What is the age distribution of adult shoplifters (21 years of age or older) in supermarkets? The following is based on information taken from the National Retail Federation. A random sample of 895 incidents of shoplifting gave the following age distribution. Estimate the mean age, sample variance, and sample standard deviation for the shoplifters. For the class 41 and over, use 45.5 as the class midpoint. (Enter your answers to one decimal place.)

Age range (years) 21-30 31-40 41 and over
Number of shoplifters 280 368 247

Step-by-step explanation:

Hello!

You have the data of 895 shoplifting incidents separated into three categories. You need to estimate the sample mean, sample variance and sample standard deviation.

Given that the information is given in intervals of age, in order to calculate the asked estimations you have to determine the classmark for each interval.

Age range 21-30

Class mark x'= (30+21)/2= 25.5

Age range 31-40

Class mark x'= (31+40)/2= 35.5

Age range 41 and over

Class mark x'= 45.5

To calculate the sample mean you have to use the following formula:

X[bar]= (∑x'fi)/n= 31442.5/895= 35.13

∑x'fi= (25.5*280)+(35.5*368)+(45.5*247)= 31442.5

The formula for the sample variance is:

S²= *[∑x'²fi- (∑x'fi)²/n]= 1/894*[1157193.75-(31442.5²/895)]= 58.81

∑x'²fi= (25.5²*280)+(35.5²*368)+(45.5²*247)= 1157193.75

The standard deviation is the square root of the variance:

S= √58.81= 7.67

I hope it helps!

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -