Question

WILL GIVE BRAINLIEST FOR CORRECT WORKING AND ANSWER!!!
Only do (b) pls
Due soon, thanks!

Answer 1

Answer:

Step-by-step explanation:

Since the total frequency = 56, then Q1 will be at 14 and Q3 will be at 42

(a)

Based on the cumulative frequency (c.f.) graph (refer to attachment2), Q1 is approximately 22 seconds and Q3 is approximately 43 seconds

$\boxed{Interquartile\ range\ (IQR)=Q3-Q1}$

$IQR=43-22$

       $=21\ seconds$

(b)

Q1 (cumulative frequency = 14) lies within 20 < t ≤ 30 which has:

• lower boundary (l.b.)= 20
• higher boundary (h.b.)= 30
• lower c.f. = 12
• higher c.f. = 23

Therefore:

$\boxed{\frac{Q_1\ c.f.-lower\ c.f.}{higher\ c.f.-lower\ c.f.} =\frac{Q_1-l.b.}{h.b.-l.b.} }$

$\frac{14-12}{23-12} =\frac{Q_1-20}{30-20}$

$\frac{2}{11} =\frac{Q_1-20}{10}$

$Q_1-20=\frac{2\times10}{11}$

$Q_1=21\frac{9}{11} \ seconds$

Q3 (cumulative frequency = 42) lies within 40 < t ≤ 50 which has:

• lower boundary (l.b.)= 40
• higher boundary (h.b.)= 50
• lower c.f. = 38
• higher c.f. = 52

Therefore:

$\boxed{\frac{Q_3\ c.f.-lower\ c.f.}{higher\ c.f.-lower\ c.f.} =\frac{Q_3-l.b.}{h.b.-l.b.} }$

$\frac{42-38}{52-38} =\frac{Q_3-40}{50-40}$

$\frac{4}{14} =\frac{Q_3-40}{10}$

$Q_3-40=\frac{4\times10}{14}$

$Q_3=42\frac{6}{7} \ seconds$

$IQR=Q_3-Q_1$

        $=42\frac{6}{7}- 21\frac{9}{11}$

        $=21\frac{3}{77} \ seconds$