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?

$a)P\left(z<\frac{0.3-0.4}{\sqrt{{\displaystyle \frac{(0.4)(0.6)}{60}}}}\right)$

$b)P\left(z<\frac{0.4-0.3}{\sqrt{{\displaystyle \frac{(0.3)(0.7)}{60}}}}\right)$

role="math" localid="1650519113387" $c)P\left(z<\frac{0.3-0.4}{{\displaystyle \frac{\sqrt{(0.4)(0.6)}}{60}}}\right)$

role="math" localid="1650519757907" $d)P\left(z<\frac{0.3-0.4}{{\displaystyle \frac{(0.4)(0.6)}{\sqrt{60}}}}\right)$

$e)P\left(z<\frac{0.4-0.3}{{\displaystyle \frac{\sqrt{(0.3)(0.7})}{60}}}\right)$

Short answer

The answer is option$a)P\left(z<\frac{0.3-0.4}{\sqrt{{\displaystyle \frac{(0.4)(0.6)}{60}}}}\right)$

Step-by-step solution

Given Information

$40\%$of the small candies are orange.

On depositing money, 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.

Simplification

The problem is about population proportions. The problem states that $40\%$are orange indicating the true population proportion being $0.4$.

The Z score formula for proportions is:

$Z=\frac{\hat{p}-p}{\sqrt{{\displaystyle \frac{p\times (1-p)}{n}}}}$

$18$of the $60$candies is $30$percent. Therefore, the correct boundary when plugging in the number is:

$Z=\frac{0.3-0.4}{\sqrt{{\displaystyle \frac{0.4\times 0.6}{60}}}}$