Answer:
[tex]Q_1 = 15.8[/tex]
[tex]Q_2 = 16[/tex]
[tex]Q_3 = 16.2[/tex]
Step-by-step explanation:
Given
[tex]\bar x = 16[/tex] ---- the average
[tex]\sigma = 0.3[/tex] -- standard deviation
Required
The 1st to 3rd quartile
Since the distribution is normal, then:
[tex]Median = Mean[/tex]
i.e.
[tex]Median = \bar x[/tex]
So, we have:
[tex]Median = 16[/tex]
Rewrite Median as Q2
[tex]Q_2 = 16[/tex]
To solve for Q1 and Q3, we use the following formula
[tex]\frac{Q_3 - Q_1}{2} =\frac{2}{3} * \sigma[/tex]
Multiply both sides by 2
[tex]Q_3 - Q_1 =\frac{4}{3} * \sigma[/tex]
Substitute [tex]\sigma = 0.3[/tex]
[tex]Q_3 - Q_1 =\frac{4}{3} * 0.3[/tex]
[tex]Q_3 - Q_1 =0.4[/tex]
Also, we have:
[tex]Q_3 - Q_2 = Q_2 - Q_1[/tex] ----- quadrants are equidistant
Rewrite as:
[tex]Q_3 + Q_1 = Q_2 + Q_2[/tex]
[tex]Q_3 + Q_1 = 2Q_2[/tex]
Substitute: [tex]Q_2 = 16[/tex]
[tex]Q_3 + Q_1 = 2*16[/tex]
[tex]Q_3 + Q_1 = 32[/tex]
Make Q3 the subject
[tex]Q_3 = 32 - Q_1[/tex]
Substitute [tex]Q_3 = 32 - Q_1[/tex] in [tex]Q_3 - Q_1 =0.4[/tex]
[tex]32 - Q_1 - Q_1 = 0.4[/tex]
Collect like terms
[tex]Q_1 + Q_1 = 32-0.4[/tex]
[tex]2Q_1 = 31.6[/tex]
Divide by 2
[tex]Q_1 = 15.8[/tex]
Substitute [tex]Q_1 = 15.8[/tex] in [tex]Q_3 = 32 - Q_1[/tex]
[tex]Q_3 = 32 - 15.8[/tex]
[tex]Q_3 = 16.2[/tex]
