This frequency distribution represents the commission earned (in dollars) by 100 salespeople employed at several branches of a large chain store. Find the mean and modal class for the data.

Frequency distribution is missing, so i have attached it.

A)x¯ = 180.28

B)class 177 - 185

Step-by-step explanation:

A) For the class of 150 - 158;

the class boundaries will be 149.5 - 158.5.

Middle value(x_m) = (150 + 158)/2 = 154

Frequency(f) = 5.

f(x_m) = 154 × 5 = 770

For class of 159 - 167;

the class boundaries will be 158.5 - 167.5

Middle value (x_m) = (159 + 167)/2 = 163

Frequency(f) = 16

f(x_m) = 163 × 16 = 2608

For class of 168 - 176;

the class boundaries will be 167.5 - 176.5

Middle value (x_m) = (168 + 176)/2 = 172

Frequency(f) = 20

f(x_m) = 172 × 20 = 3440

For class of 177 - 185;

the class boundaries will be 176.5 - 185.5

Middle value (x_m) = (177 + 185)/2 = 181

Frequency(f) = 21

f(x_m) = 181 × 21 = 3801

For class of 186 - 194;

the class boundaries will be 185.5 - 194.5

Middle value (x_m) = (186 + 194)/2 = 190

Frequency(f) = 20

f(x_m) = 190 × 20 = 3800

For class of 195 - 203;

the class boundaries will be 194.5 - 203.5

Middle value (x_m) = (195 + 203)/2 =199

Frequency(f) = 15

f(x_m) = 199 × 15 = 2985

For class of 204 - 212;

the class boundaries will be 203.5 - 212.5

Middle value (x_m) = (204 + 212)/2=208

Frequency(f) = 3

f(x_m) = 208 × 3 = 624

Now,mean is;

x¯ = Σf(x_m)/Σf

Σf(x_m) = 770 + 2608 + 3440 + 3801 + 3800 + 2985 + 624 = 18028

Σf = 5 + 16 + 20 + 21 + 20 + 15 + 3 = 100

x¯ = 18028/100

x¯ = 180.28

B) Modal class is the class with the highest occurring frequency.

In this case, it's class 177 - 185 that has frequency of 21

step-by-step explanation:

1

you first to choose the points and then move from then as the slope you have