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.”

Question

Find out the fallacy if any in the statement:&ldquo;The mean and the variance of a binomial distribution is 16.2 and 29.4 respectively.&rdquo;

Prepp Team · Accepted Answer

Understanding the Binomial Distribution and its Parameters

The binomial distribution is a discrete probability distribution that describes the number of successes in a fixed number of independent trials, each having only two possible outcomes (success or failure).

A binomial distribution is characterized by two parameters:

$n$: The number of trials.
$p$: The probability of success in a single trial.

From these, the probability of failure is given by $q = 1 - p$.

For a valid binomial distribution, the parameters $p$ and $q$ must satisfy the condition $0 \le p \le 1$ and $0 \le q \le 1$. Consequently, $p+q=1$.

Mean and Variance Formulas for Binomial Distribution

The mean ($\mu$) and variance ($\sigma^2$) of a binomial distribution are given by the following formulas:

Mean: $\mu = np$
Variance: $\sigma^2 = npq$

Analyzing the Given Statement

The statement claims that a binomial distribution has:

Mean, $\mu = 16.2$
Variance, $\sigma^2 = 29.4$

Calculating Parameters from Mean and Variance

We can use the formulas for the mean and variance to find the parameters $p$, $q$, and $n$.

We have the equations:

$np = 16.2$
$npq = 29.4$

Substitute the value of $np$ from equation (1) into equation (2):

$16.2 \times q = 29.4$

Now, solve for $q$:

$q = \dfrac{29.4}{16.2}$

To simplify the fraction, we can multiply the numerator and denominator by 10 to remove decimals:

$q = \dfrac{294}{162}$

We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor. Both numbers are divisible by 6:

$294 \div 6 = 49$
$162 \div 6 = 27$

So, we get:

$q = \dfrac{49}{27}$

Identifying the Fallacy

Now let's examine the calculated value of $q$. For $q$ to be a valid probability, it must be between 0 and 1, inclusive ($0 \le q \le 1$).

In our case, $q = \dfrac{49}{27}$.

$\dfrac{49}{27} \approx 1.81$

Since $q = \dfrac{49}{27} > 1$, this value is not possible for the probability of failure in any probability distribution, including the binomial distribution.

This inconsistency means that a binomial distribution with a mean of 16.2 and a variance of 29.4 cannot exist. The statement contains a fallacy.

Let's also find $p$ for completeness, although the fallacy is already clear from $q$.

$p = 1 - q$

$p = 1 - \dfrac{49}{27}$

$p = \dfrac{27}{27} - \dfrac{49}{27}$

$p = \dfrac{27 - 49}{27}$

$p = -\dfrac{22}{27}$

Since $p = -\dfrac{22}{27} < 0$, this value is also not possible for the probability of success.

Matching the Fallacy with Options

We found that $q = \dfrac{49}{27}$, which is not possible for a probability. Let's look at the options:

Option 1: $q=\dfrac{49}{27}$ is not possible. This matches our finding.
Option 2: The statement is true. This is incorrect as we found a contradiction.
Option 3: $p=\dfrac{47}{27}$ is not possible. Our calculated $p$ is $-\dfrac{22}{27}$. While $\dfrac{47}{27}$ is also not possible as $\dfrac{47}{27} > 1$, this is not the parameter value directly derived from the given mean and variance.
Option 4: $p=\dfrac{49}{27}$ is not possible. Our calculated $p$ is $-\dfrac{22}{27}$. Similar to Option 3, while $\dfrac{49}{27}$ is not possible as $\dfrac{49}{27} > 1$, this is not the parameter value directly derived from the given mean and variance. The direct derived value of q is 49/27.

The most direct fallacy identified from the calculation using the given mean and variance is that $q = \dfrac{49}{27}$, which is an impossible value for a probability.

Parameter	Formula	Calculated Value	Validity	Reason
Mean	$\mu = np$	$16.2$	Given	-
Variance	$\sigma^2 = npq$	$29.4$	Given	-
Probability of Failure	$q = \dfrac{\sigma^2}{\mu}$	$\dfrac{29.4}{16.2} = \dfrac{49}{27}$	Invalid	$\dfrac{49}{27} > 1$
Probability of Success	$p = 1 - q$	$1 - \dfrac{49}{27} = -\dfrac{22}{27}$	Invalid	$-\dfrac{22}{27} < 0$

Conclusion on Binomial Distribution Statement Fallacy

The calculation clearly shows that the probability of failure, $q$, must be $\dfrac{49}{27}$ if the mean is 16.2 and the variance is 29.4. Since probability cannot be greater than 1, $q=\dfrac{49}{27}$ is not possible. This directly points to the fallacy in the original statement about the binomial distribution.

Revision Table: Binomial Distribution Properties

Property	Formula	Requirement
Probability of Success	$p$	$0 \le p \le 1$
Probability of Failure	$q$	$0 \le q \le 1$
Relationship	$p + q = 1$	Must hold
Mean	$\mu = np$	-
Variance	$\sigma^2 = npq$	-
Relationship between Mean & Variance	$\sigma^2 = \mu q$ or $q = \dfrac{\sigma^2}{\mu}$	Implies $\sigma^2 \le \mu$ (since $q \le 1$)

Additional Information: Variance vs Mean in Binomial Distribution

An important property of the binomial distribution is that its variance is always less than or equal to its mean. This comes directly from the formulas:

$\sigma^2 = npq$

Since $p = np$, we have $\sigma^2 = \mu q$.

As $0 \le q \le 1$, it follows that $\sigma^2 \le \mu \times 1$, so $\sigma^2 \le \mu$.

In the given statement, the mean is 16.2 and the variance is 29.4. Here, $\sigma^2 = 29.4 > \mu = 16.2$.

This inequality $\sigma^2 > \mu$ is a strong indicator that the claimed parameters do not correspond to a valid binomial distribution. This reinforces the conclusion drawn from calculating the value of $q$. The variance of a binomial distribution cannot be greater than its mean.

Prepp Team · Answer

The statement is true.

Prepp Team · Answer

$p=\dfrac{47}{27}$ is not possible.

Prepp Team · Answer

$p=\dfrac{49}{27}$ is not possible.