Question

recently, a random sample of 13 - 18 year olds was asked, \how much do you currently have in savings?\ the data in the table represent the responses to the survey. approximate the mean and standard deviation amount of savings. click the icon to view the frequency distribution for the amount of savings. savings $0 - $199 $200 - $399 $400 - $599 $600 - $799 $800 - $999 $1000 - $1199 $1200 - $1399 frequency 347 91 57 16 9 8 1 the sample mean amount of savings is $ (round to the nearest dollar as needed.) the sample standard deviation is $ (round to the nearest dollar as needed.)

Explanation:

Step1: Calculate the mid - point of each class

For the class \(0 - 199\), the mid - point \(x_1=\frac{0 + 199}{2}=99.5\)
For the class \(200 - 399\), the mid - point \(x_2=\frac{200+399}{2}=299.5\)
For the class \(400 - 599\), the mid - point \(x_3=\frac{400 + 599}{2}=499.5\)
For the class \(600 - 799\), the mid - point \(x_4=\frac{600+799}{2}=699.5\)
For the class \(800 - 999\), the mid - point \(x_5=\frac{800 + 999}{2}=899.5\)
For the class \(1000 - 1199\), the mid - point \(x_6=\frac{1000+1199}{2}=1099.5\)
For the class \(1200 - 1399\), the mid - point \(x_7=\frac{1200+1399}{2}=1299.5\)

Step2: Calculate the sample mean \(\bar{x}\)

The formula for the sample mean of a frequency distribution is \(\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}\)
where \(f_i\) is the frequency and \(x_i\) is the mid - point.
\(\sum_{i = 1}^{n}f_ix_i=347\times99.5+91\times299.5 + 57\times499.5+16\times699.5+9\times899.5+8\times1099.5+1\times1299.5\)
\(=347\times99.5+91\times299.5+57\times499.5 + 16\times699.5+9\times899.5+8\times1099.5+1299.5\)
\(=34526.5+27254.5+28471.5+11192+8095.5+8796+1299.5\)
\(=34526.5+27254.5=61781\); \(61781+28471.5 = 90252.5\); \(90252.5+11192=101444.5\); \(101444.5+8095.5=109540\); \(109540+8796=118336\); \(118336+1299.5=119635.5\)
\(\sum_{i=1}^{n}f_i=347 + 91+57+16+9+8+1=529\)
\(\bar{x}=\frac{119635.5}{529}\approx226\)

Step3: Calculate the sample variance \(s^{2}\)

The formula for the sample variance is \(s^{2}=\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^{2}}{n - 1}\)
\((x_1-\bar{x})=(99.5 - 226)=-126.5\); \((x_2-\bar{x})=(299.5 - 226)=73.5\); \((x_3-\bar{x})=(499.5 - 226)=273.5\); \((x_4-\bar{x})=(699.5 - 226)=473.5\); \((x_5-\bar{x})=(899.5 - 226)=673.5\); \((x_6-\bar{x})=(1099.5 - 226)=873.5\); \((x_7-\bar{x})=(1299.5 - 226)=1073.5\)
\(\sum_{i = 1}^{n}f_i(x_i-\bar{x})^{2}=347\times(- 126.5)^{2}+91\times(73.5)^{2}+57\times(273.5)^{2}+16\times(473.5)^{2}+9\times(673.5)^{2}+8\times(873.5)^{2}+1\times(1073.5)^{2}\)
\(=347\times16002.25+91\times5402.25+57\times74702.25+16\times224202.25+9\times453502.25+8\times763002.25+1\times1152392.25\)
\(=347\times16002.25 = 5552770.75\); \(91\times5402.25=491604.75\); \(57\times74702.25 = 4258028.25\); \(16\times224202.25=3587236\); \(9\times453502.25 = 4081520.25\); \(8\times763002.25=6104018\); \(1\times1152392.25=1152392.25\)
\(\sum_{i = 1}^{n}f_i(x_i-\bar{x})^{2}=5552770.75+491604.75+4258028.25+3587236+4081520.25+6104018+1152392.25\)
\(=5552770.75+491604.75 = 6044375.5\); \(6044375.5+4258028.25=10302403.75\); \(10302403.75+3587236=13889639.75\); \(13889639.75+4081520.25=17971160\); \(17971160+6104018=24075178\); \(24075178+1152392.25=25227570.25\)
\(s^{2}=\frac{25227570.25}{529 - 1}=\frac{25227570.25}{528}\approx47780\)

Step4: Calculate the sample standard deviation \(s\)

\(s=\sqrt{s^{2}}=\sqrt{47780}\approx219\)

Answer:

The sample mean amount of savings is \(\$226\). The sample standard deviation is \(\$219\)