Chapter 5.2, Problem 68E

(a)

To determine

To find: The mean and standard deviation of the random variable X.

Answer to Problem 68E

Solution: The mean and standard deviation of the binomial random variable X is 900 and 15 respectively.

Explanation of Solution

Calculation: In binomial distribution, the mean can be calculated by this formula,

${\mu }_x=np$

In binomial distribution, standard deviation can be calculated by this formula,

${\sigma }_x=\sqrt{npq}$

Where, n is the number of trials and p is the probability of success in a trial. For $x=0,1,2,\mathrm{3...1200}$, the mean and standard deviation can be calculated by substituting values of n and p in the above formulas. The mean can be calculated as:

$\begin{array}{c}{\mu }_X=np\\ =1200\times 0.75\\ =900\end{array}$

The standard deviation is calculated as:

$\begin{array}{c}{\sigma }_X=\sqrt{np\left(1-p\right)}\\ =\sqrt{1200\left(0.75\right)\left(1-0.75\right)}\\ =15\end{array}$

Hence, the average value and standard deviation are 900 and 15 respectively.

(b)

To determine

To find: The probability.

Answer to Problem 68E

Solution: The probability is $P\left(X\ge 800\right)=1$.

Explanation of Solution

Calculation: The binomial random variable approximately follows the normal distribution with mean $\mu =np$ and standard deviation $\sigma =\sqrt{np\left(1-p\right)}$ if the condition $np\left(1-p\right)\ge 10$ holds true, where n is the number of trials and p is the probability of success in a trial.

$\begin{array}{c}np\left(1-p\right)=1200\times 0.75\times \left(1-0.75\right)\\ =225\\ >10\end{array}$

Therefore, the random variable X can be approximated to normal distribution because the given condition is fulfilled.

The parameters of the normal distribution are calculated as follows:

$\begin{array}{c}{\mu }_X=np\\ =1200\times 0.75\\ =900\end{array}$

And:

$\begin{array}{c}{\sigma }_X=\sqrt{np\left(1-p\right)}\\ =\sqrt{1200\left(0.75\right)\left(1-0.75\right)}\\ =15\end{array}$

The probability that the value taken by X is greater than or equal to 800, the area right to the point 800 under the normal curve. To obtain the area lying to the left of a point under the normal curve, first, convert the value of the variable into Z-score and then obtain the area lying to the right by subtracting the area left to the Z-score from 1. The Z-score for the value $X=800$ is calculated as:

$\begin{array}{c}Z=\frac{X-\mu }{\sigma }\\ =\frac{800-900}{15}\\ =-6.67\end{array}$

The area left to the particular Z-score can be obtained using Excel by using the command $\text{'}=\text{NORMSDIST}\left(\right)\text{'}$. Specify the value of z as $-6.67$.

The area left to the Z-score,$-6.67$, is obtained as 0.0000 from Excel.

So, the required probability can be calculated as:

$1-0.0000=1$

Hence, the probability is 1.

(c)

To determine

To find: The probability.

Answer to Problem 68E

Solution: $P\left(X>1000\right)=0.3676$.

Explanation of Solution

Calculation: For the probability that more than 950 candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command $\text{'}=\text{BINOMDIST}\left(\right)\text{'}$ in one of the cells and specify the value of $\text{number}=950,\text{trials = 1200, probability = 0}\text{.75,}\text{and cumulative as 'TRUE'}$. Press enter to obtain the probability that the value of the random variable X is less than or equal to 950.

$P\left(X\le 950\right)=0.9997$.

Therefore, the required probability is calculated as:

$\begin{array}{c}P\left(X>950\right)=1-P\left(X\le 950\right)\\ =1-0.9997\\ =0.0003\end{array}$

Hence, the probability is 0.0003.

(d)

To determine

To find: The probability.

Answer to Problem 68E

Solution: $P\left(X>950\right)=0.9408$.

Explanation of Solution

Calculation: The probability that more than 950candidates will accept the admission, Excel has to be used. The following procedure is followed in Excel:

Step 1: Open the Excel spreadsheet.

Step 2: Type the command $\text{'}=\text{BINOMDIST}\left(\right)\text{'}$ in one of the cell and specify the value of $\text{number}=950,\text{trials = 1300, probability = 0}\text{.75,}\text{and cumulative as 'TRUE'}$. Press enter to obtain the probability that the value of the random variable X is less than or equal to 950.

$P\left(X\le 950\right)=0.0592$

Therefore, the required probability is calculated as:

$\begin{array}{c}P\left(X>950\right)=1-P\left(X\le 950\right)\\ =1-0.0592\\ =0.9408\end{array}$

Hence, the probability is 0.9408.