 4999.5 to z-scores gives us z-values of 0.01 and -0.01 respectively. Using the z-table allows us to find the probability between these z-values, which results in approximately 0.008.
For part (b), looking for the probability of between 4900 and 5075 heads involves converting these to z-scores and using the z-table.
We find z-scores as follows: -2 for 4900 and 1.5 for 5075. According to the z-table, the cumulative probabilities give us a final probability of 0.9104.
Z-Score
A z-score is a measure that describes a value's position relative to the mean of a group of values, measured in terms of standard deviations. It allows us to standardize individual data points and make calculations easier.
The formula to find a z-score (\text{Z}) is: \(Z = \frac{(X - μ)}{σ}\), where X is the value to be determined, μ is the mean, and σ is the standard deviation.
In our example, when dealing with the normal approximation for binomial distributions, z-scores make it easier to convert normal distribution values into standard normal (mean of 0 and standard deviation of 1).
For instance, converting an X value of 5000.5, where μ=5000 and σ=50, gives us: \(Z = \frac{5000.5 - 5000}{50} = 0.01 \). Similarly, X value of 4999.5 results in \(Z = -0.01\).
This way, the resulting z-scores can be easily referenced in a standard normal distribution table (z-table), and you can find the corresponding probabilities.
The use of z-scores is a powerful tool in probability calculations because it simplifies the complex calculations of the binomial distribution into a more manageable normal distribution framework.
The formula to find a z-score (\text{Z}) is: \(Z = \frac{(X - μ)}{σ}\), where X is the value to be determined, μ is the mean, and σ is the standard deviation.
In our example, when dealing with the normal approximation for binomial distributions, z-scores make it easier to convert normal distribution values into standard normal (mean of 0 and standard deviation of 1).
For instance, converting an X value of 5000.5, where μ=5000 and σ=50, gives us: \(Z = \frac{5000.5 - 5000}{50} = 0.01 \). Similarly, X value of 4999.5 results in \(Z = -0.01\).
This way, the resulting z-scores can be easily referenced in a standard normal distribution table (z-table), and you can find the corresponding probabilities.
The use of z-scores is a powerful tool in probability calculations because it simplifies the complex calculations of the binomial distribution into a more manageable normal distribution framework.
