The example in use is calculating the distribution of a random variable \(X\) which is equal to

**k**successful shots out of

**n**trials in basket ball. The probability

*p*of making a shot is 0.5 (cell B3) and hence missing the shot

*(1-p)*is 0.5 (cell B4). Hence this is a case of a binomial distribution.

Assuming that 30 attempts are made (cell B8), the distribution of \(X\) of making

**k**successful shots (

**k**ranges from 0 to 30 populated in cells B15 through B45) is given in cells D15 through D45 using the Binomial distrubution formula

\(^nC_{k}.p^k.(1-p)^{n-k}\)

Using the formula for the expected value we know that the expected value for \(X\) is \(E(X) = \mu_{x} = n.p\), which is \(30 * 0.5 = 15\). Also, the variance \(\sigma_{X}^2 = n.p.(1-p)\) is \(30*0.5*0.5 = 7.5\) as shown in cell B10. Hence the standard deviation for \(X, \sigma_{X}\) is 2.739 as shown in cell B11.

Using the values of \(\mu_{x}\) and \(\sigma_{x}\) the values for the continuous probability distribution function can be obtained using the formula

\(p(x)=\frac{1}{\sqrt{2\pi\sigma_{x}^2}}exp\left(-\frac{1}{2}\left(\frac{x-\mu_{x}}{\sigma_{x}}\right)^2\right)\)

which can be reduced to \(p(x) = \frac{1}{\sqrt{2\pi\sigma_{x}^2exp\left(\frac{\left(x-\mu_{x}\right)}{\sigma_{x}}\right)^2}}\).

The reduced formula is used to caluclate probabilites at different values of

**k**in cells E15 through E45. To cross check the mean, variance and standard deviation i.e. \(E[X] = \mu_{x}, \sigma_{x}^2, \sigma_{x} \) are recalculated and turn out to be, as expected, equal to 15, 7.5 and 2.73 as shown in cells D9, D10 and D11.

Now, the random variable \(X\) is standardized by performing the following operation on it.

\(

Z = \frac{X-\mu_{x}}{\sigma_{x}}

\)

The standardization is done for each value of

**k**and the standardized values can be seen in cells G15 through G45. The standardized variable can be referred to \(Z\) and the values are populated in cells G15 through G45. Since standardizing a normally distributed variable leads to another distributed variable with \(\mu_{z}=0\) and \(\sigma_{z}=1\) , the probability density function further reduces to

\(p(z) = \frac{1}{\sqrt{2\pi.exp(z^2)}}\)

The \(p(z)\) values for values of Z are populated in cells H15 to H35.

Just to cross check again I calculated the \(E[Z]=\mu_{Z}, \sigma_{Z}^2, \sigma_{Z}\) values in cells F9, F10 and F11 which should be 0, 1 and 1 respectively but they are turning out to be something else. I have tried to repeatedly recheck what I have plugged into the cells but all seems fine and yet the cross checks do not work and I am very curious to know where I am going wrong. Many thanks in advance.