A die is thrown 720 times.

(a) Find the probability that3comes up exactly 125 times.

(b) Find the probability that 3 comes up between115and 130 times.

A die is thrown 720 times.

Frequency distribution of the number of successful outcomes that can be achieved in a given number of trials, each with an equal chance of success.

(a)

The formula for binomial distribution states that$f\left(x\right)=C(x,n)p^x{q}^{n-x}$.

$\begin{array}{c}n=720\\ p=\frac{1}{6}\\ q=\frac{5}{6}\\ x=125\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{l}{f}_B\left(120\right)=C(710,125){\left(\frac{1}{6}\right)}^{120}{\left(\frac{5}{6}\right)}^{720-120}\\ f_B\left(120\right)=0.0345\end{array}$

The formula for corresponding normal approximation states that .$f\left(x\right)=\frac{e^{-{(x-np)}^2/2npq}}{\sqrt{2\pi npq}}$$f\left(x\right)=\frac{e^{-{(x-np)}^2/2npq}}{\sqrt{2\pi npq}}$

$\begin{array}{c}n=720\\ p=\frac{1}{6}\\ q=\frac{5}{6}\\ x=125\end{array}$

Substitute the values mentioned above in the formula.

$\begin{array}{c}{f}_n\left(125\right)=\frac{1}{\sqrt{2\pi \times {10}^4\times 0.25}}\\ f_n\left(125\right)=0.035\end{array}$

(b)

The probability by cumulative function is given below.

$P_B(115\le x\le 130)=0.533$

Theprobability by normal distribution function is given below.

$P_n(115\le x\le 130)=0.56$

The value required for part a is given below.

$\begin{array}{c}{f}_B\left(120\right)=0.0345\\ f_n\left(125\right)=0.035\end{array}$

The value required for part b is given below.

$\begin{array}{c}{P}_B(115\le x\le 130)=0.533\\ P_n(115\le x\le 130)=0.56\end{array}$