Question

A packaging system fills boxes to an average weight of 16 ounces with a standard deviation of 0.3 ounce. It is reasonable to assume that the weights are normally distributed. Calculate the 1st, 2nd, and 3rd quartiles of the box weight.

Answer 1

Answer:

$Q_1 = 15.8$

$Q_2 = 16$

$Q_3 = 16.2$

Step-by-step explanation:

Given

$\bar x = 16$ ---- the average

$\sigma = 0.3$ -- standard deviation

Required

The 1st to 3rd quartile

Since the distribution is normal, then:

$Median = Mean$

i.e.

$Median = \bar x$

So, we have:

$Median = 16$

Rewrite Median as Q2

$Q_2 = 16$

To solve for Q1 and Q3, we use the following formula

$\frac{Q_3 - Q_1}{2} =\frac{2}{3} * \sigma$

Multiply both sides by 2

$Q_3 - Q_1 =\frac{4}{3} * \sigma$

Substitute $\sigma = 0.3$

$Q_3 - Q_1 =\frac{4}{3} * 0.3$

$Q_3 - Q_1 =0.4$

Also, we have:

$Q_3 - Q_2 = Q_2 - Q_1$ ----- quadrants are equidistant

Rewrite as:

$Q_3 + Q_1 = Q_2 + Q_2$

$Q_3 + Q_1 = 2Q_2$

Substitute: $Q_2 = 16$

$Q_3 + Q_1 = 2*16$

$Q_3 + Q_1 = 32$

Make Q3 the subject

$Q_3 = 32 - Q_1$

Substitute $Q_3 = 32 - Q_1$ in $Q_3 - Q_1 =0.4$

$32 - Q_1 - Q_1 = 0.4$

Collect like terms

$Q_1 + Q_1 = 32-0.4$

$2Q_1 = 31.6$

Divide by 2

$Q_1 = 15.8$

Substitute $Q_1 = 15.8$ in $Q_3 = 32 - Q_1$

$Q_3 = 32 - 15.8$

$Q_3 = 16.2$