Physics, 05.05.2020 23:10 lildeb2778

G During an experiment, an aerospace engineering student measures a wind tunnel’s velocity N times. The student reports the following information, based on 90 % confidence, about the finite data set: mean velocity = 25.00 m/s, velocity standard deviation = 1.50 m/s, and uncertainty in velocity = 2.61 m/s. Determine (a) N, (b) the standard deviation of the means based upon this data set (in m/s), (c) the uncertainty, at 95 % confidence, in the estimate of the true mean value of the velocity (in m/s), and (d) the interval about the sample mean over which 50 % of the data in this set will be (in m/s).

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Answer from: benwill0702

N = 4

The standard deviation is $\sigma = 0.75$

The uncertainty at 95% is $v_u__{95}} = 1.46$

The interval about the sample mean is $25 \pm 0.52$

Explanation:

From the question we are told that

The confidence level is k= 90 %

The mean velocity is v_m = 25.0 m/s

The velocity standard deviation is $v_\sigma = 1.5 m/s$

The uncertainty of the velocity v_u = 2.61 m/s

Generally uncertainty can be represented mathematically as

$\frac{v_u}{2} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.05}$

Where $Z_{0.05}$ is the z-score of 0,05 = 1.645 from the z-table

Substituting value

$\frac{2.61}{2} = \frac{1.5}{\sqrt{N } } * 1.645$

$N=( \frac{1.5 *1.645 }{1.305})^2$

N = 4

The standard deviation of the mean is mathematically represented as

$\sigma = \frac{v_{\sigma }}{\sqrt{N} }$

Substituting the values

$\sigma = \frac{1.5}{\sqrt{4} }$

$\sigma = 0.75$

The uncertainty for confidence level of 95% is mathematically represented as

$v_u__{95}} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.95}$

Where $Z_{0.95}$ is the z value of the 0.95 which is 1.96, this is obtained using the z-table

Substituting values

$v_u__{95}} = \frac{1.5}{\sqrt{4} } * 1.96$

$v_u__{95}} = 1.46$

The uncertainty for confidence level of 50 % is mathematically represented as

$v_u__{50}} = \frac{v_{\sigma }}{\sqrt{N} } * Z_{0.50 }$

Where $Z_{0.50}$ is the z value of the 0.50 which is 0.69146, this is obtained using the z-table

Substituting values

$v_u__{50}} = \frac{1.5}{\sqrt{4} } * 0.69146$

$v_u__{50}} = 0.52$

So the interval about the sample mean is

$25 \pm 0.52$

Answer from: Quest

from

explanation:

wind is defined by the direction it blows from. for example, a wind blowing from north to south is called a northerly wind.

Answer from: Quest

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles

b) 0.4 miles

c) 1 mile

d) 3.15 miles

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