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79. A normal population has a mean of 12.2 and a standard deviation of 2.5.
a. Compute the z value associated with 14.3.
Using website
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we can
compute
P(Z>14.3)=0.2
b. What proportion of the population is between 12.2 and 14.3?
Using website
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we get
P(12.2<Z<14.3)=0.3
c. What proportion of the population is less than 10.0?
P(Z<10.0)=0.18943
92. Assume that the mean hourly cost to operate a commercial airplane follows the
normal distribution with a mean $2,100 per hour and a standard deviation of $250.
What is the operating cost for the lowest 3 percent of the airplanes?
Solution. Denote the cost by X, then X~N(2100, 250^2). So,
P(X<2570)=3%
So, the operating cost for the lowest 3 percent of the airplanes is $2570
104. Shaver Manufacturing, Inc. offers dental insurance to its employees. A recent
study by the Human Resource Director shows the annual cost per employee per
year followed the normal distribution, with a mean of $1,280 and a standard
deviation of $420 per year.
a. What fraction of the employees cost more than $1,500 per year for dental expenses?
Solution. Using website,
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we
have
P(X>1500)=0.30
So, 30% of the employees cost more than $1,500 per year for dental expenses.
b. What fraction of the employees cost between $1,500 and $2,000 per year?
Using website,
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we
have
P(2000>X>1500)=0.257
So, 25.7% of the employees cost between $1,500 and $2,000 per year.
c. Estimate the percent that did not have any dental expense.
Using website,
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we
have
P(X<=0)=0.00
So, 0% of the employees did not have any dental expense
d. What was the cost for the 10 percent of employees that incurred the highest dental
expense?
Using website,
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stats/normaldist.html , we
have
P(X>1818.44)=10%
So, we know that the cost for the 10 percent of employees that incurred the highest dental
expense is $1818.44
179. A sample of 81 observations is taken from a normal population. The sample
mean is 40, and the sample standard deviation is 5.
Determine the 95 percent confidence interval for the population mean.
Solution. We use T-statistic
T  X S n ~ t (n  1)
Where n  81, X  40, S  5 . Using 1    95% , we get
t 0.025 (80)  1.99
So, the 95 percent confidence interval for the population mean is
( X  S * t 0.025 (80) / n )
i.e.,
(40  5 *1.99 / 9)
i.e.,
(40  1.106)  (38.894,41.106)
So, the 95 percent confidence interval for the population mean is
(38.894, 41.106)