Answer:
[tex]X= 19.6 cm3[/tex]
Therefore
X is not Accurate as it Falls outside of the Volume 15cm^3
Step-by-step explanation:
From the question we are told that:
Sample size [tex]n=50[/tex]
Volume Range [tex]\triangle V= 914 cm^3 to 1492cm^3[/tex]
Exact standard deviation [tex]\sigma_E 186.4 cm^3,[/tex]
Generally the equation for range rule of thumb is mathematically given by
Where
[tex]\triangle V=1492cm^3- 914 cm^3[/tex]
[tex]\triangle V=578cm^2[/tex]
Therefore
Estimate Standard deviation
[tex]\sigma_e=\frac{578}{4}[/tex]
[tex]\sigma_e=144.5cm^3[/tex]
Since
Exact standard deviation is 186.4 cm^3
Comparing
[tex]X=\sigma_E- \sigma_e[/tex]
[tex]X= ( 164.1-144.5 ) cm3[/tex]
[tex]X= 19.6 cm3[/tex]
Therefore
X is not Accurate as is Falls outside of the Hallmark 15cm^3