Question

A large machine is filled with thousands of small pieces of candy, $40\%$of which are orange. When money is deposited, the machine dispenses$60$randomly selected pieces of candy. The machine will be recalibrated if a group of $60$candies contains fewer than$18$that are orange. What is the approximate probability that this will happen if the machine is working correctly?

a. $P(z<0.3-0.4(0.4\left)\right(0.6\left)60\right)P\left(z\&1\mathrm{t};\frac{0.3-0.4}{\sqrt{\frac{(0.4)(0.6)}{60}}}\right)$

b.$P(z<0.3-0.4(0.3\left)\right(0.7\left)60\right)P\left(z\&1\mathrm{t};\frac{0.3-0.4}{\sqrt{\frac{(0.3)(0.7)}{60}}}\right)$

d. $\mathrm{P}(\mathrm{z}<0.3-0.4(0.4\left)\right(0.6\left)60\right)$$P\left(z\&lt;\frac{0.3-0.4}{\frac{(0.4)(0.6)}{\sqrt{60}}}\right)$

c. $\mathrm{P}(\mathrm{z}<0.3-0.4(0.4\left)\right(0.6\left)60\right)$

$P\left(z\&lt;\frac{0.3-0.4}{\frac{\sqrt{(0.4)(0.6)}}{60}}\right)$

e. $\mathrm{P}(\mathrm{z}<0.4-0.3(0.3\left)\right(0.7\left)60\right)P\left(z\&lt;\frac{0.4-0.3}{\frac{\sqrt{(0.3)(0.7)}}{60}}\right)$

Short answer

The approximate probability is

option (a)$P(z<0.3-0.4(0.4\left)\right(0.6\left)60\right)P\left(z\&1\mathrm{t};\frac{0.3-0.4}{\sqrt{\frac{(0.4)(0.6)}{60}}}\right)$

Step-by-step solution

Given information 

Given in the question that, a large machine is filled with thousands of small pieces of candy, $40\%$of which are orange. When money is deposited, the machine dispenses $60$randomly selected pieces of candy. The machine will be recalibrated if a group of $60$ candies contains fewer than 18 that are orange. We need to find the approximate probability that will happen if the machine is working correctly .

Explanation

Thousands of small pieces of candy are stacked in a big machine, with orange candy accounting for $40\%$of the total. The issue here is population proportions. The issue is that $40\%$of the dollars are orange, indicating that the genuine population proportion is $0.4$. For proportions, the z-score formula is:

$z=\frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$=\frac{0.3-0.4}{\sqrt{\frac{0.4(1-0.4)}{60}}}$

The answer is option (a).