In eight throws of a die, 5 or 6 is considered a success. The mean and standard deviation of total number of successes is respectively given by
Understanding the Binomial Distribution Problem
This problem involves a series of independent trials, where each trial has only two possible outcomes: success or failure. This is the definition of a binomial distribution. In this specific case, the experiment is throwing a die 8 times.
Trial: Throwing a single die once.
Number of trials (n): The die is thrown 8 times, so \(n = 8\).
Success: Getting a 5 or a 6 in a single throw.
Failure: Getting any number other than 5 or 6 (i.e., 1, 2, 3, or 4).
Calculating Probability of Success and Failure
A standard die has 6 faces, each equally likely to appear. The total possible outcomes in one throw are {1, 2, 3, 4, 5, 6}.
The outcomes considered a success are {5, 6}. There are 2 favourable outcomes.
The probability of success (p) in a single trial is the number of favourable outcomes divided by the total number of outcomes:
\(p = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}} = \frac{2}{6} = \frac{1}{3}\)
The probability of failure (q) is the complement of the probability of success:
\(q = 1 - p = 1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{2}{3}\)
We can verify that \(p + q = \frac{1}{3} + \frac{2}{3} = 1\).
Finding the Mean of the Number of Successes
For a binomial distribution with \(n\) trials and probability of success \(p\), the mean (expected value) of the number of successes is given by the formula:
\(\mu = np\)
Using the values from the problem:
\(\mu = 8 \times \frac{1}{3}\)
\(\mu = \frac{8}{3}\)
So, the mean number of successes in eight throws is \(\frac{8}{3}\).
Finding the Standard Deviation of the Number of Successes
For a binomial distribution with \(n\) trials, probability of success \(p\), and probability of failure \(q\), the variance of the number of successes is given by the formula:
\(\sigma^2 = npq\)
Using the values from the problem:
\(\sigma^2 = 8 \times \frac{1}{3} \times \frac{2}{3}\)
\(\sigma^2 = \frac{8 \times 1 \times 2}{3 \times 3}\)
\(\sigma^2 = \frac{16}{9}\)
The standard deviation (\(\sigma\)) is the square root of the variance:
\(\sigma = \sqrt{\sigma^2} = \sqrt{npq}\)
\(\sigma = \sqrt{\frac{16}{9}}\)
\(\sigma = \frac{\sqrt{16}}{\sqrt{9}}\)
\(\sigma = \frac{4}{3}\)
So, the standard deviation of the number of successes is \(\frac{4}{3}\).
Summary of Results
Based on our calculations for the binomial distribution with \(n=8\) and \(p=\frac{1}{3}\):
Mean (\(\mu\)) = \(\frac{8}{3}\)
Standard Deviation (\(\sigma\)) = \(\frac{4}{3}\)
Thus, the mean and standard deviation of the total number of successes are respectively \(\frac{8}{3}\) and \(\frac{4}{3}\).
Revision Table: Binomial Distribution Formulas
| Parameter | Formula | Calculation | Result |
|---|---|---|---|
| Number of trials (n) | - | Given as 8 throws | \(n = 8\) |
| Probability of success (p) | - | Getting 5 or 6 on a die (\(\frac{2}{6}\)) | \(p = \frac{1}{3}\) |
| Probability of failure (q) | \(1 - p\) | \(1 - \frac{1}{3}\) | \(q = \frac{2}{3}\) |
| Mean (\(\mu\)) | \(np\) | \(8 \times \frac{1}{3}\) | \(\mu = \frac{8}{3}\) |
| Variance (\(\sigma^2\)) | \(npq\) | \(8 \times \frac{1}{3} \times \frac{2}{3}\) | \(\sigma^2 = \frac{16}{9}\) |
| Standard Deviation (\(\sigma\)) | \(\sqrt{npq}\) | \(\sqrt{\frac{16}{9}}\) | \(\sigma = \frac{4}{3}\) |
Additional Information: What is Binomial Distribution?
The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each with the same probability of success. Key characteristics of a binomial experiment are:
There are a fixed number of trials, denoted by \(n\).
Each trial has only two possible outcomes: "success" or "failure".
The probability of success, denoted by \(p\), is the same for each trial.
The trials are independent of each other.
The number of successes \(X\) in a binomial experiment follows a binomial distribution, denoted as \(X \sim B(n, p)\). The probability mass function for a binomial distribution is given by:
\(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)
where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) is the binomial coefficient, representing the number of ways to choose \(k\) successes from \(n\) trials.
The mean and standard deviation are important measures that describe the center and spread of the distribution, as calculated in this problem.
Important Questions from Mean and Variance of Random variables
A continuous random variable x has a probability density function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {6x\left( {1 - x} \right),}&{0 < x \le 1}\\ {0,}&{otherwise} \end{array}} \right.\) Then the variance of x is:
X is a non-negative integer valued random variable with
\( P(X = x) =\left\{ \begin{matrix} \dfrac {x+1}{2^{(x+2)}} & x = 0, 1, 2... \\\ 0 & \rm{otherwise} \end{matrix} \right.\)
Then, mean and variance of X are respectively
A random variable X has the distribution law as given below:
X
1
2
3
P(X = x)
0.3
0.4
0.3
The variance of the distribution is:
