the standard deviation is approximately??

Answer:
[tex]SD = \$ 239[/tex]
Step-by-step explanation:
Given
[tex]\$277.36, \$269.29, \$307.98, \$361.63, \$407.97,\\ \$442.99, \$601.71, \$693.64, \$870.21, \$970.60[/tex]
Required
Determine the standard deviation
First, we calculate the mean.
[tex]Mean = \frac{\$277.36+ \$269.29+ \$307.98+ \$361.63+ \$407.97+ \$442.99+ \$601.71+ \$693.64+ \$870.21+ \$970.60}{10}[/tex]
[tex]Mean = \frac{\$5203.38}{10}[/tex]
[tex]Mean = \$520.338[/tex]
[tex]Mean = \$520.34[/tex] --- approximately
Next, subtract the mean from each data
[tex]\$277.36 - \$520.34 =-\$242.98[/tex]
[tex]\$269.29 - \$520.34 = -\$251.05[/tex]
[tex]\$307.98 - \$520.34 = -\$212.36[/tex]
[tex]\$361.63 - \$520.34 = -\$158.71[/tex]
[tex]\$407.97 - \$520.34 = -\$112.37[/tex]
[tex]\$442.99 - \$520.34 = -\$77.35[/tex]
[tex]\$601.71 - \$520.34 = \$81.37[/tex]
[tex]\$693.64 - \$520.34 = \$173.3[/tex]
[tex]\$870.21 - \$520.34 =\$349.87[/tex]
[tex]\$970.60 - \$520.34 =\$450.26[/tex]
Square the results above
[tex](-\$242.98)^2 = \$59039.2804[/tex]
[tex](-\$251.05)^2 =\$63026.1025[/tex]
[tex](-\$212.36)^2 =\$45096.7696[/tex]
[tex](-\$158.71)^2 =\$25188.8641[/tex]
[tex](-\$112.37)^2 =\$12627.0169[/tex]
[tex](-\$77.35)^2 =\$5983.0225[/tex]
[tex](-\$81.37)^2 =\$6621.0769[/tex]
[tex](-\$173.3)^2 =\$30032.89[/tex]
[tex](-\$349.87)^2 =\$122409.0169[/tex]
[tex](-\$450.26)^2 =\$202734.0676[/tex]
Add the squared results
[tex]\$59039.2804 + \$63026.1025 + \$45096.7696 + \$25188.8641 + \$12627.0169 + \$5983.0225 +\$6621.0769 + \$30032.89 + \$122409.0169 + \$202734.0676[/tex]
[tex]= \$572758.1074[/tex]
Divide by number of data, to get the variance
[tex]Variance = \frac{\$572758.1074}{10}[/tex]
[tex]Variance = \$57275.81074[/tex]
Square the above, to get the standard deviation
[tex]SD = \sqrt{\$57275.81074}[/tex]
[tex]SD = \$ 239.323652697[/tex]
[tex]SD = \$ 239[/tex] --- approximated