Question

Miriam took a survey, asking each family in her neighborhood how many pets they have. She made a graph showing the results: [asy] size(5.5cm); draw((0,30)--(0,0)--(41,0)); for(int i=1; i<7; i+=1) {draw((0.6,5*i)--(-0.6,5*i)); draw((0.6,5*i)--(41,5*i),black+0.5+dashed); }; path a=(3,0)--(3,30)--(9,30)--(9,0)--cycle; path b=(9,0)--(9,15)--(15,15)--(15,0)--cycle; path c=(15,0)--(15,15)--(21,15)--(21,0)--cycle; path d=(21,0)--(21,30)--(27,30)--(27,0)--cycle; path e=(27,0)--(27,5)--(33,5)--(33,0)--cycle; path f=(33,0)--(33,5)--(39,5)--(39,0)--cycle; fill(a,gray); fill(b,gray); fill(c,gray); fill(d,gray); fill(e,gray); fill(f,gray); draw(a); draw(b); draw(c); draw(d); draw(e); draw(f); label(rotate(-45)*scale(0.8)*"0 pets",(5,-1),SE); label(rotate(-45)*scale(0.8)*"1 pet",(11,-1),SE); label(rotate(-45)*scale(0.8)*"2 pets",(17,-1),SE); label(rotate(-45)*scale(0.8)*"3 pets",(23,-1),SE); label(rotate(-45)*scale(0.8)*"4 pets",(29,-1),SE); label(rotate(-45)*scale(0.8)*"5 pets",(35,-1),SE); label(scale(0.8)*"1",(-0.6,5),W); label(scale(0.8)*"2",(-0.6,10),W); label(scale(0.8)*"3",(-0.6,15),W); label(scale(0.8)*"4",(-0.6,20),W); label(scale(0.8)*"5",(-0.6,25),W); label(scale(0.8)*"6",(-0.6,30),W); label(scale(0.8)*rotate(90)*"Number of families",(-3.5,15),W); [/asy] What is the average number of pets per family?

Answer 1

Answer:

1.8

Step-by-step explanation:

The graph of the problem is plotted using Asymptote and attached.

From the graph:

• 6 Families had 0 Pets
• 3 Families had 1 Pets
• 3 Families had 2 Pets
• 6 Families had 3 Pets
• 1 Family had 4 Pets
• 1 Family had 5 Pets

There are a total of 20 families.

Average Number of Pets

$=\dfrac{6*0+3*1+3*2+6*3+1*4+1*5}{20} \\\\=\dfrac{36}{20}\\\\=1.8$

The average number of pets per family is 1.8.