For Binomial distribution, n = 10 and p = 0.6, E(X 2) (second moment about origin) is:
38.4
Calculating the Second Moment about Origin for Binomial Distribution
The question asks for the second moment about the origin, denoted as $E(X^2)$, for a Binomial distribution with parameters n = 10 and p = 0.6.
For a Binomial distribution, the parameters are:
- Number of trials, \(n = 10\)
- Probability of success, \(p = 0.6\)
We need to find the second moment about the origin, \(E(X^2)\).
For a Binomial distribution, we know the formulas for the mean \(E(X)\) and the variance \(Var(X)\):
- Mean, \(E(X) = np\)
- Variance, \(Var(X) = np(1-p)\) or \(Var(X) = npq\), where \(q = 1-p\) is the probability of failure.
Let's first calculate the mean \(E(X)\):
\[E(X) = np = 10 \times 0.6 = 6\]
Next, let's calculate the variance \(Var(X)\). First, find \(q\):
\[q = 1 - p = 1 - 0.6 = 0.4\]
Now, calculate the variance:
\[Var(X) = npq = 10 \times 0.6 \times 0.4 = 6 \times 0.4 = 2.4\]
The variance \(Var(X)\) is related to the moments about the origin by the formula:
\[Var(X) = E(X^2) - [E(X)]^2\]
We can rearrange this formula to solve for \(E(X^2)\):
\[E(X^2) = Var(X) + [E(X)]^2\]
Now, substitute the calculated values for \(Var(X)\) and \(E(X)\):
\[E(X^2) = 2.4 + (6)^2\]
\[E(X^2) = 2.4 + 36\]
\[E(X^2) = 38.4\]
Therefore, the second moment about the origin for the given Binomial distribution is 38.4.
Revision Table: Binomial Distribution Properties
| Property | Formula | Value (for n=10, p=0.6) |
|---|---|---|
| Number of trials | \(n\) | 10 |
| Probability of success | \(p\) | 0.6 |
| Probability of failure | \(q = 1-p\) | 0.4 |
| Mean | \(E(X) = np\) | \(10 \times 0.6 = 6\) |
| Variance | \(Var(X) = npq\) | \(10 \times 0.6 \times 0.4 = 2.4\) |
| Relation between Variance and Moments | \(Var(X) = E(X^2) - [E(X)]^2\) | \(2.4 = E(X^2) - 6^2\) |
| Second Moment about Origin | \(E(X^2)\) | \(2.4 + 36 = 38.4\) |
Additional Information: Understanding Moments
Moments are important descriptive measures of a probability distribution. They provide information about the shape and characteristics of the distribution.
- Moments about the origin: These are the expected values of powers of the random variable. The \(k^{th}\) moment about the origin is \(E(X^k)\).
- The first moment about the origin \(E(X^1) = E(X)\) is the mean.
- The second moment about the origin \(E(X^2)\) is used in calculating the variance.
- Central moments: These are the expected values of powers of the deviation of the random variable from its mean. The \(k^{th}\) central moment is \(E[(X - E(X))^k]\).
- The first central moment is always 0.
- The second central moment is the variance, \(E[(X - E(X))^2] = Var(X)\).
- Higher central moments are used to describe skewness (third central moment) and kurtosis (fourth central moment).
Understanding the relationship between moments about the origin and central moments is key to analyzing probability distributions.
Important Questions from Binomial Distribution
Indicate the correct answer for the combination from the following regarding the conditions for the applicability of a binominal distribution:
(a) There are n independent trials
(b) Each trial has only two possible outcomes
(c) The probabilities of two outcomes do not remain constant
(d) The trials are independent
Which of the following options is correct?
In which of the following practical situations, Poisson Distribution can be used?
A. Number of customers arriving at the super markets per hour.
B. Number of typographical errors per page in a typed material.
C. Number of accidents taking place per day on a busy road.
D. Dice throwing problems.
E. Number of defective material say, blades, etc. in a packing manufactured by a good concern.
Choose the most appropriate answer from the options given below:
The mean and variance of binomial distribution B (x, n, p) are 4 and \(\dfrac{4}{3}\) respectively. What is the probability of getting 2 successes?
Find out the fallacy if any in the statement:
“The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.”
If log e(x) is normally distributed with mean 1 and variance 4, then P(0.5 < x < 2) is:
(where the area between z = 0 and z = 0.25 is 0.0987 and the area between z = 0 and z = 0.5 is 0.1915)
