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MA23 Exam 3
Spring 2008
Name_________________________________
Directions: Show all your work in the blue books provided. Show all steps and
make a sketch for each problem.
1. Use the standard normal distribution to find each of the following probabilities:
2.
3.
4.
5.
a) P0  z  2.16 Ans. 0.4846
b) P 1.87  z  1.32 Ans. 0.4692+0.4066 = 0.8759
c) P1.25  z  2.17 Ans. 0.4850-0.3944 = 0.0906
d) Pz  2.03 Ans. 0.5 + 0.4788 = 0.9788
A national survey found that Americans drink an average of 6 glasses of water
daily. Assuming that the number of glasses of water is approximately normally
distributed with a standard deviation of 1.7 glasses, what is the probability that a
randomly selected American drink more than the amount of 8 glasses of water?
Round your z-value to the nearest hundredth.
Let X = glasses of water consumed daily
Ans. P(x>8) = P(z > 1.18) = 0.5 - .3810 = 0.119
At a ski area in Vermont, the daytime high temperature is normally distributed
during January, with a mean of 22ºF and a standard deviation of 10ºF (U.S.
Department of Commerce, Environmental Data Services). You are planning to
ski there this January. What is the probability that the daytime high temperature
will be between 29ºF and 40ºF?
Ans. P(29  x  40) = P(0.7  z  1.8) = 0.4641 – 0.2580 = 0.2061
Let X = daytime high temperature
Final averages are typically approximately normally distributed with a mean of 72
and a standard deviation of 12.5. Your professor says that the top 8% of the class
will receive an A. What average must you exceed to obtain an A?
Let X = final average
Find Z such that the probability to the right of z is 8%. That is z = 1.41.
x    z  72  (1.41)(12.5)  89.63
Based on 52 years of data compiled by the National Climatic Data Center
(http://lwf.ncdc.noaa.gov/oa/climate/online/ccd/avgwind.html), the average speed
of winds in Honolulu is 11.3 miles per hour. Assume that wind speeds are
normally distributed with a standard deviation of 3.5 miles per hour. Find the
probability that the mean speed of a random sample of nine readings exceeds 13.5
mph.
Ans. P( x  13.5)  P( z  1.89)  0.5  0.4706 Here the Central Limit Theorem
is used where the standard deviation of sampling distribution is the
population standard deviation divided by n .
MA23 Exam 3
Spring 2008
Name_________________________________
6. Overproduction of uric acid in the body can be an indication of cell breakdown.
This may be an advance indication of illness such as gout, leukemia, or
lymphoma (Reference: Manual of Laboratory and Diagnostic Tests, F.
Fischbach). Over a period of months, an adult male patient has taken eight blood
tests for uric acid. The sample mean concentration was 5.35 mg/dl. The
distribution of uric acid in healthy adult males can be assumed to be normal with
σ=1.85 mg/dl. Find a 95% confidence interval for the population mean
concentration of uric acid in this patient’s blood.
Ans. This is a confidence interval. Since you know the population standard
deviation, you can use the Z-table to find the z. 5.35  1.282    5.35  1.282
or 4.068    6.632 .
7. For a group of 10 men subjected to a stress situation, the mean number of
heartbeats per minutes was 126, and the standard deviation was 4. Find the 95%
confidence interval of the true mean.
Ans. This is to find confidence interval. Since we do not know the population
standard deviation and the sample size n is less than 30 (It is 10) , we use the
t-table. The degreed of freedom is 9. The t is equal to 2.262. So the answer is
123.14    128.86