Chapter 11, Problem 11PT

(a)

To determine

To find: The number of body builders between 180 to 190 pounds.

Answer to Problem 11PT

Thenumber of bodybuilders are 645.

Explanation of Solution

Given:

The weight of the 1500 bodybuilders is normally distributed with the mean of 1906 pounds and the standard deviation of 5.8 pounds.

Calculation:

Consider the random variable that represents the weight of the body builder is X .

Then, the range of the events is,

  $180\le X\le 190$

Then,

  $\begin{array}{l}\frac{180-190.6}{5.8}\le \frac{X-190.6}{5.8}\le \frac{190-190.6}{5.8}\\ -1.83\le Z\le -0.10\end{array}$

Here Z is the standard normal distributions.

Then, consider the probability as,

  $P\left(180\le X\le 190\right)=P\left(-1.83\le Z\le -0.10\right)$

On the calculator Ti83.

Press 2^nd VARS and select the draw as shown in Figure 1

  

Figure 1

Press 1 to select the shade Norm and enter the lower and upper bound to determine the probability as shown I Figure 2

  

Figure 2

Then,

  $\begin{array}{c}P\left(180\le X\le 190\right)=P\left(-1.83\le Z\le -0.10\right)\\ =0.43\end{array}$

Thus, 43% of the bodybuilder out of 1500 is in the interval 180 pounds to 190 pounds.

Therefore, the number of required body builder is,

  $43\%\left(1500\right)=645$

Thus, $645$ are bodybuilders are between 180 pounds to 190 pounds.

(b)

To determine

To find: The probability that the body that a bodybuilder selected at random has a weight greater than 195 pounds.

Answer to Problem 11PT

Theprobability that the body builders selected at random has the weigh greater than 195 pounds is 0.23.

Explanation of Solution

Consider the event is given by,

  $\begin{array}{l}X>195\\ \frac{X-190.6}{5.8}>\frac{195-190.6}{5.8}\\ Z>0.75\end{array}$

The graph for the probability is shown in Figure 3

  

Figure 3

Then,

  $\begin{array}{c}P\left(X>195\right)=P\left(Z>0.75\right)\\ =0.23\end{array}$

Therefore, the probability that the body builders selected at random has the weigh greater than 195 pounds is 0.23.