Answer :
Answer:
a
[tex]N = 4[/tex]
b
The standard deviation is [tex]\sigma = 0.75[/tex]
c
The uncertainty at 95% is [tex]v_u__{95}} = 1.46[/tex]
d
The interval about the sample mean is [tex]25 \pm 0.52[/tex]
Explanation:
From the question we are told that
The confidence level is [tex]k= 90[/tex]%
The mean velocity is [tex]v_m = 25.0 m/s[/tex]
The velocity standard deviation is [tex]v_\sigma = 1.5 m/s[/tex]
The uncertainty of the velocity [tex]v_u = 2.61 m/s[/tex]
Generally uncertainty can be represented mathematically as
[tex]\frac{v_u}{2} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.05}[/tex]
Where [tex]Z_{0.05}[/tex] is the z-score of 0,05 = 1.645 from the z-table
Substituting value
[tex]\frac{2.61}{2} = \frac{1.5}{\sqrt{N } } * 1.645[/tex]
[tex]N=( \frac{1.5 *1.645 }{1.305})^2[/tex]
[tex]N = 4[/tex]
The standard deviation of the mean is mathematically represented as
[tex]\sigma = \frac{v_{\sigma }}{\sqrt{N} }[/tex]
Substituting the values
[tex]\sigma = \frac{1.5}{\sqrt{4} }[/tex]
[tex]\sigma = 0.75[/tex]
The uncertainty for confidence level of 95% is mathematically represented as
[tex]v_u__{95}} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.95}[/tex]
Where [tex]Z_{0.95}[/tex] is the z value of the 0.95 which is 1.96, this is obtained using the z-table
Substituting values
[tex]v_u__{95}} = \frac{1.5}{\sqrt{4} } * 1.96[/tex]
[tex]v_u__{95}} = 1.46[/tex]
The uncertainty for confidence level of 50 % is mathematically represented as
[tex]v_u__{50}} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.50 }[/tex]
Where [tex]Z_{0.50}[/tex] is the z value of the 0.50 which is 0.69146, this is obtained using the z-table
Substituting values
[tex]v_u__{50}} = \frac{1.5}{\sqrt{4} } * 0.69146[/tex]
[tex]v_u__{50}} = 0.52[/tex]
So the interval about the sample mean is
[tex]25 \pm 0.52[/tex]