Download PHYS1140 SP07 HW #3 due March 1/2 (Much as Taylor 5.4) A

PHYS1140 SP07 HW #3 due March 1/2
1. (Much as Taylor 5.4)
A student measures the time it takes for a ball to
drop some distance in oil, and records how many times she gets results
within 0.20 seconds of each time point listed.
Time <6.6 7.0 7.4
N
0
0. 2
7.8
5
8.2
9
8.6
12
9.0
7
9.4 9.8
4
1
10.2 >10.6 sec
0
0 times
a) make a histogram with an appropriate scale to show these data
You all got this ok.
b) compute the mean time tavg for this experiment
tavg=(7.0x0+7.4x2+7.8x5+8.2x9+8.6x12+9.0x7+9.4x4+9.8x1+10.2x0)/40=8.530
(do not round off yet—the average will be known better than any one time)
c) compute the standard deviation t of the results of this experiment
0.5648690.56
d) compute the standard deviation of the mean for these data, and quote the
final answer to summarize these data
SDOM=0.564869/sqrt40=0.0893140.09
So tavg+/-tavg = 8.53+/-0.09 seconds [be sure to include the units!]
e) On your histogram, sketch the shape, at least somewhat to scale, of a
normal distribution that is similar to these data.
2. (much as Taylor 5.19)
A student measures some distance y many times
to find that yavg=22 cm and the standard deviation y is 2 cm. What
fraction of the readings might this student expect to find between
a) 22 cm and 24 cm?
integral from 22 to 24=integral from X to X+1=0.3413 [ok to keep all digits
from the table]
b) 22 cm and 26 cm ?
integral from 22 to 26 = integral from X to X+2
c) 26 cm and 28cm?
integral from 26 to 28=integral from 22 to 28(t=3)-integral from 22 to 26(t=2)
=0.4987-0.4772=0.0215
d) 19cm and 23cm?
integral from 19 to 23 = integral from 22 to 19 (t=-1.5)+integral from 22 to
23(t=0.5)
=integral from 22 to 25(t=+1.5)+integral from 22 to 23(t=0.5) [by symmetry]
=0.4332+0.1915=0.6247
3. (much as Taylor 5.21)
The CU men’s basketball team is in desperate
need of help, and seeks to recruit tall young men from our students. The
14,000 male CU students average height is 1.70 m, with a standard
deviation of 20 cm. These heights are randomly distributed and follow a
normal distribution.
[I use Table B, but some of you have calculators to do these areas. Did you
check that the results are the same?]
a) About how many CU men stand at least 1.85 m tall?
The difference from the mean is 1.85-1.70=0.15m=0.75The probability for
this requirement is the area from 1.70m to infinity – the area from 1.70m to
1.85m(t=0.75)
=0.5000-0.2734=0.2266.
The students number 14,000x0.2266=3172 (rounded to the nearest entire person)
b) About how many candidates for center, at least 2.0 meters tall, are
available?
The difference is 2.00-1.70=0.30m=1.5The probability for this is the area from
1.70m to infinity-the area from1.70m to 2.00m
=0.5000-0.4332=0.0668.
The tall students number 14,000x0.0668=935
c) About how many candidates for center, at least 2.0 m tall, are expected
from among only 10% of the male students? This might estimate the fraction
of tall guys with adequate athletic skills.
The probability fo being athletic is the same for tall students as the average,
so we find 93 suitable candidates.
e) Estimate the uncertainty in your answer to part c).
For a random count of 94, the uncertainty is sqrt94=10.