In a certain plant batteries are being produced with a life expectancy
that has a variance of 5.76 months squared. Suppose the average life
expectancy in a sample of 64 batteries is 12.35 months. Estimate the
average life expectancy of all the batteries produced at this plant.
for samples of size 64 we know
that thesample means are
normally distributed with a
standard deviation of
s = 5.76 = 0.3 . At a 90%
sx = n 64 confidence level , the population
mean is 12.35 ± 1.645 ⎛⎜⎝ 0.3 ⎞⎟⎠ .
Therefore, we can be certain that
the mean life expectancy of the batteries is between 11.8565 and
12.8435 months.
A government investigator plans to test the mean quantity of a
particular pollutant that a manufacturer is dumping per day into a
river. He needs an estimate that is within 50 grams at the 95%
confidence level. If previous measurements indicate that the variance
is approximately 21,800 grams squared, how many days should he
include in the sample?
zσ 2
n = error
⎛
⎜
⎜
⎜
⎝
2
z0.95 = 1.96 , so
1.96 ⋅ 147.65 ⎞
n = ⎛⎜
= 33.49 ⎟
50
⎝
⎠
⎞
⎟
⎟
⎟
⎠
• polyval(p,x)
• roots(p)
EDU>> p=[1 0 0 0 0 0 -1];
EDU>> disp(roots(p))
-1.000000000000000 + 0.000000000000000i
-0.500000000000000 + 0.866025403784439i
-0.500000000000000 - 0.866025403784439i
0.500000000000000 + 0.866025403784438i
0.500000000000000 - 0.866025403784438i
1.000000000000000 + 0.000000000000000i
• poly(r)
•
•
•
•
•
conv(p1,p2)
[q r] = deconv(u,v)
k = polyder(p)
k = polyder(p1,p2)
[n d] = polyder(u,v)
• polyfit(x,y,n)
%
P(X < 0.33)
EDU>> normcdf(0.33,mean(v),std(v))
ans =
0.0226890673554349
%
P(X > 0.33)
EDU>> 1-normcdf(0.33,mean(v),std(v))
ans =
0.977310932644565
%
P(0.33 < X < 0.34)
ans =
0.213520415461181
EDU>>
ans =
norminv(0.0226890673554349,mean(v),std(v))
0.33
An engineer measured the number of pounds per square inch (psi)
required to crack a certain type of concrete slab and found it to
be 2,500 psi.
Is it possible that these slabs crack at values that are
from a normal distribution with µ = 2300 and σ 2 =6400?
EDU>> 1
ans =
-
normcdf(2500,2300,sqrt(6400))
0.006208665322577616
% Possible, but extremely improbable
EDU>> norminv(1 - 0.005,2300,sqrt(6400))
ans =
2506.06634428391