Question

A packaging system fills boxes to an average weight of 17 ounces with a standard deviation of 0.5 ounce. It is reasonable to assume that the weights are normally distributed. Calculate the 1st, 2nd, and 3rd quartiles of the box weight. (You may find it useful to reference the z table. Round "z" value to 3 decimal places and final answers to 2 decimal places.)

1st Quartile=

2nd Quartile=

3rd Quartile=

Answer 1

Solution:-

Given that,

mean = = 17

standard deviation = = 0.5

Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.675 ) = 0.25

z = -0.675

Using z-score formula,

x = z * +

x = -0.675 * 0.5 + 17

x = 16.66

First quartile =Q1 = 16.66

The z dist'n Second quartile is,

P(Z < z) = 50%

= P(Z < z) = 0.50  

= P(Z < 0) = 0.50  

z = 0

Using z-score formula,

x = z * +

x = 0 * 0.5 + 17

x = 17.00

Second quartile =Q1 = 17.00

The z dist'n Third quartile is,

P(Z < z) = 75%

= P(Z < z) = 0.75  

= P(Z < 0.675 ) = 0.75

z = 0.675

Using z-score formula,

x = z * +

x = 0.675 * 0.5 + 17

x = 17.34

Third quartile =Q3 = 17.34