SMAM 314
Exam 2
Name_____________
1. The flow of current in milliamperes in wire strips of a certain type can be modeled as a
normal distribution with mean 20 and standard deviation 1.
A. What proportion of strips have a current flow between 18.5 and 22 milliamps?(10
points)
" 18.5 ! 20
22 ! 20 %
P(18.5 < X < 22) = P $
<Z<
1
1 '&
#
= P(!1.5 < Z < 2) = .9772 ! .0668 = .9104
normalcdf(18.5,22,20,1) = .9104
B. How large must a current flow be to be among the largest 5% of all flows?(10 points)
P(X < c) = .95
"
c ! 20 %
P$ Z <
= .95
1 '&
#
c ! 20
= 1.645
1
c = 21.645
invNorm(,95,20,1) = 21.645
2. Suppose that only 40% of all drivers in a certain state wear a seatbelt. A random
sample of 500 drivers is selected. Use the normal approximation to the binomial
distribution with the continuity correction to determine the probability that the number of
drivers in the sample that wear a seat belt is at most 190. (15 points)
µ = 500(.4) = 200
! = 500(.4)(.6) = 10.95
$
190.5 # 200 '
P(X " 190) = P & Z "
= P(Z < #.87) = .1922
10.95 )(
%
normalcdf(-10^99,190.5,200,10.95)=.1928
3. The Rockwell hardness of certain metal pins is known to have a mean of 50 and a
standard deviation of 1.5. What is the approximate probability that the average hardness
of a random sample of thirty-six pins is at least 50.4?Use the Central Limit Theorem. (15
points)
µ x = 50,! x = 1.5 / 36 = .25
$
50.4 # 50 '
P(x " 50.4) = P & Z >
+ P(Z > 1.6)
.25 )(
%
= 1# .9452 = .0548
normalcdf(50.4,10^99,50, .25)=.0548
4. The modulus of rupture (MOR) for a particular grade of pencil lead is known to have a
standard deviation of 250 psi. A random sample from a normal population of 16 pencil
leads yielded a sample mean of 6490.
A. Construct a 95% confidence interval for the true mean MOR. (10 points)
x±
z!
n
6490 ±
1.96(250)
16
6490 ± 122.5
(6367.5,6612.5)
B. Interpret the confidence interval obtained in part A. (4 points)
The interval in A has probability .95 of containing the true MOR.
C. Find the sample size required to estimate the true MOR to within ±100 using a 99%
confidence interval. (8 points)
2.576(250)
n
= 100
100 n = 2.576(250)
n = 6.44
n = 41.47
n = 42
D. Suppose a test of hypothesis is performed on the population mean MOR where
H 0 µ=6300
H1 µ ! 6300
Based only on the confidence interval in part A tell whether you would reject or not reject
H0 at α =.05. Explain your answer.(5 points)
You would reject the null hypothesis at α =.05 because 6300 is not in the confidence
interval.
5. Medical researchers have developed a new artificial heart constructed primarily of
titanium and plastic. The heart will last and operate almost indefinitely once it is
implanted in the patient’s body, but the battery pack needs to be recharged every about
four hours. A random sample of 50 battery packs is selected and subjected to a life test.
The average life of these batteries is 4.05 hours. Assume the battery life is normally
distributed with standard deviation σ = 0.2 hour.
A. Perform a test of hypothesis at α =.05 to determine whether the true mean battery life
is significantly greater than 4 hours. Write the complete report. (15 points)
H0 µ =4
H1 µ >4
Assumptions
Normal population with known standard deviation
Region of rejection
Z >1.645
Z=
Computation
Z=
x!µ
"/ n
4.05 ! 4
.2 / 50
= 1.7677
Reject or do not reject H0
reject H0 at α =.05
Conclusion and statement
There is reason to believe that the life of these batteries is significantly greater than 4
hours.
B. What is the p value of the test? (3 points)
Pvalue = P(Z>1.77)=.0384
C. Would you reject or not reject H0 at α =.01?. Explain your answer. (4 points)
You would not reject the null hypothesis at α =.01 for two reasons. First the p value is
.0384 >.01. Second for a test at α =.01 the critical value is 2.33<1.77.