Question

You are trying to determine if you should accept a shipment of eggs for a local grocery store. About 4% of all cartons which are shipped have had an egg crack while traveling. You are instructed to accept the shipment if no more than 10 cartons out of the 300 you inspect have cracked eggs. What is the probability that you accept the shipment? (In other words, what is the probability that, at the most, you had 16 cartons with cracked eggs?)
a. 1012.
b. 3669.
c. 5319.
d. 4681.
e. 5832.
f. 6331.

Answer 1

Answer:

0.8809

Step-by-step explanation:

Given that:

The population proportion p = 4% = 4/100 = 0.04

Sample mean x = 16

The sample size n = 300

The sample proportion $\hat p =\dfrac{x}{n}$

= 16/300

= 0.0533

∴

$P(\hat p \leq 0.0533) = P\bigg ( \dfrac{\hat p - p}{\sqrt{\dfrac{P(1-P)}{n}}}\leq\dfrac{0.0533 - 0.04}{\sqrt{\dfrac{0.04(1-0.04)}{300}}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{\dfrac{0.04(0.96)}{300}}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{\dfrac{0.0384}{300}}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{\sqrt{1.28 \times 10^{-4}}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq\dfrac{0.0133}{0.0113}}\bigg )$

$P(\hat p \leq 0.0533) = P\bigg ( Z\leq1.18}\bigg )$

From the z tables;

= 0.8809

OR

Let X be the random variation that follows a normal distribution;

Then;

population mean $\mu$ = n × p

population mean $\mu$ = 300 × 0.04

population mean $\mu$ = 12

The standard deviation $\sigma = \sqrt{np(1-p)}$

The standard deviation  $\sigma = \sqrt{300 \times 0.04(1-0.04)}$

The standard deviation $\sigma = \sqrt{11.52}$

The standard deviation  $\sigma = 3.39$

The z -score can be computed as:

$z = \dfrac{x - \mu}{\sigma}$

$z = \dfrac{16 -12}{3.39}$

$z = \dfrac{4}{3.39}$

z = 1.18

The required probability is:

P(X ≤ 10) = Pr (z  ≤ 1.18)

= 0.8809