Chapter 8.2, Problem 59E

Solution

a.

To find: The probability that the circumference of the randomly selected orange is more than 22 cm.

The probability is 0.3085.

Given:

  $\begin{array}{l}\text{Population mean}\left(\mu \right)=21.1\\ \text{Standard deviation}\left(\sigma \right)=1.8\\ \text{Sample size}\left(n\right)=20\end{array}$

Calculation:

Consider, X be the random variable that represents the circumference of orange follows the normal distribution with mean = 21.1 cm and standard deviation = 1.8 cm.

The probability that the circumference of the randomly selected orange is more than 22 cm can be calculated as:

  $\begin{array}{c}P\left(X>22\right)=P\left(\frac{x-\mu }{\sigma }>\frac{22-21.1}{1.8}\right)\\ =P\left(Z>0.5\right)\\ =1-P\left(Z<0.5\right)\\ =0.3085\end{array}$

Thus, the required probability is 0.3085.

b.

To find: The probability that the mean circumference of the randomly selected 20 oranges is more than 22 cm.

The probability is 0.012676.

The probability that the mean circumference of the randomly selected 20 oranges is more than 22 cm can be calculated as:

  $\begin{array}{c}P\left(\overline{X}>22\right)=P\left(\frac{\overline{x}-\mu }{\frac{\sigma }{\sqrt{n}}}>\frac{22-21.1}{\frac{1.8}{\sqrt{20}}}\right)\\ =P\left(Z>2.236\right)\\ =1-P\left(Z<2.236\right)\\ =0.012676\end{array}$

Thus, the required probability is 0.012676.