Chapter 12: Q. 8 (page 800)

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?
$a) P (z < \frac{0.3 - 0.4}{\sqrt{\frac{(0.4) (0.6)}{60}}})$
$b) P (z < \frac{0.4 - 0.3}{\sqrt{\frac{(0.3) (0.7)}{60}}})$
role="math" localid="1650519113387" $c) P (z < \frac{0.3 - 0.4}{\frac{\sqrt{(0.4) (0.6)}}{60}})$
role="math" localid="1650519757907" $d) P (z < \frac{0.3 - 0.4}{\frac{(0.4) (0.6)}{\sqrt{60}}})$
$e) P (z < \frac{0.4 - 0.3}{\frac{\sqrt{(0.3) (0.7})}{60}})$

Expert verified

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

Step by step solution

$40 %$ 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.

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:

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

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

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

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x_i:	0	1	2	3	4
p_i :	0.1	0.2	0.35	0.25	0.15