Question

Blood diamonds. According to Global Research News (March 4, 2014), one-fourth of all rough diamonds produced in the world are blood diamonds, i.e., diamonds mined to finance war or an insurgency. (See Exercise 3.81, p. 200.) In a random sample of 700 rough diamonds purchased by a diamond buyer, let x be the number that are blood diamonds.

a. Find the mean of x.

b. Find the standard deviation of x.

c. Find the z-score for the value x = 200.

d. Find the approximate probability that the number of the 700 rough diamonds that are blood diamonds is less than or equal to 200.

Short answer

$$\begin{gathered} a.\mu =175 \\ b.\sigma =11.4567 \\ c.z=2.1822 \\ d.\mathrm{The}\mathrm{approximate}\mathrm{probability}\mathrm{is}0.9854 \end{gathered}$$

Step-by-step solution

Given information

According to Global Research News (March 4,2014), one-fourth of all rough diamonds produced in the world are blooddiamonds.

The blood diamond x be a binomial distribution with n = 700 and p = 0.25

Calculation of mean

$$\begin{gathered} a. \\ \mu =np \\ =700\times \frac{1}{4} \\ =700\times 0.25 \\ =175 \\ \mu =175 \end{gathered}$$

Thus, the mean=175.

Calculation of standard deviation

$$\begin{gathered} b. \\ \sigma =\sqrt{np(1-p)} \\ =\sqrt{700\times 0.25\times (1-0.25)} \\ =11.4564 \\ \sigma =11.4564 \end{gathered}$$

Therefore, the variance of x is 11.4564.

Calculation of z-score

$$\begin{gathered} c. \\ \mu =175 \\ \sigma =11.4564 \\ x=200 \end{gathered}$$

The z-score is,

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{200-175}{11.4564} \\ =2.1822 \\ z=2.1822 \end{gathered}$$

Therefore, the z-score is 2.1822.

Finding the approximate probability

$$\begin{gathered} d. \\ P(x\le 200)=P(x\le 200) \\ =P(z\le 2.1822) \\ =0.9853713 \\ \approx 0.9854 \\ P(x\le 200)=0.9854 \end{gathered}$$

Therefore, the approximate probability is 0.9854.