Question

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).

Answer 1

a

`N = 4`

b

The standard deviation is
`\sigma = 0.75`

c

The uncertainty at 95% is
`v_u__(95)} = 1.46`

d

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

Step-by-step 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

`(v_u)/(2) = (v_(\sigma ))/(√(N) ) * Z_(0.05)`

Where
`Z_(0.05)` is the z-score of 0,05 = 1.645 from the z-table

Substituting value

`(2.61)/(2) = (1.5)/(√(N ) ) * 1.645`

`N=( (1.5 *1.645 )/(1.305))^2`

`N = 4`

The standard deviation of the mean is mathematically represented as

`\sigma = (v_(\sigma ))/(√(N) )`

Substituting the values

`\sigma = (1.5)/(√(4) )`

`\sigma = 0.75`

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

`v_u__(95)} = (v_(\sigma ))/(√(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)} = (1.5)/(√(4) ) * 1.96`

`v_u__(95)} = 1.46`

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

`v_u__(50)} = (v_(\sigma ))/(√(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)} = (1.5)/(√(4) ) * 0.69146`

`v_u__(50)} = 0.52`

So the interval about the sample mean is

`25 \pm 0.52`