Answer: [tex]\mu=849[/tex]
[tex]\sigma=18.43[/tex]
Minimum usual value[tex]=\mu-2\sigma=849-2(18.43)=849-36.86=812.14[/tex]
Maximum usual value[tex]=\mu+2\sigma=849+2(18.43)=849+36.86=885.86[/tex]
Step-by-step explanation:
The Normal approximation can be used to binomial distribution, when n is large and p is near to 0.5.
[tex]X\sim N(np,\sqrt{np(1-p)})[/tex]
here, Mean = [tex]\mu=np[/tex]
Standard deviation :[tex]\sigma=\sqrt{np(1-p)}[/tex]
Given : [tex]n= 1415,\ p=\dfrac{3}{5}[/tex]
Then, [tex]\mu=1415\times\dfrac{3}{5}=849[/tex]
[tex]\sigma=\sqrt{1415(\dfrac{3}{5})(1-\dfrac{3}{5})}\\\\=\sqrt{339.6}\approx18.43[/tex]
Range rule of thumb : It says that the range is approximately four times the standard deviation.
Minimum usual value[tex]=\mu-2\sigma=849-2(18.43)=849-36.86=812.14[/tex]
Maximum usual value[tex]=\mu+2\sigma=849+2(18.43)=849+36.86=885.86[/tex]
