Estimate the mean age, sample variance, and sample standard deviation

Question

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. (Round your answers to one decimal place.)

1. Age range (years): 21-30, 31-40, 41 and over
2. Number of shoplifters: 244, 357, 294

x=

s²=

s=

Answer 1

To estimate the mean age , the formula is x = ∑ (x_i•f_i) /∑f_i

_{x = (6222+ 126735 +13377) / (244+357+294)}

_{x = 146344/895}

_{x= 163.5}

_{To find the sample variance, I use the formula s}² = ∑ [(x_i-x)² •f_i]

------------------

( ∑f_i )-1

To find the sample standard deviation s, we take the square root of the sample variance.

Let us complete the following setup in order to be able to use the formula.

class midpoint x_i f_i x_i•f_i x_i-x (x_i - x)² (x_i - x)²•f_i

_{21 - 30 25.5 244 6222 -138 19044 4646736}

_{31 - 40 35.5 357 126735 128 16384 5849088}

≥41 45.5 294 13377 -118 13924 4093656

s² =(4646736+5849088+4093656)

-----------------------------------------

895-1

=14589480/ 894

s² =16319.3

Sample standard deviation s = √16319.3

= 127.7

Please try the calculations to check for accuracy.