Chapter 7.8, Problem 97E

a.

To determine

Approximate the probability $P\left(x=30\right)$ using the normal approximation with the continuity correction.

Answer to Problem 97E

The approximate probability $P\left(x=30\right)$ using the normal approximation with the continuity correction is 0.1114.

Explanation of Solution

Calculation:

It is given that the random variable x follows binomial distribution with $n=50\text{ and }p=0.6$ such that $np\left(\mu \right)=30\text{ and }\sigma =\sqrt{np\left(1-p\right)}=3.464$1.

Conditions for binomial to follow normal distribution (approximately):

• $np\ge 10$
• $n\left(1-p\right)\ge 10$

Substitute $n=50\text{ and }p=0.6$

$\begin{array}{c}np=\left(50\right)\left(0.6\right)\\ =30\\ >10\end{array}$

$\begin{array}{c}n\left(1-p\right)=\left(50\right)\left(1-0.6\right)\\ =\left(50\right)\left(0.4\right)\\ =20\\ >10\end{array}$

Thus, both the conditions are satisfied. Hence, the distribution of x is approximately normal.

The desired probability is approximately the area under the normal curve between 29.5 and 30.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(x=30\right)\approx P\left(29.5\le x\le 30.5\right)\\ =P\left(\frac{29.5-30}{3.4641}\le z\le \frac{30.5-30}{3.4641}\right)\\ =P\left(\frac{-0.5}{3.4641}\le z\le \frac{0.5}{3.4641}\right)\\ =P\left(-0.14\le z\le 0.14\right)\\ =P\left(z\le 0.14\right)-P\left(z\le -0.14\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -0.14\right)$.

Procedure:

• In the $z*$ row locate −0.1.
• In column locate .04.
• The intersection of the row −0.1 and column .04 gives 0.4443.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 0.14\right)$.

Procedure:

• In the $z*$ row locate 0.1.
• In column locate .04.
• The intersection of the row 0.1 and column .04 gives 0.5557.

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 0.14\right)-P\left(z\le -0.14\right)=0.5557-0.4443\\ =\text{0}\text{.1114}\end{array}$

Thus, the approximate probability $P\left(x=30\right)$ using the normal approximation with the continuity correction is 0.1114.

b.

To determine

Approximate the probability $P\left(x=25\right)$ using the normal approximation with the continuity correction.

Answer to Problem 97E

The approximate probability $P\left(x=25\right)$ using the normal approximation with the continuity correction is 0.0409.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve between 24.5 and 25.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(x=25\right)\approx P\left(24.5\le x\le 25.5\right)\\ =P\left(\frac{24.5-30}{3.4641}\le z\le \frac{25.5-30}{3.4641}\right)\\ =P\left(\frac{-5.5}{3.4641}\le z\le \frac{-4.5}{3.4641}\right)\\ =P\left(-1.59\le z\le -1.30\right)\\ =P\left(z\le -1.30\right)-P\left(z\le -1.59\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.59\right)$.

Procedure:

• In the $z*$ row locate −1.5
• In column locate .09
• The intersection of the row −1.5 and column .09 gives 0.0559

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.30\right)$.

Procedure:

• In the $z*$ row locate −1.3.
• In column locate .00.
• The intersection of the row −1.3 and column .00 gives 0.0968.

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le -1.30\right)-P\left(z\le -1.59\right)=0.0968-0.0559\\ =\text{0}\text{.0409}\end{array}$

Thus, the approximate probability $P\left(x=25\right)$ using the normal approximation with the continuity correction is 0.0409.

c.

To determine

Approximate the probability $P\left(x\le 25\right)$ using the normal approximation with the continuity correction.

Answer to Problem 97E

The approximate probability $P\left(x\le 25\right)$ using the normal approximation with the continuity correction is 0.0968.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve less than or equal to 25.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(x\le 25\right)\approx P\left(x\le 25.5\right)\\ =P\left(z\le \frac{25.5-30}{3.4641}\right)\\ =P\left(z\le \frac{-4.5}{3.4641}\right)\\ =P\left(z\le -1.30\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.30\right)$.

Procedure:

• In the $z*$ row locate −1.3.
• In column locate .00.
• The intersection of the row −1.3 and column .00 gives 0.0968.

The approximated probability is obtained as given below:

$P\left(z\le -1.30\right)=0.0968$

Thus, the approximate probability $P\left(x\le 25\right)$ using the normal approximation with the continuity correction is 0.0968.

d.

To determine

Approximate the probability $P\left(25\le x\le 40\right)$ using the normal approximation with the continuity correction.

Answer to Problem 97E

The approximate probability $P\left(25\le x\le 40\right)$ using the normal approximation with the continuity correction is 0.9429.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve between 24.5 and 40.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(25\le x\le 40\right)\approx P\left(24.5\le x\le 40.5\right)\\ =P\left(\frac{24.5-30}{3.4641}\le z\le \frac{40.5-30}{3.4641}\right)\\ =P\left(\frac{-5.5}{3.4641}\le z\le \frac{10.5}{3.4641}\right)\\ =P\left(-1.59\le z\le 3.03\right)\\ =P\left(z\le 3.03\right)-P\left(z\le -1.59\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.59\right)$.

Procedure:

• In the $z*$ row locate −1.5.
• In column locate .09.
• The intersection of the row −1.5 and column .09 gives 0.0559.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 3.03\right)$.

Procedure:

• In the $z*$ row locate 3.0.
• In column locate .03.
• The intersection of the row 3.0 and column .03 gives 0.9988.

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 3.03\right)-P\left(z\le -1.59\right)=0.9988-0.0559\\ =\text{0}\text{.9429}\end{array}$

Thus, the approximate probability $P\left(25\le x\le 40\right)$ using the normal approximation with the continuity correction is 0.9429.

e.

To determine

Approximate the probability $P\left(25<x<40\right)$ using the normal approximation with the continuity correction.

Answer to Problem 97E

The approximate probability $P\left(25<x<40\right)$ using the normal approximation with the continuity correction is 0.9001.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve between 25.5 and 39.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(25<x<40\right)\approx P\left(25.5\le x\le 39.5\right)\\ =P\left(\frac{25.5-30}{3.4641}\le z\le \frac{39.5-30}{3.4641}\right)\\ =P\left(\frac{-4.5}{3.4641}\le z\le \frac{9.5}{3.4641}\right)\\ =P\left(-1.30\le z\le 2.74\right)\\ =P\left(z\le 2.74\right)-P\left(z\le -1.30\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.30\right)$.

Procedure:

• In the $z*$ row locate −1.3.
• In column locate .00.
• The intersection of the row −1.3 and column .00 gives 0.0968.

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 2.74\right)$.

Procedure:

• In the $z*$ row locate 2.7.
• In column locate .04.
• The intersection of the row 2.74 and column .04 gives 0.9969.

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 2.74\right)-P\left(z\le -1.30\right)=0.9969-0.0968\\ =\text{0}\text{.9001}\end{array}$

Thus, the approximate probability $P\left(25<x<40\right)$ using the normal approximation with the continuity correction is 0.9001.