If kurtosis is defined as \[ \left(\frac{m_4}{\sigma^4}-3\right), \]
where $m_4$ is the $4^{th}$ order central moment and $\sigma$ is the standard deviation, then its value is positive for

Question

There are 3 chess players in the 1st year, 4 chess players in the 2nd year, and 5 chess players in the 3rd year of college. What is the probability of selecting a team of 4 chess players if at least one player is from the 1st year?

Answer 1

Step 1: Read what the formula means.
The value here is the excess kurtosis, written as $\dfrac{m_4}{\sigma^4}-3$. The part $\dfrac{m_4}{\sigma^4}$ is the plain kurtosis, and we take away $3$ to compare the curve with a normal one.

Step 2: Fix the normal curve as the bench mark.
For a normal distribution the plain kurtosis is exactly $3$, so its excess kurtosis is
\[ 3-3=0 \]
This is the reason we subtract $3$. The normal curve sits at zero.

Step 3: Decide when the value is positive.
A positive value needs $\dfrac{m_4}{\sigma^4}>3$, which means a sharper peak and fatter tails than the normal curve. Such a curve is called leptokurtic.

Step 4: Check the other names.
A mesokurtic curve and the normal curve both give zero. A platykurtic curve is flatter, so it gives a negative value. None of these are positive.

Step 5: Pick the answer.
Only the sharp peaked case is positive.
\[ \boxed{\text{Leptokurtic distribution}} \]

If kurtosis is defined as \[ \left(\frac{m_4}{\sigma^4}-3\right), \]
where \(m_4\) is the \(4^{th}\) order central moment and \(\sigma\) is the standard deviation, then its value is positive for

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The Correct Option is A

Solution and Explanation

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If kurtosis is defined as \[ \left(\frac{m_4}{\sigma^4}-3\right), \] where \(m_4\) is the \(4^{th}\) order central moment and \(\sigma\) is the standard deviation, then its value is positive for